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\begin{document}

\title{Poincar\'{e}-Birkhoff-Witt type results for inclusions of Lie algebras}
\author{Darij Grinberg}
\date{Version 1.5 (\today)}
\maketitle
\tableofcontents

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%\begin{vershort}
%\Large
%Mathematisches Institut \\
%der Ludwig-Maximilians-Universit\"at M\"unchen
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%\LARGE
%\textbf{Poincar\'{e}-Birkhoff-Witt type results for inclusions of Lie algebras}\\[15mm]
%\Large
%\begin{verlong}
%Darij Grinberg
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%\begin{vershort}
%vorgelegt von Darij Grinberg
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%betreut von Prof. Dr. Andreas Rosenschon
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\begin{noncompile}
\textbf{TO-DO\ LIST:}

[exterminate wrong $\otimes$ instead of $\cdot$]

[\textbf{END} fix all footnotes]

[\textbf{6} Leibniz: write up proofs]

[\textbf{6} Clifford: write up proofs]

[\textbf{6} supercase]

[\textbf{6} restricted case - does it really work? ask MO]

[\textbf{6} character case]

[try more for Clifford]

[search for $\mathfrak{h}$ not defined in Sect 3]

[proofreading: sloppily done until definition of $\mathfrak{g}$-module
$V\otimes W$; also prop.h\symbol{94}(1) pf sloppily done except for the lemmata]

[\textbf{END} check for commdiags fighting footnotes]

[SEARCH for graded algebras]
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%Ich erkl\"{a}re hiermit, dass ich die vorliegende Arbeit selbstst\"{a}ndig
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%\bigskip
%M\"{u}nchen, den 15. Juni 2011
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%Darij Grinberg
%$\ $


\newpage

\begin{center}
{\Huge \textbf{\textsf{Poincar\'{e}-Birkhoff-Witt type results for inclusions
of Lie algebras}} }
\end{center}

\begin{noncompile}
\textbf{Remark on this version:}

This is the short version of this paper. In contrast to the long (detailed)
version\footnote{avaliable at \texttt{http://www.cip.ifi.lmu.de/\symbol{126}%
grinberg/pbwlong.pdf}}, it has some proofs omitted and some other proofs
shortened, so the long version should be consulted in case some step in a
proof is unclear. However, I hope this will not be necessary.\newline At the
moment, Sections \ref{sect.pbw} and \ref{sect.gen} are almost identical in the
long and short versions. This will probably be changed when I have some more time.
\end{noncompile}

\begin{verlong}
\textbf{Remark on this version:}

This is the long (detailed) version of this paper. Its goal is to provide
proofs for most of the results involved, even if the results are well-known or
their proof is an almost verbatim imitation of another (already given) proof.
As a consequence, it is very long (even longer than the supposedly "short"
version), so I recommend the short version\footnote{avaliable at
\texttt{http://www.cip.ifi.lmu.de/\symbol{126}grinberg/pbw.pdf}} for
reading.\newline At the moment, Sections \ref{sect.pbw} and \ref{sect.gen} are
almost identical in the long and short versions. This will probably be changed
when I have some more time.
\end{verlong}

\subsection{Introduction}

The goal of this paper is to give an elementary and self-contained proof of
the relative Poincar\'{e}-Birkhoff-Witt theorem that was formulated and proved
by Calaque, C\u{a}ld\u{a}raru and Tu in \cite{calaque1}. While our proof
passes the same landmarks as the one given in \cite{calaque1}, it will often
take a different path in between. In particular, it will completely avoid the
use of Koszul algebras and Hopf algebras in the proofs of two crucial lemmata.
It will be completely elementary except for applying the (standard,
non-relative) Poincar\'{e}-Birkhoff-Witt theorem - something I was not able to eschew.

Besides the elementarity, an advantage of our approach is that it applies to a
slightly more general setting than the one given in \cite{calaque1}. The
proofs of the first two main lemmata still hold true for Lie algebras which
are modules over an arbitrary commutative ring $k$ (rather than vector spaces
over a field $k$), as long as a weak splitting condition (which is always
satisfied in the case of a field) is satisfied (an inclusion of Lie algebras
is supposed to split \textit{as a }$k$\textit{-module inclusion}).
Unfortunately this generality is lost in the proof of the third main lemma,
but it still applies to some rather broad cases encompassing that of $k$ being
a field.

Let us sketch the course of action of \cite{calaque1}, and meanwhile point out
where \textit{our} course of action is going to differ:\footnote{Note that the
aim of this Introduction is to give an overview of the results some of which
we are going to prove in the following, \textit{not} to define and formulate
everything in full detail. The reader can safely skip this Introduction: Every
notion we define in it will be defined in greater detail (and often in greater
generality) in one of the subsequent Sections (unless it will not ever be used
outside this Introduction). The situation we consider in this Introduction (a
Lie algebra $\mathfrak{g}$ over a field $k$, and a Lie subalgebra
$\mathfrak{h}$ of $\mathfrak{g}$) will \textit{not} be the situation we
consider in the rest of this paper; instead we will consider slightly more
general situations in the rest of this paper.}

\begin{vershort}
One of the many (albeit not the strongest or most general) avatars of the
Poincar\'{e}-Birkhoff-Witt theorem states that if $k$ is a field of
characteristic $0$, and $\mathfrak{g}$ is a $k$-Lie algebra, then the
universal enveloping algebra $U\left(  \mathfrak{g}\right)  $ is isomorphic to
the symmetric algebra $\operatorname{Sym}\mathfrak{g}$ as a $\mathfrak{g}%
$-module.\footnote{See Remark \ref{rmk.g-mod-U(g).short} for the right
definition of the $\mathfrak{g}$-module structure on $U\left(  \mathfrak{g}%
\right)  $.} Even dropping the characteristic $0$ condition, we still know
that the canonical filtration of $U\left(  \mathfrak{g}\right)  $ (the one
obtained from the degree filtration of the tensor algebra $\otimes
\mathfrak{g}$) results in an associated graded algebra $\operatorname{gr}%
\left(  U\left(  \mathfrak{g}\right)  \right)  $ which is isomorphic to the
symmetric algebra $\operatorname{Sym}\mathfrak{g}$ as a $\mathfrak{g}%
$-algebra\footnote{See Definition \ref{defs.g-alg} for the definition of the
notion of a $\mathfrak{g}$-algebra. (It is a very natural notion and probably
known in literature under a similar name. Hopf algebraists can translate it as
"$U\left(  \mathfrak{g}\right)  $-module algebra".)}.
\end{vershort}

\begin{verlong}
One of the many (albeit not the strongest or most general) avatars of the
Poincar\'{e}-Birkhoff-Witt theorem states that if $k$ is a field of
characteristic $0$, and $\mathfrak{g}$ is a $k$-Lie algebra, then the
universal enveloping algebra $U\left(  \mathfrak{g}\right)  $ is isomorphic to
the symmetric algebra $\operatorname{Sym}\mathfrak{g}$ as a $\mathfrak{g}%
$-module.\footnote{See Remark \ref{rmk.g-mod-U(g).long} for the right
definition of the $\mathfrak{g}$-module structure on $U\left(  \mathfrak{g}%
\right)  $.} Even dropping the characteristic $0$ condition, we still know
(from another "Poincar\'{e}-Birkhoff-Witt theorem") that the canonical
filtration of $U\left(  \mathfrak{g}\right)  $ (the one obtained from the
degree filtration of the tensor algebra $\otimes\mathfrak{g}$) results in an
associated graded algebra $\operatorname{gr}\left(  U\left(  \mathfrak{g}%
\right)  \right)  $ which is isomorphic to the symmetric algebra
$\operatorname{Sym}\mathfrak{g}$ as a $\mathfrak{g}$-algebra\footnote{See
Definition \ref{defs.g-alg} for the definition of the notion of a
$\mathfrak{g}$-algebra. (It is a very natural notion and probably known in
literature under a similar name. Hopf algebraists can translate it as
"$U\left(  \mathfrak{g}\right)  $-module algebra".)}.
\end{verlong}

The paper \cite{calaque1} is concerned with generalizing these properties to a
relative situation, in which we are given a Lie algebra $\mathfrak{g}$ and a
Lie subalgebra $\mathfrak{h}\subseteq\mathfrak{g}$, and we consider the
$\mathfrak{h}$-modules $U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ and $\operatorname{Sym}%
\left(  \mathfrak{g}\diagup\mathfrak{h}\right)  $ instead of $U\left(
\mathfrak{g}\right)  $ and $\operatorname{Sym}\mathfrak{g}$. (Here, $U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}$ means the right ideal of $U\left(
\mathfrak{g}\right)  $ generated by the image of $\mathfrak{h}\subseteq
\mathfrak{g}$ under the canonical map $\mathfrak{g}\rightarrow U\left(
\mathfrak{g}\right)  $.) In this relative situation, we do not get much for
free anymore, but \cite{calaque1} proves the following results:

\begin{itemize}
\item If $k$ is a field of arbitrary characteristic, then we have an
isomorphism $U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \cong\operatorname{Sym}\left(
\mathfrak{g}\diagup\mathfrak{h}\right)  $ of filtered $k$-modules (here, the
filtration on $U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ comes from the canonical
filtration on $U\left(  \mathfrak{g}\right)  $), even if not necessarily of
$\mathfrak{h}$-modules. This isomorphism needs not be canonical. However,
there is a canonical isomorphism of associated graded $\mathfrak{h}$-modules
$\operatorname{gr}_{n}\left(  U\left(  \mathfrak{g}\right)  \diagup\left(
U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \right)
\cong\operatorname{Sym}^{n}\left(  \mathfrak{g}\diagup\mathfrak{h}\right)  $
for every $n\in\mathbb{N}$.

\item If $k$ is a field of characteristic $0$, then we \textit{do} have a
canonical isomorphism $U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \cong\operatorname{Sym}\left(
\mathfrak{g}\diagup\mathfrak{h}\right)  $ of filtered $\mathfrak{h}$-modules
\textit{if and only if a certain Lie-algebraic condition on }$\mathfrak{g}$
\textit{and }$\mathfrak{h}$ \textit{is fulfilled}. This condition takes three
equivalent forms (Assertions 2, 3 and 4 in Theorem \ref{intro.MT}), is
(comparably) easy to check and is rather often fulfilled in classical cases.
\end{itemize}

We will now come to the exact statements and strengthenings of these results.

Theorem 1.3 of \cite{calaque1} (the main result of the paper) states:

\begin{theorem}
[Relative Poincar\'{e}-Birkhoff-Witt theorem]\label{intro.MT}Let $k$ be a
field of characteristic $0$, and let $\mathfrak{g}$ be a $k$-Lie algebra. Let
$\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $\mathfrak{n}$
denote the quotient $\mathfrak{h}$-module $\mathfrak{g}\diagup\mathfrak{h}%
$.\newline\textbf{Preparations:}\newline\textbf{1.} Consider the universal
enveloping algebra $U\left(  \mathfrak{g}\right)  $ of $\mathfrak{g}$. By
using the canonical embedding $\mathfrak{g}\rightarrow U\left(  \mathfrak{g}%
\right)  $ (this is an embedding due to the standard
Poincar\'{e}-Birkhoff-Witt theorem), we can consider $\mathfrak{g}$ a subset
of $U\left(  \mathfrak{g}\right)  $, and thus $\mathfrak{h}\subseteq
\mathfrak{g}\subseteq U\left(  \mathfrak{g}\right)  $.\newline\textbf{2.} Now,
define a new $k$-Lie algebra $\mathfrak{h}^{(1)}$ as follows (see Proposition
\ref{prop.h^(1)} for a more detailed definition): Let $\operatorname*{FreeLie}%
\mathfrak{g}$ denote the free Lie algebra on the $k$-module $\mathfrak{g}$,
and let $\iota:\mathfrak{g}\rightarrow\operatorname*{FreeLie}\mathfrak{g}$ be
the corresponding embedding. Let $\mathfrak{h}^{(1)}$ denote the $k$-Lie
algebra obtained by factoring the free Lie algebra $\operatorname*{FreeLie}%
\mathfrak{g}$ by the Lie ideal generated by its $k$-submodule $\left\langle
\left[  \iota\left(  v\right)  ,\iota\left(  w\right)  \right]  -\iota\left(
\left[  v,w\right]  \right)  \ \mid\ \left(  v,w\right)  \in\mathfrak{h}%
\times\mathfrak{g}\right\rangle $.\newline We have a canonical injective
$k$-Lie algebra homomorphism $\mathfrak{h}\rightarrow\mathfrak{h}^{(1)}$ (see
Proposition \ref{prop.h^(1).restr} \textbf{(a)} for its construction).\newline%
\textbf{3.} Let $\widetilde{\beta}:\mathfrak{h}\otimes\mathfrak{n}%
\rightarrow\mathfrak{n}$ be the $k$-linear map defined by%
\[
\left(
\begin{array}
[c]{c}%
\widetilde{\beta}\left(  h\otimes n\right)  =\left(  \text{the action of }%
h\in\mathfrak{h}\text{ on the element }n\text{ of the }\mathfrak{h}%
\text{-module }\mathfrak{n}\right) \\
\ \ \ \ \ \ \ \ \ \ \text{for every }h\in\mathfrak{h}\text{ and }%
n\in\mathfrak{n}%
\end{array}
\right)  .
\]
\newline\textbf{4.} Consider the exact sequence
$\xymatrixcolsep{4pc}\xymatrix{
0 \ar[r] & \mathfrak h \ar[r]^{\text{inclusion}} & \mathfrak g \ar[r]^{\text{projection}} & \mathfrak n \ar[r] & 0
}$ of $\mathfrak{h}$-modules. Tensoring this exact sequence with
$\mathfrak{n}$, we obtain an exact sequence $\xymatrix{
0 \ar[r] & \mathfrak h \otimes \mathfrak n \ar[r] & \mathfrak g \otimes \mathfrak n \ar[r] & \mathfrak n \otimes \mathfrak n \ar[r] & 0
}$ of $\mathfrak{h}$-modules. This exact sequence gives rise to an element of
$\operatorname{Ext}_{\mathfrak{h}}^{1}\left(  \mathfrak{n}\otimes
\mathfrak{n},\mathfrak{h}\otimes\mathfrak{n}\right)  $. Applying the map
$\operatorname{Ext}_{\mathfrak{h}}^{1}\left(  \mathfrak{n}\otimes
\mathfrak{n},\mathfrak{h}\otimes\mathfrak{n}\right)
\overset{\operatorname{Ext}_{\mathfrak{h}}^{1}\left(  \operatorname{id}%
,\widetilde{\beta}\right)  }{\longrightarrow}\operatorname{Ext}_{\mathfrak{h}%
}^{1}\left(  \mathfrak{n}\otimes\mathfrak{n},\mathfrak{n}\right)  $ (this map
is owed to the functoriality of $\operatorname{Ext}_{\mathfrak{h}}^{1}$) to
this element, we obtain an element of $\operatorname{Ext}_{\mathfrak{h}}%
^{1}\left(  \mathfrak{n}\otimes\mathfrak{n},\mathfrak{n}\right)  $ which we
call $\alpha$.\newline\textbf{Statement of the theorem:}\newline The following
assertions are equivalent:\newline\textit{Assertion 1:} The natural filtration
on the $\mathfrak{h}$-module $U\left(  \mathfrak{g}\right)  \diagup\left(
U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ (the one obtained by
quotienting from the natural filtration on $U\left(  \mathfrak{g}\right)  $
which, in turn, is obtained by quotienting from the degree filtration on
$\otimes\mathfrak{g}$) is $\mathfrak{h}$-split. (By "$\mathfrak{h}$-split" we
mean "split as a filtration of $\mathfrak{h}$-modules", i. e., the splitting
must be $\mathfrak{h}$-linear.)\newline\textit{Assertion 2:} Considering the
natural filtration on the $\mathfrak{h}$-module $U\left(  \mathfrak{g}\right)
\diagup\left(  U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $,
there exists an isomorphism $U\left(  \mathfrak{g}\right)  \diagup\left(
U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \cong%
\operatorname{Sym}\mathfrak{n}$ of filtered $\mathfrak{h}$-modules. (Here, an
"isomorphism of filtered $\mathfrak{h}$-modules" means an isomorphism of
$\mathfrak{h}$-modules which respects the filtration, as does its
inverse.)\newline\textit{Assertion 3:} The class $\alpha\in\operatorname{Ext}%
_{\mathfrak{h}}^{1}\left(  \mathfrak{n}\otimes\mathfrak{n},\mathfrak{n}%
\right)  $ is trivial.\newline\textit{Assertion 4:} The $\mathfrak{h}$-module
$\mathfrak{n}$ is the restriction of an $\mathfrak{h}^{(1)}$-module to
$\mathfrak{h}$ (via the above-mentioned $k$-Lie algebra homomorphism
$\mathfrak{h}\rightarrow\mathfrak{h}^{(1)}$).
\end{theorem}

Before we proceed any further, let us note that the equivalence of Assertions
3 and 4 in this theorem is rather easy and was proven in \cite{calaque1} (even
in greater generality). More precisely, it is a particular case of the
following lemma (\cite[Lemma 2.3]{calaque1}):

\begin{lemma}
\label{intro.lem2.3}Let $k$ be a field, and let $\mathfrak{g}$ be a $k$-Lie
algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $E$ be
an $\mathfrak{h}$-module. Let $\mathfrak{n}$ denote the quotient
$\mathfrak{h}$-module $\mathfrak{g}\diagup\mathfrak{h}$.\newline%
\textbf{Preparations:}\newline\textbf{1.} Define a Lie algebra $\mathfrak{h}%
^{(1)}$ as in Theorem \ref{intro.MT}. \newline\textbf{2.} Let
$\widetilde{\beta}_{E}:\mathfrak{h}\otimes E\rightarrow E$ be the $k$-linear
map defined by%
\[
\left(
\begin{array}
[c]{c}%
\widetilde{\beta}_{E}\left(  h\otimes E\right)  =\left(  \text{the action of
}h\in\mathfrak{h}\text{ on the element }e\text{ of the }\mathfrak{h}%
\text{-module }E\right) \\
\ \ \ \ \ \ \ \ \ \ \text{for every }h\in\mathfrak{h}\text{ and }e\in E
\end{array}
\right)  .
\]
\newline\textbf{3.} Consider the exact sequence
$\xymatrixcolsep{4pc}\xymatrix{
0 \ar[r] & \mathfrak h \ar[r]^{\text{inclusion}} & \mathfrak g \ar[r]^{\text{projection}} & \mathfrak n \ar[r] & 0
}$ of $\mathfrak{h}$-modules. Tensoring this exact sequence with $E$, we
obtain an exact sequence $\xymatrix{
0 \ar[r] & \mathfrak h \otimes E \ar[r] & \mathfrak g \otimes E \ar[r] & \mathfrak n \otimes E \ar[r] & 0
}$ of $\mathfrak{h}$-modules. This exact sequence gives rise to an element of
$\operatorname{Ext}_{\mathfrak{h}}^{1}\left(  \mathfrak{n}\otimes
E,\mathfrak{h}\otimes E\right)  $. Applying the map $\operatorname{Ext}%
_{\mathfrak{h}}^{1}\left(  \mathfrak{n}\otimes E,\mathfrak{h}\otimes E\right)
\overset{\operatorname{Ext}_{\mathfrak{h}}^{1}\left(  \operatorname{id}%
,\widetilde{\beta}_{E}\right)  }{\longrightarrow}\operatorname{Ext}%
_{\mathfrak{h}}^{1}\left(  \mathfrak{n}\otimes E,E\right)  $ (this map is owed
to the functoriality of $\operatorname{Ext}_{\mathfrak{h}}^{1}$) to this
element, we obtain an element of $\operatorname{Ext}_{\mathfrak{h}}^{1}\left(
\mathfrak{n}\otimes E,E\right)  $ which we call $\alpha_{E}$.\newline%
\textbf{Statement of the lemma:}\newline The class $\alpha_{E}\in
\operatorname{Ext}_{\mathfrak{h}}^{1}\left(  \mathfrak{n}\otimes E,E\right)  $
is trivial if and only if the $\mathfrak{h}$-module $E$ is the restriction of
an $\mathfrak{h}^{(1)}$-module to $\mathfrak{h}$ (via the $k$-Lie algebra
homomorphism $\mathfrak{h}\rightarrow\mathfrak{h}^{(1)}$ mentioned in Theorem
\ref{intro.MT}).
\end{lemma}

This lemma is proven in \cite[Lemma 2.3]{calaque1}. The proof generalizes to
the case when $k$ is a commutative ring, as long as we require the inclusion
$\mathfrak{h}\rightarrow\mathfrak{g}$ to split as a $k$-module
inclusion.\footnote{This is a reasonable requirement, as we also need it for
Lemma \ref{intro.lem2.3} to make sense: If we do not require it, it is no
longer clear why the sequence $\xymatrix{
0 \ar[r] & \mathfrak h \otimes E \ar[r] & \mathfrak g \otimes E \ar[r] & \mathfrak n \otimes E \ar[r] & 0
}$ is exact, but we need this sequence to be exact in order to define the
class $\alpha_{E}\in\operatorname{Ext}_{\mathfrak{h}}^{1}\left(
\mathfrak{n}\otimes E,\mathfrak{h}\otimes E\right)  $.} We are not going to
repeat the proof here.

We are actually going to \textit{avoid the use of the Lie algebra
}$\mathfrak{h}^{(1)}$ in this paper. While it is a very natural construction,
it is rather cumbersome to deal with, and it is nowhere actually used in
\cite{calaque1}; the only things used are the notion of an $\mathfrak{h}%
^{(1)}$-module and the universal enveloping algebra $U\left(  \mathfrak{h}%
^{(1)}\right)  $. Instead of the notion of an $\mathfrak{h}^{(1)}$-module, we
will use the equivalent notion of a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule (a notion we define in Definition \ref{defs.semimod}, and whose
equivalence to that of an $\mathfrak{h}^{(1)}$-module we prove in Proposition
\ref{prop.h^(1)}). Instead of $U\left(  \mathfrak{h}^{(1)}\right)  $, we will
use a $k$-algebra $U\left(  \mathfrak{g},\mathfrak{h}\right)  $ that we define
in Definition \ref{defs.U.semi}, and which turns out to be isomorphic to
$U\left(  \mathfrak{h}^{(1)}\right)  $ (Proposition \ref{prop.U(g,h)=U(h^(1))}%
). Thus, Assertion 4 of Theorem \ref{intro.MT} will rewrite as follows:

\textit{Assertion 4:} The $\mathfrak{h}$-module $\mathfrak{n}$ is the
restriction of a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule to
$\mathfrak{h}$.

In a nutshell, a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule is
the same as a $\mathfrak{g}$-module, except that we no longer require
\[
\left[  a,b\right]  \rightharpoonup v=a\rightharpoonup\left(  b\rightharpoonup
v\right)  -b\rightharpoonup\left(  a\rightharpoonup v\right)
\]
to hold for all $a\in\mathfrak{g}$ and $b\in\mathfrak{g}$ (where
$\rightharpoonup$ denotes the action of the Lie algebra $\mathfrak{g}$ on the
$\mathfrak{g}$-module/$\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule), but only require it to hold for all $a\in\mathfrak{h}$ and
$b\in\mathfrak{g}$. This is a rather down-to-earth notion, and in my opinion
it is much more primordial than that of $\mathfrak{h}^{(1)}$. It actually
gives a justification for the interest in $\mathfrak{h}^{(1)}$ - as the Lie
algebra whose module category is equivalent to the category of $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodules.

The next step in the proof of Theorem \ref{intro.MT} is showing the following
lemma (\cite[Lemma 3.4]{calaque1}):

\begin{lemma}
\label{intro.lem3.2}Let $k$ be a field, and let $\mathfrak{g}$ be a $k$-Lie
algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let
$\mathfrak{n}$ denote the quotient $\mathfrak{h}$-module $\mathfrak{g}%
\diagup\mathfrak{h}$.\newline Let $J$ be the two-sided ideal%
\[
\left(  \otimes\mathfrak{g}\right)  \cdot\left\langle v\otimes w-w\otimes
v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}%
\times\mathfrak{h}\right\rangle \cdot\left(  \otimes\mathfrak{g}\right)
\]
of the $k$-algebra $\otimes\mathfrak{g}$. The degree filtration of the tensor
$k$-algebra $\otimes\mathfrak{g}$ descends to a filtration of the quotient
algebra $\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $, which we denote by
$\left(  F_{n}\right)  _{n\geq0}$. This is actually a filtration of the
$\mathfrak{h}$-module $\left(  \otimes\mathfrak{g}\right)  \diagup\left(
J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $.\newline
Then, for every $n\in\mathbb{N}$, the $n$-th associated graded $\mathfrak{h}%
$-module of $\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ with this filtration
is isomorphic to $\mathfrak{n}^{\otimes n}$ as $\mathfrak{h}$-module. In other
words, every $n\in\mathbb{N}$ satisfies $F_{n}\diagup F_{n-1}\cong%
\mathfrak{n}^{\otimes n}$ as $\mathfrak{h}$-modules.
\end{lemma}

This lemma is proven using the theory of Koszul algebras in \cite{calaque1}.
We are going to prove it elementarily (by recursive construction of an
isomorphism and its inverse) in Section \ref{sect.L25}. Our elementary
approach has the advantage of not depending on homological algebra and thus
not requiring $k$ to be a field; we only need the inclusion $\mathfrak{h}%
\rightarrow\mathfrak{g}$ to split as a $k$-module inclusion. It would not
surprise me if this generality could also be attained by means of the argument
from \cite{calaque1} using relative homology, but this would require redoing
the theory of Koszul algebras in the relative setting, which was too time
consuming a task for me (although probably a rewarding one).

Note that the above statement of Lemma \ref{intro.lem3.2} is not exactly what
this lemma wants to state. Just knowing that $F_{n}\diagup F_{n-1}%
\cong\mathfrak{n}^{\otimes n}$ as $\mathfrak{h}$-modules is not enough for us;
we need to know that a \textbf{very particular} homomorphism $F_{n}\diagup
F_{n-1}\rightarrow\mathfrak{n}^{\otimes n}$ is well-defined and an
isomorphism. This is what Lemma \ref{intro.lem3.2} actually should tell, if we
would allow it to be twice as long. We refer the reader to Theorem
\ref{thm.l25} \textbf{(c)} below for the "right" statement of this lemma. This
"right" statement actually shows that we have a \textit{canonical} isomorphism
$F_{n}\diagup F_{n-1}\rightarrow\mathfrak{n}^{\otimes n}$. However, we are
going to \textit{construct} it by means of a \textit{non-canonical}
isomorphism $\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \rightarrow
\otimes\mathfrak{n}$ (which, however, is non-canonical only by virtue of
depending on the choice of a $k$-vector space complement for $\mathfrak{h}$ in
$\mathfrak{g}$); this will be the isomorphism $\overline{\varphi}$ in
Proposition \ref{prop.l25.Ker-alpha}. The canonicity of the resulting
isomorphism $F_{n}\diagup F_{n-1}\rightarrow\mathfrak{n}^{\otimes n}$ will
come as a surprise.

Lemma \ref{intro.lem3.2} tells us what the associated graded $\mathfrak{h}%
$-modules of the filtered $\mathfrak{h}$-module $\left(  \otimes
\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  $ are isomorphic to, but it does not directly show
how the filtered $\mathfrak{h}$-module $\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
$ \textit{itself} looks; in fact, passing from a filtered $\mathfrak{h}%
$-module to its associated graded $\mathfrak{h}$-modules entails loss of
information (even though a lot of important properties are preserved).
However, when a filtration on a filtered $\mathfrak{h}$-module is
$\mathfrak{h}$-split, then it is determined up to isomorphism by its
associated graded $\mathfrak{h}$-modules. We therefore can ask ourselves when
the filtration $\left(  F_{n}\right)  _{n\geq0}$ on the filtered
$\mathfrak{h}$-module $\left(  \otimes\mathfrak{g}\right)  \diagup\left(
J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ is
$\mathfrak{h}$-split. This is answered by the next lemma, which is \cite[Lemma
3.9]{calaque1}:

\begin{lemma}
\label{intro.lem3.7}Let $k$ be a field, and let $\mathfrak{g}$ be a $k$-Lie
algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$.\newline Let
$\left(  F_{n}\right)  _{n\geq0}$ be defined as in Lemma \ref{intro.lem3.2}.
Let $\alpha$ be defined as in Theorem \ref{intro.MT}.\newline Then, the
filtration $\left(  F_{n}\right)  _{n\geq0}$ is $\mathfrak{h}$-split if and
only if the class $\alpha$ is trivial.
\end{lemma}

Note that one direction of this lemma is more or less straightforward: Namely,
if the filtration $\left(  F_{n}\right)  _{n\geq0}$ is $\mathfrak{h}$-split,
then abstract nonsense (of the trivial sort) shows that the short exact
sequence
$\xymatrixcolsep{4pc}\xymatrix{ 0 \ar[r] & F_1 \diagup F_0 \ar[r]^{\text{inclusion}} & F_2 \diagup F_0 \ar[r]^{\text{projection}} & F_2 \diagup F_1 \ar[r] & 0 }$
must also be $\mathfrak{h}$-split, and thus the class $\alpha$ is trivial
(because it is, up to isomorphism, the class of this sequence, as \cite[Lemma
3.4]{calaque1} shows). We are not going to delve in the details of this argument.

The interesting part is the other direction: to assume that the class $\alpha$
is trivial, and then to show that the filtration $\left(  F_{n}\right)
_{n\geq0}$ is $\mathfrak{h}$-split. In \cite{calaque1}, this is proven using a
Lie-algebraic analogue of the famous projection formula from representation
theory (\cite[Lemma 3.8]{calaque1}). The proof uses Hopf algebras (although
only as a language - no nontrivial facts are used; as opposed to the proof of
Lemma \ref{intro.lem3.2}, this one is completely elementary). Here we are
going to give a different proof (somewhat similar to our proof of Lemma
\ref{intro.lem3.2}) in Section \ref{sect.split} (more precisely, our Theorem
\ref{thm.l37ba} \textbf{(d)} yields that the filtration $\left(  F_{n}\right)
_{n\geq0}$ is $\mathfrak{h}$-split even in a more general context than Lemma
\ref{intro.lem3.7} claims it). Both our proof and the proof given in
\cite{calaque1} begin by applying the equivalence of Assertions 3 and 4 in
Theorem \ref{intro.MT}, so that we know that $\mathfrak{n}$ is the restriction
of an $\mathfrak{h}^{(1)}$-semimodule (i. e., of a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule) to $\mathfrak{h}$, and we want to prove
that the filtration $\left(  F_{n}\right)  _{n\geq0}$ is $\mathfrak{h}$-split.
Both proofs hold true for $k$ being an arbitrary ring as long as the inclusion
$\mathfrak{h}\rightarrow\mathfrak{g}$ splits as a $k$-module inclusion.
Actually, it seems to me that the proofs are kindred (as opposed to the proofs
for Lemma \ref{intro.lem3.2}), although written in different lingos.

The next step is the passage from $\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
$ to $U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(  \mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  $. This is done in \cite[Lemma
4.3]{calaque1}. While the precise assertion of \cite[Lemma 4.3]{calaque1} is
contained in our Theorem \ref{thm.l42} \textbf{(d)}, its actual significance
to the proof lies within the following consequence of \cite[Lemma
4.3]{calaque1}:

\begin{lemma}
\label{intro.lem4.2}Let $k$ be a field, and let $\mathfrak{g}$ be a $k$-Lie
algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let
$\mathfrak{n}$ denote the quotient $\mathfrak{h}$-module $\mathfrak{g}%
\diagup\mathfrak{h}$.\newline Let $n\in\mathbb{N}$. Then, there exists a
canonical $\mathfrak{h}$-module isomorphism $\Theta_{n}:\operatorname{gr}%
_{n}\left(  U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \right)  \rightarrow
\operatorname{Sym}^{n}\mathfrak{n}$ for which the diagram%
\[
\xymatrixcolsep{7pc}\xymatrix{
\operatorname{gr}_n\left(\otimes \mathfrak g\right) \ar[r]^-{\operatorname*{gr}_n\left(\rho\circ\psi\right)} \ar[d]_{\operatorname*{gr}_n\left(\otimes \pi\right)} & \operatorname*{gr}_n\left( U\left(\mathfrak g\right) \diagup \left(U\left(\mathfrak g\right)\cdot\mathfrak h\right) \right) \ar[dd]^{\Theta_n} \\ \operatorname{gr}_n\left(\otimes\mathfrak n\right) \ar[d]_{\operatorname*{grad}_{\mathfrak n,n}^{-1}}^{\cong} & \\ \mathfrak n^{\otimes n} \ar[r]_{\operatorname*{sym}_{\mathfrak n,n}} & \operatorname*{Sym}^n \mathfrak n }
\]
commutes. Here, $\pi$ denotes the canonical projection $\mathfrak{g}%
\rightarrow\mathfrak{g}\diagup\mathfrak{h}=\mathfrak{n}$, while $\psi$ denotes
the canonical projection $\otimes\mathfrak{g}\rightarrow U\left(
\mathfrak{g}\right)  $, while $\rho$ denotes the canonical projection
$U\left(  \mathfrak{g}\right)  \rightarrow U\left(  \mathfrak{g}\right)
\diagup\left(  U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $,
while $\operatorname{grad}_{\mathfrak{n},n}$ denotes the canonical isomorphism
$\mathfrak{n}^{\otimes n}\rightarrow\operatorname{gr}_{n}\left(
\otimes\mathfrak{n}\right)  $, and while $\operatorname{sym}_{\mathfrak{n},n}$
denotes the canonical projection $\mathfrak{n}^{\otimes n}\rightarrow
\operatorname{Sym}^{n}\mathfrak{n}$.
\end{lemma}

This will be proven in parts \textbf{(c)} and \textbf{(d)} of our Corollary
\ref{cor.l42}. The proof is identic to that in \cite[proof of Lemma
4.3]{calaque1}, except that we give more details (as usual) and replace the
"$k$ is a field" condition by something more general - albeit not as general
as for the results before. In Subsection \ref{subsect.goodwillie}, we will
somewhat improve this condition.

So what remains is the proof of Theorem \ref{intro.MT} using all of these
lemmata. We already know that Assertions 3 and 4 are equivalent, which allows
us to forget Assertion 3. Assertions 1 and 2 are also easily seen to be
equivalent (by Proposition \ref{prop.split.filtr-g-2}; see the proof of
Proposition \ref{prop.l49} for how this is used). So we only need to show the
equivalence between Assertions 1 and 4. We will \textit{not} show that
Assertion 1 implies Assertion 4 as this is not difficult and well-explained in
\cite[proof of Theorem 4.5 (c) $\Rightarrow$ (a)]{calaque1} (and is,
apparently, not of too much use: Assertion 1 is much harder to check than
Assertion 4). We \textit{will} show that Assertion 4 implies Assertion 1 in
Theorem \ref{thm.l45}.

\subsection{Remarks on the structure of this paper}

The plan of this paper is as follows:

\begin{vershort}
In \textbf{Section \ref{sect.intro}}, we define a number of notions related to
Lie algebras and their modules, and prove some basic theorems that will later
be used. Every statement in this Section is either well-known or follows
easily from well-known facts; most proofs are only given for the sake of
completeness and would be more appropriate as solutions to homework exercises
in a first course of algebra. Therefore Section \ref{sect.intro} can be safely
skipped by anyone acquainted to Lie algebra theory, except for Definition
\ref{defs.g-alg} (this is a well-known notion, but I am not sure whether it is
well-known under this exact name) and Remark \ref{rmk.g-mod-U(g).short} (for
the disambiguation of the $\mathfrak{g}$-module structure on $U\left(
\mathfrak{g}\right)  $ that I will be using - as it is one of two different,
but equally natural structures).
\end{vershort}

\begin{verlong}
In \textbf{Section \ref{sect.intro}}, we define a number of notions related to
Lie algebras and their modules, and prove some basic theorems that will later
be used. Every statement in this Section is either well-known or follows
easily from well-known facts; most proofs are only given for the sake of
completeness and would be more appropriate as solutions to homework exercises
in a first course of algebra. Therefore Section \ref{sect.intro} can be safely
skipped by anyone acquainted to Lie algebra theory, except for Definition
\ref{defs.g-alg} (this is a well-known notion, but I am not sure whether it is
well-known under this exact name) and Remark \ref{rmk.g-mod-U(g).long} (for
the disambiguation of the $\mathfrak{g}$-module structure on $U\left(
\mathfrak{g}\right)  $ that I will be using - as it is one of two different,
but equally natural structures).
\end{verlong}

\textbf{Section \ref{sect.L25}} is devoted to the proof of Lemma
\ref{intro.lem3.2} in a more general setting. The proof is based on a
$k$-module homomorphism $\varphi:\otimes\mathfrak{g}\rightarrow\otimes N$
(where $N$ is a $k$-module complement to $\mathfrak{h}$ in $\mathfrak{g}$)
which is constructed recursively in Definition \ref{defs.l25.core}. This
homomorphism $\varphi$ was obtained by educated guessing (which, I believe, is
the main contribution of this paper) based on experience with similarly
constructed maps for Clifford algebras (see Subsection \ref{subsect.cliff} for them).

\begin{vershort}
\textbf{Section \ref{sect.(g,h)}} defines the notion of a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule. This is my replacement for the
notion of an $\mathfrak{h}^{(1)}$-module used in \cite{calaque1} (the
equivalence to this latter notion is proven in Proposition \ref{prop.h^(1)})
and shares many properties with the familiar notion of $\mathfrak{g}$-module.
We will not explicitly state most of these properties, as they are direct
analogues of properties of $\mathfrak{g}$-modules given in Section
\ref{sect.intro}.
\end{vershort}

\begin{verlong}
\textbf{Section \ref{sect.(g,h)}} defines the notion of a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule. This is my replacement for the
notion of an $\mathfrak{h}^{(1)}$-module used in \cite{calaque1} (the
equivalence to this latter notion is proven in Proposition \ref{prop.h^(1)})
and shares many properties with the familiar notion of $\mathfrak{g}$-module.
We will state some of these properties; many of them are analogous to (and
often more general than) similar properties of $\mathfrak{g}$-modules given in
Section \ref{sect.intro}.
\end{verlong}

\textbf{Section \ref{sect.split}} proves Lemma \ref{intro.lem3.7} in the
equivalent form given above (instead of assuming that $\alpha$ is trivial, we
assume that $\mathfrak{n}$ is the restriction of a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule). Again, the proof is given in more
generality than Lemma \ref{intro.lem3.7} itself. The idea of the proof - a
recursive construction of a homomorphism $\gamma:\otimes\mathfrak{g}%
\rightarrow\otimes\mathfrak{n}$ (this time, as opposed to Section
\ref{sect.L25}, we use $\otimes\mathfrak{n}$ instead of $\otimes N$, albeit
these two $k$-algebras are isomorphic), which is an $\mathfrak{h}$-module
homomorphism this time - is similar to that of Section \ref{sect.L25}, and so
are some further steps of the proof.

\textbf{Section \ref{sect.pbw}} then states the Poincar\'{e}-Birkhoff-Witt
theorem in several versions, and completes the proofs of Lemma
\ref{intro.lem4.2} and Theorem \ref{intro.MT}. Again, the situation considered
in Section \ref{sect.pbw} is more general than that of Lemma
\ref{intro.lem4.2} and Theorem \ref{intro.MT}, although not as general as that
of Sections \ref{sect.L25} and \ref{sect.split}.

The final \textbf{Section \ref{sect.gen}} is a kind of odds-and-ends section.
It begins with Subsection \ref{subsect.general}, which tries to squeeze out
some additional generality from the results of Sections \ref{sect.L25} and
\ref{sect.split}. Subsection \ref{subsect.cliff} discusses analogues of the
results of \cite{calaque1} in the Clifford algebra of a quadratic space
(instead of the universal enveloping algebra of a Lie algebra). Subsection
\ref{subsect.super} is devoted to generalization to Lie superalgebras.
Subsection \ref{subsect.goodwillie} extends Theorem \ref{thm.l42} to a less
restrictive case (rather than requiring $\mathfrak{h}$ and $N$ to be free
$k$-modules, we only demand $\mathfrak{g}\diagup\mathfrak{h}$ to be a flat
$k$-module), whose proof is due to Thomas Goodwillie.

\begin{noncompile}
Subsection \ref{subsect.character} finally discusses a slightly more general
situation, which is reduced to the situation considered before by an
automorphism of $U\left(  \mathfrak{g}\right)  $. Subsection
\ref{subsect.restr} gives a remark about a version of Theorem \ref{intro.MT}
for fields of characteristic $p>0$, where the universal enveloping algebra
$U\left(  \mathfrak{g}\right)  $ is replaced by the restricted universal
enveloping algebra (where $\mathfrak{g}$ is supposed to be a restricted
$p$-Lie algebra).
\end{noncompile}

Here is a graph depicting the dependencies of the sections of this note on
each other:%
\[%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{5pc}
%\xymatrix{
%& \text{Section 1} & \\
%\text{Section 2} \ar[ru] & \text{Section 3} \ar[u] & \text{Section 4}
%\ar[lu] \ar[l] \\
%\text{Section 5} \ar[ruu] \ar[u] \ar[rru] \ar@{.>}[ru] & & \text{Section 6}
%\ar[ll] \ar[llu] \ar[lu] \ar[u] \ar[luu]
%}}}%
%BeginExpansion
\xymatrixcolsep{5pc}
\xymatrix{
& \text{Section 1} & \\
\text{Section 2} \ar[ru] & \text{Section 3} \ar[u] & \text{Section 4}
\ar[lu] \ar[l] \\
\text{Section 5} \ar[ruu] \ar[u] \ar[rru] \ar@{.>}[ru] & & \text{Section 6}
\ar[ll] \ar[llu] \ar[lu] \ar[u] \ar[luu]
}%
%EndExpansion
\]
(arrow means dependency; dotted arrow means very minor dependency).

I have tried to keep this paper as detailed and unambiguous as possible. In
particular, I have abdicated many of the common abuses of notation, like
silently identifying things which are actually only isomorphic rather than
equal\footnote{I have not abdicated this completely. I \textit{do} make such
identifications in certain places: For example, I identify $V^{\otimes
n}\otimes V^{\otimes m}$ with $V^{\otimes\left(  n+m\right)  }$ for any
$k$-module $V$ and $n\in\mathbb{N}$ and $m\in\mathbb{N}$. And I identify
$V^{\otimes n}$ with a submodule of $\bigoplus\limits_{n\in\mathbb{N}%
}V^{\otimes n}$. However, I try to keep these identifications to a minimum; in
particular I never identify $F_{n}\diagup F_{n-1}$ with $\mathfrak{n}^{\otimes
n}$ in Theorem \ref{thm.l25} (although $F_{n}\diagup F_{n-1}\cong%
\mathfrak{n}^{\otimes n}$ canonically), and I never identify
$\operatorname{gr}_{n}\left(  U\left(  \mathfrak{g}\right)  \right)  $ with
$\operatorname{Sym}^{n}\mathfrak{g}$ in Proposition \ref{prop.pbw} (despite
the isomorphism $\operatorname{gr}_{n}\left(  U\left(  \mathfrak{g}\right)
\right)  \cong\operatorname{Sym}^{n}\mathfrak{g}$ when the $n$-PBW condition
is satisfied), and I do not even identify $\operatorname{gr}_{p}\left(
\otimes V\right)  $ with $V^{\otimes p}$ in Proposition \ref{defs.filtr.grad}
(although $\operatorname{gr}_{p}\left(  \otimes V\right)  \cong V^{\otimes p}$
rather trivially).}, or saying "$U$ and $V$ are isomorphic" when actually
meaning the stronger assertion "a very particular homomorphism $U\rightarrow
V$ is an isomorphism". This noticeably contributes to the length of this
paper, but hopefully does so to its readability as well.

Also I have tried to keep theorems self-contained. This means that all
notations used in a theorem are defined there, or the places where they are
defined are referenced in the theorem. Unsurprisingly, this has stretched the
lengths of theorems, but again I hope it was not a vain endeavour.

\subsection{Acknowledgements}

This paper grew out of a work \cite{calaque1} by Damien Calaque, Andrei
C\u{a}ld\u{a}raru, Junwu Tu. I am indebted to Giovanni Felder for acquainting
me with this work and to Giovanni Felder and Damien Calaque for inviting me to
a research stay at the ETH Z\"{u}rich.

Andreas Rosenschon advised this thesis and helped out with many valuable
discussions. I have learned much of what I am using in this paper from
Hans-J\"{u}rgen Schneider's lectures on Hopf algebra and Pavel Etingof's texts
and e-mails.

Further thanks go to my parents for starting off my mathematical education.

\subsection{Basic conventions}

Before we come to the actual body of this note, let us fix some conventions to
prevent misunderstandings from happening:

\begin{condition}
In this note, $\mathbb{N}$ will mean the set $\left\{  0,1,2,3,...\right\}  $
(rather than the set $\left\{  1,2,3,...\right\}  $, which is denoted by
$\mathbb{N}$ by various other authors).
\end{condition}

\begin{condition}
\label{defs.ring}In this note, a \textit{ring} will always mean a ring with
$1$. If $k$ is a ring, a $k$-\textit{algebra} will mean a (not necessarily
commutative, but necessarily associative) $k$-algebra with $1$. Sometimes we
will use the word "algebra" as an abbreviation for "$k$-algebra". If $L$ is a
$k$-algebra, then a \textit{left }$L$\textit{-module} is always supposed to be
a left $L$-module on which the unity of $L$ acts as the identity. Whenever we
use the tensor product sign $\otimes$ without an index, we mean $\otimes_{k}$.
Similarly, whenever we use the $\operatorname*{Hom}$ and $\operatorname*{End}$
signs without index, we mean $\operatorname*{Hom}\nolimits_{k}$ and
$\operatorname*{End}\nolimits_{k}$, respectively.
\end{condition}

\section{\label{sect.intro}Basics about Lie algebras and their modules}

First we are going to recollect the most fundamental definitions and results
(and, sometimes, even proofs) from the theory of Lie algebras. While most of
these appear in literature, we will recapitulate them already in order to
introduce all of the notations that we are going to use.

\begin{vershort}
Almost all results in Section \ref{sect.intro} are classical and well-known,
so I am not going to give their proofs.
\end{vershort}

\begin{verlong}
Almost all results in Section \ref{sect.intro} are classical and well-known,
so there is no need to prove all of them. However, I will prove \textit{some}
of them, in order to counter the unfortunate trend in literature of leaving
such proofs to the reader. For example, our Proposition
\ref{prop.lie(X).bracketing}, which more or less says that the tensor product
of several $\mathfrak{g}$-modules does not depend on the bracketing (in the
sense that different bracketings produce different but canonically isomorphic
$\mathfrak{g}$-modules), is known to and considered obvious by everyone
working in the field. But I have not seen a detailed proof anywhere in a text.
While the proof is straightforward, I did not consider my text complete until
the proof was written out.
\end{verlong}

\subsection{Lie algebras}

First we recall the basic properties of Lie algebras. Some fundamental definitions:

\begin{definition}
\label{defs.lie}Let $k$ be a commutative ring. A $k$\textit{-Lie algebra} will
mean a $k$-module $\mathfrak{g}$ together with a $k$-bilinear map
$\beta:\mathfrak{g}\times\mathfrak{g}\rightarrow\mathfrak{g}$ satisfying the
conditions%
\begin{align}
&  \left(  \beta\left(  v,v\right)  =0\text{ for every }v\in\mathfrak{g}%
\right)  \ \ \ \ \ \ \ \ \ \ \text{and}\label{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{beta-2}%
\end{align}
\newline This $k$-bilinear map $\beta:\mathfrak{g}\times\mathfrak{g}%
\rightarrow\mathfrak{g}$ will be called the \textit{Lie bracket} of the
$k$-Lie algebra $\mathfrak{g}$. We will often use the square brackets notation
for $\beta$, which means that we are going to abbreviate $\beta\left(
v,w\right)  $ by $\left[  v,w\right]  $ for any $v\in\mathfrak{g}$ and
$w\in\mathfrak{g}$. Using this notation, the equations (\ref{beta-1}) and
(\ref{beta-2}) rewrite as%
\begin{align}
&  \left(  \left[  v,v\right]  =0\text{ for every }v\in\mathfrak{g}\right)
\ \ \ \ \ \ \ \ \ \ \text{and}\label{[]-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{[]-2}%
\end{align}
\newline The equation (\ref{beta-2}) (or its equivalent version (\ref{[]-2}))
is called the \textit{Jacobi identity}.\newline Also, we will abbreviate the
notion "$k$-Lie algebra" as "Lie algebra", as long as the underlying ring $k$
will be obvious from the context.
\end{definition}

Note that Lie algebras are not algebras (in our nomenclature), since we
require algebras to be associative.

\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 $\mathfrak{g}$ (it
needs not be actually called $\mathfrak{g}$; I only refer to it by
$\mathfrak{g}$ here in this Convention), and we are given two elements $v$ and
$w$ of $\mathfrak{g}$ (they need not be actually called $v$ and $w$; I only
refer to them by $v$ and $w$ here in this Convention), 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).
\end{condition}

\begin{proposition}
\label{prop.[v,w]=-[w,v]}Let $k$ be a commutative ring. Every $k$-Lie algebra
$\mathfrak{g}$ satisfies%
\begin{equation}
\left(  \left[  v,w\right]  =-\left[  w,v\right]  \text{ for every }%
v\in\mathfrak{g}\text{ and }w\in\mathfrak{g}\right)  . \label{[]-3}%
\end{equation}
(Here, according to Convention \ref{conv.[]}, we denote by $\left[
v,w\right]  $ the Lie bracket of the Lie algebra $\mathfrak{g}$, applied to
$\left(  v,w\right)  $, and we denote by $\left[  w,v\right]  $ the Lie
bracket of the Lie algebra $\mathfrak{g}$, applied to $\left(  w,v\right)  $.)
In other words, if $\beta$ denotes the Lie bracket of $\mathfrak{g}$, then we
have
\begin{equation}
\left(  \beta\left(  v,w\right)  =-\beta\left(  w,v\right)  \text{ for every
}v\in\mathfrak{g}\text{ and }w\in\mathfrak{g}\right)  . \label{beta-3}%
\end{equation}

\end{proposition}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.[v,w]=-[w,v]}.} Let $v\in\mathfrak{g}$
and $w\in\mathfrak{g}$ be arbitrary.
\end{verlong}

\begin{verlong}
Let $\beta$ denote the Lie bracket of $\mathfrak{g}$. Applying (\ref{beta-1})
to $v+w$ instead of $v$, we obtain $\beta\left(  v+w,v+w\right)  =0$. Thus,%
\begin{align*}
0  &  =\beta\left(  v+w,v+w\right)  =\underbrace{\beta\left(  v+w,v\right)
}_{\substack{=\beta\left(  v,v\right)  +\beta\left(  w,v\right)
\\\text{(since }\beta\text{ is bilinear)}}}+\underbrace{\beta\left(
v+w,w\right)  }_{\substack{=\beta\left(  v,w\right)  +\beta\left(  w,w\right)
\\\text{(since }\beta\text{ is bilinear)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\beta\text{ is bilinear}\right) \\
&  =\underbrace{\beta\left(  v,v\right)  }_{=0\text{ (by (\ref{beta-1}))}%
}+\beta\left(  w,v\right)  +\beta\left(  v,w\right)  +\underbrace{\beta\left(
w,w\right)  }_{\substack{=0\text{ (by (\ref{beta-1}),}\\\text{applied to
}w\text{ instead of }v\text{)}}}=0+\beta\left(  w,v\right)  +\beta\left(
v,w\right)  +0\\
&  =\beta\left(  w,v\right)  +\beta\left(  v,w\right)  .
\end{align*}
This rewrites as $\beta\left(  v,w\right)  =-\beta\left(  w,v\right)  $. Thus
(\ref{beta-3}) is proven.
\end{verlong}

\begin{verlong}
According to Convention \ref{conv.[]}, we denote by $\left[  v,w\right]  $ the
Lie bracket of $\mathfrak{g}$, applied to $\left(  v,w\right)  $. Since the
Lie bracket of $\mathfrak{g}$ is $\beta$, this becomes $\left[  v,w\right]
=\beta\left(  v,w\right)  $. Similarly, $\left[  w,v\right]  =\beta\left(
w,v\right)  $. Thus, $\left[  v,w\right]  =\beta\left(  v,w\right)
=-\underbrace{\beta\left(  w,v\right)  }_{=\left[  w,v\right]  }=-\left[
w,v\right]  $. This proves (\ref{[]-3}). We have thus proven Proposition
\ref{prop.[v,w]=-[w,v]}.
\end{verlong}

\subsection{Lie subalgebras and Lie algebra homomorphisms}

The following definition of the notion of a Lie subalgebra holds little surprise:

\begin{definition}
\label{defs.liesub}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $\mathfrak{h}$ be a $k$-submodule of $\mathfrak{g}$.
Then, we say that $\mathfrak{h}$ is a $k$\textit{-Lie subalgebra} of
$\mathfrak{g}$ if every $u\in\mathfrak{h}$ and $v\in\mathfrak{h}$ satisfy
$\left[  u,v\right]  \in\mathfrak{h}$.\newline We will abbreviate "$k$-Lie
subalgebra" as "Lie subalgebra" when $k$ is clear from the context.
\end{definition}

This Definition \ref{defs.liesub} is fundamental to this paper, as we are
going to study the interplay between the universal enveloping algebra
$U\left(  \mathfrak{g}\right)  $ (defined in Definition \ref{defs.U}) with a
Lie subalgebra $\mathfrak{h}$ of $\mathfrak{g}$.

As opposed to this, the following four definitions will only be used
marginally (namely, in Subsection \ref{subsect.U(g,h)}):

\begin{definition}
\label{defs.liehom}Let $k$ be a commutative ring. Let $\mathfrak{g}$ and
$\mathfrak{h}$ be two $k$-Lie algebras. Let $f:\mathfrak{g}\rightarrow
\mathfrak{h}$ be a map. This map $f$ is said to be a \textit{Lie algebra
homomorphism} if and only if it is $k$-linear and satisfies%
\[
\left(  f\left(  \left[  v,w\right]  \right)  =\left[  f\left(  v\right)
,f\left(  w\right)  \right]  \ \ \ \ \ \ \ \ \ \ \text{for every }%
v\in\mathfrak{g}\text{ and }w\in\mathfrak{g}\right)  .
\]
(In this equation, according to Convention \ref{conv.[]}, the term $\left[
v,w\right]  $ denotes the Lie bracket of $\mathfrak{g}$ applied to $\left(
v,w\right)  $, whereas the term $\left[  f\left(  v\right)  ,f\left(
w\right)  \right]  $ denotes the Lie bracket of $\mathfrak{h}$ applied to
$\left(  f\left(  v\right)  ,f\left(  w\right)  \right)  $.)
\end{definition}

\begin{definition}
\label{defs.gener}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $S$ be a subset of $\mathfrak{g}$.\newline\textbf{(1)}
Consider the subset of $\mathfrak{g}$ which consists of every element which
can be obtained by repeated addition, scalar multiplication (i. e.,
multiplication with elements of $k$) and forming the Lie bracket from elements
of $S$. This subset is called the \textit{Lie subalgebra of }$\mathfrak{g}$
\textit{generated by }$S$. (It is easy to see that this subset indeed is a Lie
subalgebra of $\mathfrak{g}$, so this name is justified. It is also easy to
see that this subset is the smallest Lie subalgebra of $\mathfrak{g}$ which
contains $S$ as a subset, where "smallest" means "smallest with respect to
inclusion".)\newline\textbf{(2)} The Lie algebra $\mathfrak{g}$ is said to be
\textit{generated (as a Lie algebra) by the subset }$S$ (or also
\textit{generated (as a Lie algebra) by the elements of }$S$) if and only if
$\mathfrak{g}$ is identical with the Lie subalgebra of $\mathfrak{g}$
generated by $S$.\newline(Note that assertions like "The Lie algebra
$\mathfrak{g}$ is generated (as a Lie algebra) by the subset $S$" should never
be confused with assertions like "The $k$-module $\mathfrak{g}$ is generated
(as a $k$-module) by the subset $S$", even though every Lie algebra is a
$k$-module. If we have a Lie algebra $\mathfrak{g}$ and we know that the
$k$-module $\mathfrak{g}$ is generated (as a $k$-module) by some subset $S$,
then we can conclude that the Lie algebra $\mathfrak{g}$ is generated (as a
Lie algebra) by $S$ as well; but the converse direction does not hold.)
\end{definition}

\begin{definition}
\label{defs.lieideal}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $\mathfrak{i}$ be a subset of $\mathfrak{g}$%
.\newline\textbf{(a)} We say that $\mathfrak{i}$ is a \textit{Lie ideal} of
$\mathfrak{g}$ if and only if $\mathfrak{i}$ is a $k$-submodule of
$\mathfrak{g}$ satisfying%
\[
\left(  \left[  v,x\right]  \in\mathfrak{i}\ \ \ \ \ \ \ \ \ \ \text{for every
}v\in\mathfrak{g}\text{ and }x\in\mathfrak{i}\right)  .
\]
\newline\textbf{(b)} If $\mathfrak{i}$ is a Lie ideal of $\mathfrak{g}$, then
the $k$-module $\mathfrak{g}\diagup\mathfrak{i}$ can be made into a $k$-Lie
algebra by setting%
\[
\left(  \left[  \overline{v},\overline{w}\right]  =\overline{\left[
v,w\right]  }\ \ \ \ \ \ \ \ \ \ \text{for every }v\in\mathfrak{g}\text{ and
}w\in\mathfrak{g}\right)
\]
(where for every $t\in\mathfrak{g}$, the residue class of $t$ modulo
$\mathfrak{i}$ is denoted by $\overline{t}$). This $k$-Lie algebra is indeed
well-defined, as can easily be seen.
\end{definition}

\begin{definition}
\label{defs.lieideal.gener}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie algebra. Let $S$ be a subset of $\mathfrak{g}$.\newline For any
$a\in\mathfrak{g}$, define a map $\operatorname*{ad}\nolimits_{a}%
:\mathfrak{g}\rightarrow\mathfrak{g}$ by $\left(  \operatorname*{ad}%
\nolimits_{a}\left(  x\right)  =\left[  a,x\right]  \text{ for every }%
x\in\mathfrak{g}\right)  $.\newline Consider the subset of $\mathfrak{g}$
which consists of every element which can be obtained by repeated addition,
scalar multiplication (i. e., multiplication with elements of $k$) and
application of the maps $\operatorname*{ad}\nolimits_{a}$ (where
$a\in\mathfrak{g}$ can be arbitrarily chosen and does not have to be the same
each time we apply $\operatorname*{ad}\nolimits_{a}$) from elements of $S$.
This subset is called the \textit{Lie ideal of }$\mathfrak{g}$
\textit{generated by }$S$. (It is easy to see that this subset indeed is a Lie
ideal of $\mathfrak{g}$, so this name is justified. It is also easy to see
that this subset is the smallest Lie ideal of $\mathfrak{g}$ which contains
$S$ as a subset, where "smallest" means "smallest with respect to inclusion".)
\end{definition}

\subsection{Modules over Lie algebras}

While Lie algebras are interesting for themselves, they are often better
understood through their modules. Here is a definition of this notion:

\begin{definition}
\label{defs.liemod}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $V$ be a $k$-module. Let $\mu:\mathfrak{g}\times
V\rightarrow V$ be a $k$-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{L-mod}%
\end{equation}
If $\left(  V,\mu\right)  $ is a $\mathfrak{g}$-module, then the $k$-bilinear
map $\mu:\mathfrak{g}\times V\rightarrow V$ is called the \textit{Lie action}
of the $\mathfrak{g}$-module $V$.\newline 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{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{[L-mod]}%
\end{equation}
\newline Also, an abuse of notation allows us to write "$V$ is a
$\mathfrak{g}$-module" instead of "$\left(  V,\mu\right)  $ is a
$\mathfrak{g}$-module" 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 $k$-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. (Thus, in order to make a $k$-module into a
$\mathfrak{g}$-module, we must define a $\mathfrak{g}$-module structure on it.)
\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 $V$ (they need not be actually called $\mathfrak{g}$ and $V$; I only
refer to them as $\mathfrak{g}$ and $V$ here in this Convention), and we are
given two elements $a\in\mathfrak{g}$ and $v\in V$ (they need not be actually
called $a$ and $v$; I only refer to them by $a$ and $v$ here in this
Convention), 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).\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 brackets in this case to make the correct interpretation of
the formula even more obvious.
\end{condition}

\textbf{Notational remark.} Most authors abbreviate the term $\mu\left(
a,v\right)  $ (where $\mu$ is the Lie action of a $\mathfrak{g}$-module) by
$a\cdot v$ or (even shorter) by $av$, wherever $a$ is an element of a Lie
algebra $\mathfrak{g}$ and $v$ is an element of a $\mathfrak{g}$-module $V$.
However, we cannot afford using this abbreviation, since we will define a
$\mathfrak{g}$-module structure on $\otimes\mathfrak{g}$ which is \textit{not}
the left multiplication, so, if we would abbreviate the term $\mu\left(
a,v\right)  $ by $a\cdot v$, we would risk confusing it with the product of
$a$ and $v$ in the tensor algebra $\otimes\mathfrak{g}$. This is why I prefer
the abbreviation $a\rightharpoonup v$.

\begin{remark}
The notion of a "$\mathfrak{g}$-module" that we defined in Definition
\ref{defs.liemod} is often referred to as a "left $\mathfrak{g}$-module".
There is also a similar notion of a "right $\mathfrak{g}$-module". However,
there is not much difference between left $\mathfrak{g}$-modules and right
$\mathfrak{g}$-modules (in particular, every left $\mathfrak{g}$-module can be
canonically made a right $\mathfrak{g}$-module and vice versa).{\footnotemark}
When we speak of $\mathfrak{g}$-modules, we will always mean left
$\mathfrak{g}$-modules.
\end{remark}

\footnotetext{This situation is contradistinctive to the situation for
$A$-modules, where $A$ is an associative algebra. In fact, when $A$ is a
non-commutative associative algebra, there is (in general) no way to transform
left $A$-modules into right $A$-modules.}

Now that we have defined a $\mathfrak{g}$-module, let us do the next logical
step and define a $\mathfrak{g}$-module homomorphism:

\begin{definition}
\label{defs.liemodhom}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $V$ and $W$ be two $\mathfrak{g}$-modules. Let
$f:V\rightarrow W$ be a $k$-linear map. Then, $f$ is said to be a
$\mathfrak{g}$\textit{-module homomorphism} if and only if%
\[
\left(  f\left(  a\rightharpoonup v\right)  =a\rightharpoonup\left(  f\left(
v\right)  \right)  \ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{
and }v\in V\right)  .
\]
\newline Often, we will use the words "$\mathfrak{g}$-module map" or the words
"homomorphism of $\mathfrak{g}$-modules" or the words "$\mathfrak{g}$-linear
map" as synonyms for "$\mathfrak{g}$-module homomorphism".
\end{definition}

It is easy to see that for every commutative ring $k$ and every $k$-Lie
algebra $\mathfrak{g}$, there is a category whose objects are $\mathfrak{g}%
$-modules and whose morphisms are $\mathfrak{g}$-module homomorphisms. We
further define a $\mathfrak{g}$-module isomorphism as an isomorphism in this
category; this is equivalent to the following definition:

\begin{definition}
\label{defs.liemodiso}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $V$ and $W$ be two $\mathfrak{g}$-modules. Let
$f:V\rightarrow W$ be a $k$-linear map. Then, $f$ is said to be a
$\mathfrak{g}$\textit{-module isomorphism} if and only if $f$ is an invertible
$\mathfrak{g}$-module homomorphism whose inverse $f^{-1}$ is also a
$\mathfrak{g}$-module homomorphism.
\end{definition}

We can easily prove that this definition is somewhat redundant, viz., the
condition that $f^{-1}$ be also a $\mathfrak{g}$-module homomorphism can be omitted:

\begin{proposition}
\label{prop.liemodiso}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $V$ and $W$ be two $\mathfrak{g}$-modules. Let
$f:V\rightarrow W$ be a $k$-linear map. Then, $f$ is a $\mathfrak{g}$-module
isomorphism if and only if $f$ is an invertible $\mathfrak{g}$-module
homomorphism. In other words, $f$ is a $\mathfrak{g}$-module isomorphism if
and only if $f$ is a $\mathfrak{g}$-module homomorphism and a $k$-module
isomorphism at the same time.
\end{proposition}

\begin{verlong}
The proof of this proposition uses the same idea as the standard proof that
the inverse of an invertible $A$-linear map is $A$-linear (where $A$ is a ring).
\end{verlong}

It is easy to see that kernels and images of $\mathfrak{g}$-module
isomorphisms are $\mathfrak{g}$-submodules.

\subsection{Restriction of $\mathfrak{g}$-modules}

If $\mathfrak{h}$ is a Lie subalgebra of a $k$-Lie algebra $\mathfrak{g}$,
then we can canonically make every $\mathfrak{g}$-module into an
$\mathfrak{h}$-module according to the following definition:

\begin{definition}
\label{defs.g-as-h-module}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a $k$-Lie algebra, and let $\mathfrak{h}$ be a Lie subalgebra of
$\mathfrak{g}$. Then, every $\mathfrak{g}$-module $V$ canonically becomes an
$\mathfrak{h}$-module (by restricting its Lie action $\mu:\mathfrak{g}\times
V\rightarrow V$ to $\mathfrak{h}\times V$). This $\mathfrak{h}$-module is
called the \textit{restriction of }$V$ \textit{to }$\mathfrak{h}$, and denoted
by $\operatorname*{Res}\nolimits_{\mathfrak{h}}^{\mathfrak{g}}V$. However,
when there is no possibility of confusion, we will denote this $\mathfrak{h}%
$-module by $V$, and we will distinguish it from the original $\mathfrak{g}%
$-module $V$ by means of referring to the former one as "the $\mathfrak{h}%
$-module $V$" and referring to the latter one as "the $\mathfrak{g}$-module
$V$".
\end{definition}

\subsection{The $\mathfrak{g}$-modules $\mathfrak{g}$ and $k$}

Now we notice that the Lie algebra $\mathfrak{g}$ itself is a $\mathfrak{g}$-module:

\begin{proposition}
\label{prop.g-is-g-module}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a Lie algebra. Let $\beta$ be the Lie bracket of $\mathfrak{g}$ (so that
$\beta\left(  v,w\right)  =\left[  v,w\right]  $ for all $v\in\mathfrak{g}$
and $w\in\mathfrak{g}$). Then, $\left(  \mathfrak{g},\beta\right)  $ is a
$\mathfrak{g}$-module. This $\mathfrak{g}$-module satisfies%
\begin{equation}
v\rightharpoonup w=\left[  v,w\right]  \ \ \ \ \ \ \ \ \ \ \text{for all }%
v\in\mathfrak{g}\text{ and }w\in\mathfrak{g}\text{.}
\label{prop.g-is-g-module.form}%
\end{equation}

\end{proposition}

\begin{definition}
\label{defs.g-mod-g}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a Lie
algebra. Whenever we speak of "the $\mathfrak{g}$-module $\mathfrak{g}$"
without specifying the $\mathfrak{g}$-module structure, we mean the
$\mathfrak{g}$-module $\left(  \mathfrak{g},\beta\right)  $ defined in
Proposition \ref{prop.g-is-g-module}.
\end{definition}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.g-is-g-module}.} According to
Definition \ref{defs.liemod}, in order to prove that $\left(  \mathfrak{g}%
,\beta\right)  $ is a $\mathfrak{g}$-module, we need to show that $\beta$ is a
$k$-bilinear map, and that the equation (\ref{L-mod}) holds with $\mu$ and $V$
replaced by $\beta$ and $\mathfrak{g}$.
\end{verlong}

\begin{verlong}
Every $a\in\mathfrak{g}$, $b\in\mathfrak{g}$ and $v\in\mathfrak{g}$ satisfy%
\begin{align*}
\beta\left(  \left[  a,b\right]  ,v\right)   &  =\left[  \left[  a,b\right]
,v\right]  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\beta\left(  v,w\right)
=\left[  v,w\right]  \text{ for any }v\in\mathfrak{g}\text{ and }%
w\in\mathfrak{g}\right) \\
&  =-\left[  v,\left[  a,b\right]  \right]  \ \ \ \ \ \ \ \ \ \ \left(
\text{due to (\ref{[]-3}), applied to }\left[  a,b\right]  \text{ and }v\text{
instead of }v\text{ and }w\right) \\
&  =\left[  a,\left[  b,v\right]  \right]  +\left[  b,\underbrace{\left[
v,a\right]  }_{\substack{=-\left[  a,v\right]  \\\text{(due to (\ref{[]-3}),
applied}\\\text{to }a\text{ instead of }w\text{)}}}\right] \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since (\ref{[]-2}) (applied to }v\text{, }a\text{, }b\text{ instead of
}u\text{, }v\text{, }w\text{) yields}\\
\left[  v,\left[  a,b\right]  \right]  +\left[  a,\left[  b,v\right]  \right]
+\left[  b,\left[  v,a\right]  \right]  =0
\end{array}
\right) \\
&  =\left[  a,\left[  b,v\right]  \right]  +\underbrace{\left[  b,-\left[
a,v\right]  \right]  }_{\substack{=-\left[  b,\left[  a,v\right]  \right]
\\\text{(since the Lie bracket is }k\text{-bilinear)}}}=\left[  a,\left[
b,v\right]  \right]  -\left[  b,\left[  a,v\right]  \right] \\
&  =\beta\left(  a,\beta\left(  b,v\right)  \right)  -\beta\left(
b,\beta\left(  a,v\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left[  v,w\right]  =\beta\left(
v,w\right)  \text{ for any }v\in\mathfrak{g}\text{ and }w\in\mathfrak{g}%
\right)  .
\end{align*}
In other words, the equation (\ref{L-mod}) holds with $\mu$ and $V$ replaced
by $\beta$ and $\mathfrak{g}$. Since $\beta$ is a $k$-bilinear map (since
$\beta$ is the Lie bracket of $\mathfrak{g}$, and the definition of a Lie
algebra requires that the Lie bracket is a $k$-bilinear map), Definition
\ref{defs.liemod} therefore yields that $\left(  \mathfrak{g},\beta\right)  $
is a $\mathfrak{g}$-module. This module satisfies $v\rightharpoonup
w=\beta\left(  v,w\right)  =\left[  v,w\right]  $ for all $v\in\mathfrak{g}$
and $w\in\mathfrak{g}$. Thus, Proposition \ref{prop.g-is-g-module} is proven.
\end{verlong}

There is one yet simpler $\mathfrak{g}$-module:

\begin{proposition}
\label{prop.k-is-g-module}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a Lie algebra. Then, $\left(  k,0\right)  $ is a $\mathfrak{g}$-module (where
$0$ denotes the map $\mathfrak{g}\times k\rightarrow k$ which sends everything
to zero). This $\mathfrak{g}$-module satisfies%
\[
v\rightharpoonup\lambda=0\ \ \ \ \ \ \ \ \ \ \text{for all }v\in
\mathfrak{g}\text{ and }\lambda\in k\text{.}%
\]

\end{proposition}

\begin{definition}
\label{defs.k-mod-g}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a Lie
algebra. Whenever we speak of "the $\mathfrak{g}$-module $k$" without
specifying the $\mathfrak{g}$-module structure, we mean the $\mathfrak{g}%
$-module $\left(  k,0\right)  $ defined in Proposition
\ref{prop.k-is-g-module}. This $\mathfrak{g}$-module is called \textit{the
trivial }$\mathfrak{g}$\textit{-module}.
\end{definition}

\subsection{Submodules, factors and direct sums of $\mathfrak{g}$-modules}

There are more interesting $\mathfrak{g}$-module structures around. One way to
obtain them is to factor existing $\mathfrak{g}$-modules by submodules:

\begin{definition}
\label{defs.submod}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a Lie
algebra. Let $V$ be a $\mathfrak{g}$-module.\newline\textbf{(a)} A
$k$-submodule $W$ of $V$ is said to be a $\mathfrak{g}$\textit{-submodule} of
$V$ if and only if%
\[
\left(  a\rightharpoonup w\in W\text{ for every }a\in\mathfrak{g}\text{ and
}w\in W\right)  .
\]
In other words, a $k$-submodule $W$ of $V$ is said to be a $\mathfrak{g}%
$\textit{-submodule} of $V$ if and only if $\mu\left(  \mathfrak{g}\times
W\right)  \subseteq W$, where $\mu$ denotes the Lie action of $V$. (We remind
ourselves that the Lie action of $V$ means the $k$-bilinear map $\mu
:\mathfrak{g}\times V\rightarrow V$ from Definition \ref{defs.liemod}%
.)\newline\textbf{(b)} If $W$ is a $\mathfrak{g}$-submodule of $V$, then the
quotient $k$-module $V\diagup W$ becomes a $\mathfrak{g}$-module by setting%
\[
\left(  a\rightharpoonup\overline{v}=\overline{a\rightharpoonup v}\text{ for
every }a\in\mathfrak{g}\text{ and }v\in V\right)
\]
(where $\overline{u}$ denotes the residue class of $u$ modulo $W$ for every
$u\in V$). (This $\mathfrak{g}$-module structure is indeed well-defined, as
can be easily seen.)
\end{definition}

We can also add $\mathfrak{g}$-modules via the direct sum:

\begin{proposition}
\label{prop.lie(+)}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a Lie
algebra. Let $V$ and $W$ be two $\mathfrak{g}$-modules. Define a map
$\mu_{V\oplus W}:\mathfrak{g}\times\left(  V\oplus W\right)  \rightarrow
V\oplus W$ by%
\begin{equation}
\left(  \mu_{V\oplus W}\left(  a,\left(  v,w\right)  \right)  =\left(
a\rightharpoonup v,a\rightharpoonup w\right)  \ \ \ \ \ \ \ \ \ \ \text{for
every }a\in\mathfrak{g}\text{, }v\in V\text{ and }w\in W\right)  .
\label{prop.lie(+).cond}%
\end{equation}
Then, this map $\mu_{V\oplus W}$ is $k$-bilinear, and $\left(  V\oplus
W,\mu_{V\oplus W}\right)  $ is a $\mathfrak{g}$-module satisfying%
\begin{equation}
a\rightharpoonup\left(  v,w\right)  =\left(  a\rightharpoonup
v,a\rightharpoonup w\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }%
a\in\mathfrak{g}\text{, }v\in V\text{ and }w\in W. \label{prop.lie(+).form}%
\end{equation}

\end{proposition}

This proposition is straightforward to prove, so we are not going to elaborate
on its proof. Anyway it allows a definition:

\begin{definition}
\label{defs.lie(+)}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a Lie
algebra. Let $V$ and $W$ be two $\mathfrak{g}$-modules.\newline The
$\mathfrak{g}$-module $\left(  V\oplus W,\mu_{V\oplus W}\right)  $ constructed
in Proposition \ref{prop.lie(+)} is called the \textit{direct sum} of the
$\mathfrak{g}$-modules $V$ and $W$. We are going to denote this $\mathfrak{g}%
$-module $\left(  V\oplus W,\mu_{V\oplus W}\right)  $ simply by $V\oplus W$.
\end{definition}

We can similarly define the direct sum of several (not necessarily just two)
$\mathfrak{g}$-modules:

\begin{proposition}
\label{prop.lie(++)}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a Lie
algebra. Let $S$ be a set. For every $s\in S$, let $V_{s}$ be a $\mathfrak{g}%
$-module. Define a map $\mu_{\oplus}:\mathfrak{g}\times\left(  \bigoplus
\limits_{s\in S}V_{s}\right)  \rightarrow\bigoplus\limits_{s\in S}V_{s}$ by%
\begin{equation}
\left(  \mu_{\oplus}\left(  a,\left(  v_{s}\right)  _{s\in S}\right)  =\left(
a\rightharpoonup v_{s}\right)  _{s\in S}\ \ \ \ \ \ \ \ \ \ \text{for every
}a\in\mathfrak{g}\text{ and every family }\left(  v_{s}\right)  _{s\in S}%
\in\bigoplus\limits_{s\in S}V_{s}\right)  . \label{prop.lie(++).cond}%
\end{equation}
Then, this map $\mu_{\oplus}$ is $k$-bilinear, and $\left(  \bigoplus
\limits_{s\in S}V_{s},\mu_{\oplus}\right)  $ is a $\mathfrak{g}$-module
satisfying%
\begin{equation}
a\rightharpoonup\left(  v_{s}\right)  _{s\in S}=\left(  a\rightharpoonup
v_{s}\right)  _{s\in S}\ \ \ \ \ \ \ \ \ \ \text{for every }a\in
\mathfrak{g}\text{ and every family }\left(  v_{s}\right)  _{s\in S}%
\in\bigoplus\limits_{s\in S}V_{s}. \label{prop.lie(++).form}%
\end{equation}

\end{proposition}

\begin{definition}
\label{defs.lie(++)}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a Lie
algebra. Let $S$ be a set. For every $s\in S$, let $V_{s}$ be a $\mathfrak{g}%
$-module.\newline The $\mathfrak{g}$-module $\left(  \bigoplus\limits_{s\in
S}V_{s},\mu_{\oplus}\right)  $ constructed in Proposition \ref{prop.lie(++)}
is called the \textit{direct sum} of the $\mathfrak{g}$-modules $V_{s}$ over
all $s\in S$. We are going to denote this $\mathfrak{g}$-module $\left(
\bigoplus\limits_{s\in S}V_{s},\mu_{\oplus}\right)  $ simply by $\bigoplus
\limits_{s\in S}V_{s}$.
\end{definition}

Again, there is nothing substantial to prove here. Notice that if
$S=\varnothing$, then $\bigoplus\limits_{s\in S}V_{s}$ is to be understood as
$0$.

Working with direct sums is greatly simplified by using the following convention:

\begin{condition}
\label{conv.(++).inj}Let $k$ be a commutative ring. Let $S$ be a set. For
every $s\in S$, let $V_{s}$ be a $k$-module. For every $t\in S$, we are going
to identify the $k$-module $V_{t}$ with the image of $V_{t}$ under the
canonical injection $V_{t}\rightarrow\bigoplus\limits_{s\in S}V_{s}$. This is
an abuse of notation, but a relatively harmless one. It allows us to consider
$V_{t}$ as a $k$-submodule of the direct sum $\bigoplus\limits_{s\in S}V_{s}$.
\end{condition}

The same applies for $\mathfrak{g}$-modules:

\begin{proposition}
\label{prop.lie(++).inj}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
Lie algebra. Let $S$ be a set. For every $s\in S$, let $V_{s}$ be a
$\mathfrak{g}$-module. In Convention \ref{conv.(++).inj}, we have identified
the $k$-module $V_{t}$ with the image of $V_{t}$ under the canonical injection
$V_{t}\rightarrow\bigoplus\limits_{s\in S}V_{s}$ for every $t\in S$. Thus, by
means of this identification, $V_{t}$ becomes a $k$-submodule of the direct
sum $\bigoplus\limits_{s\in S}V_{s}$. But actually, something stronger holds:
By means of this identification, $V_{t}$ becomes a $\mathfrak{g}$-submodule of
the direct sum $\bigoplus\limits_{s\in S}V_{s}$.
\end{proposition}

We notice an important, even if trivial, fact, which will often be silently used:

\begin{proposition}
\label{prop.lie.(+)vsRes}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$.\newline\textbf{(a)} If $V$ and $W$ are two $\mathfrak{g}$-modules, then
$\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}\left(  V\oplus W\right)
=\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V\right)
\oplus\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}W\right)  $ as
$\mathfrak{h}$-modules. This allows us to speak of "the $\mathfrak{h}$-module
$V\oplus W$" without having to worry whether we mean $\operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}\left(  V\oplus W\right)  $ or $\left(
\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V\right)  \oplus\left(
\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}W\right)  $ (because it does
not matter, since $\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}\left(
V\oplus W\right)  =\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}%
}V\right)  \oplus\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}%
}W\right)  $).\newline\textbf{(b)} If $S$ is a set, and if $V_{t}$ is a
$\mathfrak{g}$-module for every $t\in S$, then $\operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}\left(  \bigoplus\limits_{s\in S}V_{s}\right)
=\bigoplus\limits_{s\in S}\left(  \operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}V_{s}\right)  $ as $\mathfrak{h}$-modules. This allows us to
speak of "the $\mathfrak{h}$-module $\bigoplus\limits_{s\in S}V_{s}$" without
having to worry whether we mean $\operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}\left(  \bigoplus\limits_{s\in S}V_{s}\right)  $ or
$\bigoplus\limits_{s\in S}\left(  \operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}V_{s}\right)  $ (because it does not matter, since
$\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}\left(  \bigoplus
\limits_{s\in S}V_{s}\right)  =\bigoplus\limits_{s\in S}\left(
\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V_{s}\right)  $).\newline%
\textbf{(c)} The $\mathfrak{h}$-module $k$ is identical with the restriction
$\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}k$ of the $\mathfrak{g}%
$-module $k$ to $\mathfrak{h}$.\newline\textbf{(d)} If $V$ is a $\mathfrak{g}%
$-module, and if $W$ is a $\mathfrak{g}$-submodule of $V$, then
$\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}W$ is an $\mathfrak{h}%
$-submodule of the $\mathfrak{h}$-module $\operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}V$ and satisfies $\operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}\left(  V\diagup W\right)  =\left(  \operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}V\right)  \diagup\left(  \operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}W\right)  $ as $\mathfrak{h}$-modules. This
allows us to speak of "the $\mathfrak{h}$-module $V\diagup W$" without having
to worry whether we mean $\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}%
}\left(  V\diagup W\right)  $ or $\left(  \operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}V\right)  \diagup\left(  \operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}W\right)  $ (because it does not matter, since
$\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}\left(  V\diagup W\right)
=\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V\right)
\diagup\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}W\right)  $).
\end{proposition}

\subsection{A convention regarding $k$-spans}

Before we proceed any further, let us fix one convention that we are going to
use several times in this text:

\begin{condition}
\label{conv.<>}\textbf{(a)} Whenever $k$ is a commutative ring, $M$ is a
$k$-module, and $S$ is a subset of $M$, we denote by $\left\langle
S\right\rangle $ the $k$-submodule of $M$ generated by the elements of $S$.
This $k$-submodule $\left\langle S\right\rangle $ is called the $k$%
\textit{-linear span} (or simply the $k$\textit{-span}) of $S$.\newline%
\textbf{(b)} Whenever $k$ is a commutative ring, $M$ is a $k$-module, $\Phi$
is a set, and $P:\Phi\rightarrow M$ is a map (not necessarily a linear map),
we denote by $\left\langle P\left(  v\right)  \ \mid\ v\in\Phi\right\rangle $
the $k$-submodule $\left\langle \left\{  P\left(  v\right)  \ \mid\ v\in
\Phi\right\}  \right\rangle $ of $M$. (In other words, $\left\langle P\left(
v\right)  \ \mid\ v\in\Phi\right\rangle $ is the $k$-submodule of $M$
generated by the elements $P\left(  v\right)  $ for all $v\in\Phi$.)
\end{condition}

Note that some authors use the notation $\left\langle S\right\rangle $ for
various other things (e. g., the two-sided ideal generated by $S$, or the Lie
subalgebra generated by $S$), but \textit{we will only use it for the }%
$k$\textit{-submodule generated by }$S$ \textit{(as defined in Convention
\ref{conv.<>} \textbf{(a)})}.

\begin{vershort}
Let us record a trivial fact that we are going to use very often (without
explicit mention):
\end{vershort}

\begin{verlong}
Let us record a trivial fact for later use:
\end{verlong}

\begin{proposition}
\label{prop.<>}Let $k$ be a commutative ring. Let $M$ be a $k$-module. Let $S$
be a subset of $M$.\newline\textbf{(a)} Let $Q$ be a $k$-submodule of $M$ such
that $S\subseteq Q$. Then, $\left\langle S\right\rangle \subseteq Q$%
.\newline\textbf{(b)} Let $R$ be a $k$-module, and $f:M\rightarrow R$ be a
$k$-module homomorphism. Then, $f\left(  \left\langle S\right\rangle \right)
=\left\langle f\left(  S\right)  \right\rangle $.
\end{proposition}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.<>}.} \textbf{(a)} Let $\alpha
\in\left\langle S\right\rangle $. Then, $\alpha$ is a $k$-linear combination
of the elements of $S$. This means that there exists some $n\in\mathbb{N}$,
some elements $\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{n}$ of $k$, and
some elements $s_{1}$, $s_{2}$, $...$, $s_{n}$ of $S$ such that $\alpha
=\sum\limits_{i=1}^{n}\lambda_{i}s_{i}$. But $s_{i}\in Q$ for every
$i\in\left\{  1,2,...,n\right\}  $ (since $s_{i}\in S\subseteq Q$). Thus,
$\alpha=\sum\limits_{i=1}^{n}\lambda_{i}\underbrace{s_{i}}_{\in Q}\in
\sum\limits_{i=1}^{n}\lambda_{i}Q\subseteq Q$ (since $Q$ is a $k$-module).
Thus we have proven that every $\alpha\in\left\langle S\right\rangle $
satisfies $\alpha\in Q$. In other words, $\left\langle S\right\rangle
\subseteq Q$. Thus, Proposition \ref{prop.<>} \textbf{(a)} is proven.
\end{verlong}

\begin{verlong}
\textbf{(b)} First let us prove that $f\left(  \left\langle S\right\rangle
\right)  \subseteq\left\langle f\left(  S\right)  \right\rangle $.
\end{verlong}

\begin{verlong}
In fact, let $\beta\in f\left(  \left\langle S\right\rangle \right)  $. Then,
there exists some $\alpha\in\left\langle S\right\rangle $ such that
$\beta=f\left(  \alpha\right)  $. Since $\alpha\in\left\langle S\right\rangle
$, we know that $\alpha$ is a $k$-linear combination of the elements of $S$.
This means that there exists some $n\in\mathbb{N}$, some elements $\lambda
_{1}$, $\lambda_{2}$, $...$, $\lambda_{n}$ of $k$, and some elements $s_{1}$,
$s_{2}$, $...$, $s_{n}$ of $S$ such that $\alpha=\sum\limits_{i=1}^{n}%
\lambda_{i}s_{i}$. Thus, $f\left(  \alpha\right)  =f\left(  \sum
\limits_{i=1}^{n}\lambda_{i}s_{i}\right)  =\sum\limits_{i=1}^{n}\lambda
_{i}\underbrace{f\left(  s_{i}\right)  }_{\in f\left(  S\right)  \text{ (since
}s_{i}\in S\text{)}}$ (since $f$ is a $k$-linear map), so that $f\left(
\alpha\right)  $ is a $k$-linear combination of the elements of $f\left(
S\right)  $. In other words, $f\left(  \alpha\right)  \in\left\langle f\left(
S\right)  \right\rangle $. Since $f\left(  \alpha\right)  =\beta$, this
becomes $\beta\in\left\langle f\left(  S\right)  \right\rangle $. We have thus
proven that every $\beta\in f\left(  \left\langle S\right\rangle \right)  $
satisfies $\beta\in\left\langle f\left(  S\right)  \right\rangle $. In other
words, $f\left(  \left\langle S\right\rangle \right)  \subseteq\left\langle
f\left(  S\right)  \right\rangle $.
\end{verlong}

\begin{verlong}
Let us now show that $\left\langle f\left(  S\right)  \right\rangle \subseteq
f\left(  \left\langle S\right\rangle \right)  $.
\end{verlong}

\begin{verlong}
In fact, let $\gamma\in\left\langle f\left(  S\right)  \right\rangle $. Then,
$\gamma$ is a $k$-linear combination of the elements of $f\left(  S\right)  $.
This means that there exists some $n\in\mathbb{N}$, some elements $\lambda
_{1}$, $\lambda_{2}$, $...$, $\lambda_{n}$ of $k$, and some elements $t_{1}$,
$t_{2}$, $...$, $t_{n}$ of $f\left(  S\right)  $ such that $\gamma
=\sum\limits_{i=1}^{n}\lambda_{i}t_{i}$. For every $i\in\left\{
1,2,...,n\right\}  $, there exists some $s_{i}\in S$ such that $t_{i}=f\left(
s_{i}\right)  $ (because $t_{i}\in f\left(  S\right)  $). Thus, $\gamma
=\sum\limits_{i=1}^{n}\lambda_{i}\underbrace{t_{i}}_{=f\left(  s_{i}\right)
}=\sum\limits_{i=1}^{n}\lambda_{i}f\left(  s_{i}\right)  =f\left(
\sum\limits_{i=1}^{n}\lambda_{i}s_{i}\right)  $ (since $f$ is a $k$-linear
map). Since $\sum\limits_{i=1}^{n}\lambda_{i}s_{i}\in\left\langle
S\right\rangle $ (because $s_{i}\in S$ for every $i\in\left\{
1,2,...,n\right\}  $, and thus $\sum\limits_{i=1}^{n}\lambda_{i}s_{i}$ is a
$k$-linear combination of the elements of $S$), this yields $\gamma\in
f\left(  \left\langle S\right\rangle \right)  $. Thus we have proven that
every $\gamma\in\left\langle f\left(  S\right)  \right\rangle $ satisfies
$\gamma\in f\left(  \left\langle S\right\rangle \right)  $. In other words, we
have proven that $\left\langle f\left(  S\right)  \right\rangle \subseteq
f\left(  \left\langle S\right\rangle \right)  $. Combined with $f\left(
\left\langle S\right\rangle \right)  \subseteq\left\langle f\left(  S\right)
\right\rangle $, this yields $f\left(  \left\langle S\right\rangle \right)
=\left\langle f\left(  S\right)  \right\rangle $. This proves Proposition
\ref{prop.<>} \textbf{(b)}.
\end{verlong}

\subsection{Tensor products of two $\mathfrak{g}$-modules}

\begin{verlong}
But now let us go back to methods of obtaining new $\mathfrak{g}$-modules from
given $\mathfrak{g}$-modules. We already know factor modules and direct sums.
Another way to construct $\mathfrak{g}$-modules is by tensor multiplication.
This is based upon the following fact (which will be proven further below):
\end{verlong}

\begin{vershort}
But now let us go back to methods of obtaining new $\mathfrak{g}$-modules from
given $\mathfrak{g}$-modules. We already know factor modules and direct sums.
Another way to construct $\mathfrak{g}$-modules is by tensor multiplication.
This is based upon the following fact:
\end{vershort}

\begin{proposition}
\label{prop.lie(X)}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a Lie
algebra. Let $V$ and $W$ be two $\mathfrak{g}$-modules. Then, there exists one
and only one $k$-bilinear map $m:\mathfrak{g}\times\left(  V\otimes W\right)
\rightarrow V\otimes W$ which satisfies%
\begin{equation}
\left(  m\left(  a,v\otimes w\right)  =\left(  a\rightharpoonup v\right)
\otimes w+v\otimes\left(  a\rightharpoonup w\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{, }v\in V\text{ and
}w\in W\right)  . \label{prop.lie(X).cond}%
\end{equation}
If we denote this map $m$ by $\mu_{V\otimes W}$, then $\left(  V\otimes
W,\mu_{V\otimes W}\right)  $ is a $\mathfrak{g}$-module. This $\mathfrak{g}%
$-module satisfies%
\begin{equation}
a\rightharpoonup\left(  v\otimes w\right)  =\left(  a\rightharpoonup v\right)
\otimes w+v\otimes\left(  a\rightharpoonup w\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{, }v\in V\text{ and
}w\in W. \label{prop.lie(X).form}%
\end{equation}

\end{proposition}

\begin{definition}
\label{defs.lie(X)}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a Lie
algebra. Let $V$ and $W$ be two $\mathfrak{g}$-modules.\newline The
$\mathfrak{g}$-module $\left(  V\otimes W,\mu_{V\otimes W}\right)  $
constructed in Proposition \ref{prop.lie(X)} is called the \textit{tensor
product} of the $\mathfrak{g}$-modules $V$ and $W$. We are going to denote
this $\mathfrak{g}$-module $\left(  V\otimes W,\mu_{V\otimes W}\right)  $
simply by $V\otimes W$.
\end{definition}

Thus, for any two $\mathfrak{g}$-modules $V$ and $W$, the $\mathfrak{g}%
$-module $V\otimes W$ satisfies (\ref{prop.lie(X).form}).

\begin{verlong}
\textit{Proof of Proposition \ref{prop.lie(X)}.}
\end{verlong}

\begin{verlong}
In order to prove Proposition \ref{prop.lie(X)}, we must verify the following assertions:
\end{verlong}

\begin{verlong}
\textit{Assertion }$\alpha$\textit{:} There exists a $k$-bilinear map
$m:\mathfrak{g}\times\left(  V\otimes W\right)  \rightarrow V\otimes W$ which
satisfies (\ref{prop.lie(X).cond}).
\end{verlong}

\begin{verlong}
\textit{Assertion }$\beta$\textit{:} There exists one and only one
$k$-bilinear map $m:\mathfrak{g}\times\left(  V\otimes W\right)  \rightarrow
V\otimes W$ which satisfies (\ref{prop.lie(X).cond}).
\end{verlong}

\begin{verlong}
\textit{Assertion }$\gamma$\textit{:} If we denote by $\mu_{V\otimes W}$ the
$k$-bilinear map $m:\mathfrak{g}\times\left(  V\otimes W\right)  \rightarrow
V\otimes W$ which satisfies (\ref{prop.lie(X).cond}), then $\left(  V\otimes
W,\mu_{V\otimes W}\right)  $ is a $\mathfrak{g}$-module.
\end{verlong}

\begin{verlong}
\textit{Assertion }$\delta$\textit{:} The $\mathfrak{g}$-module $\left(
V\otimes W,\mu_{V\otimes W}\right)  $ defined in Assertion $\gamma$ satisfies
(\ref{prop.lie(X).form}).
\end{verlong}

\begin{verlong}
\textit{Proof of Assertion }$\alpha$\textit{:} First, we note that for every
$a\in\mathfrak{g}$, the map $V\rightarrow V,$ $v\mapsto a\rightharpoonup v$ is
$k$-linear.{\footnotemark}\footnotetext{This is because $a\rightharpoonup v$
is the image of $\left(  a,v\right)  $ under the Lie action of $\mathfrak{g}$,
and because the Lie action is $k$-bilinear.} Similarly, for every
$a\in\mathfrak{g}$, the map $W\rightarrow W,$ $w\mapsto a\rightharpoonup w$ is
$k$-linear.
\end{verlong}

\begin{verlong}
For every $a\in\mathfrak{g}$, let $m_{a}$ denote the map $V\times W\rightarrow
V\otimes W$ given by%
\[
\left(  m_{a}\left(  v,w\right)  =\left(  a\rightharpoonup v\right)  \otimes
w+v\otimes\left(  a\rightharpoonup w\right)  \ \ \ \ \ \ \ \ \ \ \text{for
every }v\in V\text{ and }w\in W\right)  .
\]
This map $m_{a}:V\times W\rightarrow V\otimes W$ is $k$-bilinear (since the
maps $V\rightarrow V,$ $v\mapsto a\rightharpoonup v$ and $W\rightarrow W,$
$w\mapsto a\rightharpoonup w$ are $k$-linear). Hence, the universal property
of the tensor product yields that there exists one and only one $k$-linear map
$\mu_{a}:V\otimes W\rightarrow V\otimes W$ satisfying $\left(  \mu_{a}\left(
v\otimes w\right)  =m_{a}\left(  v,w\right)  \text{ for every }v\in V\text{
and }w\in W\right)  $. Denoting this map $\mu_{a}$ by $\mathfrak{m}_{a}$, we
thus have%
\[
\left(  \mathfrak{m}_{a}\left(  v\otimes w\right)  =m_{a}\left(  v,w\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }v\in V\text{ and }w\in W\right)  .
\]
Thus, every $v\in V$ and $w\in W$ satisfy%
\begin{equation}
\mathfrak{m}_{a}\left(  v\otimes w\right)  =m_{a}\left(  v,w\right)  =\left(
a\rightharpoonup v\right)  \otimes w+v\otimes\left(  a\rightharpoonup
w\right)  . \label{mu_a}%
\end{equation}

\end{verlong}

\begin{verlong}
It is easy to see that%
\begin{equation}
\left(  \mathfrak{m}_{\lambda a+\kappa b}=\lambda\mathfrak{m}_{a}%
+\kappa\mathfrak{m}_{b}\ \ \ \ \ \ \ \ \ \ \text{for every }a\in
\mathfrak{g}\text{, }b\in\mathfrak{g}\text{, }\lambda\in k\text{ and }%
\kappa\in k\right)  . \label{mu_ab}%
\end{equation}
{\footnotemark}\footnotetext{\textit{Proof.} Let $a\in\mathfrak{g}$,
$b\in\mathfrak{g}$, $\lambda\in k$ and $\kappa\in k$ be arbitrary. For every
$v\in V$ and $w\in W$, we have%
\[
\mathfrak{m}_{a}\left(  v\otimes w\right)  =\left(  a\rightharpoonup v\right)
\otimes w+v\otimes\left(  a\rightharpoonup w\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{mu_a})}\right)  ,
\]%
\[
\mathfrak{m}_{b}\left(  v\otimes w\right)  =\left(  b\rightharpoonup v\right)
\otimes w+v\otimes\left(  b\rightharpoonup w\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{mu_a}), applied to }b\text{ instead
of }a\right)
\]
and%
\begin{align*}
\mathfrak{m}_{\lambda a+\kappa b}\left(  v\otimes w\right)   &  =\left(
\underbrace{\left(  \lambda a+\kappa b\right)  \rightharpoonup v}%
_{\substack{=\lambda a\rightharpoonup v+\kappa b\rightharpoonup
v\\\text{(since the Lie action}\\\text{of }V\text{ is }k\text{-bilinear)}%
}}\right)  \otimes w+v\otimes\left(  \underbrace{\left(  \lambda a+\kappa
b\right)  \rightharpoonup w}_{\substack{=\lambda a\rightharpoonup w+\kappa
b\rightharpoonup w\\\text{(since the Lie action}\\\text{of }W\text{ is
}k\text{-bilinear)}}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{mu_a}), applied to }\lambda
a+\kappa b\text{ instead of }a\right) \\
&  =\left(  \lambda a\rightharpoonup v+\kappa b\rightharpoonup v\right)
\otimes w+v\otimes\left(  \lambda a\rightharpoonup w+\kappa b\rightharpoonup
w\right) \\
&  =\lambda\left(  a\rightharpoonup v\right)  \otimes w+\kappa\left(
b\rightharpoonup v\right)  \otimes w+\lambda v\otimes\left(  a\rightharpoonup
w\right)  +\kappa v\otimes\left(  b\rightharpoonup w\right) \\
&  =\underbrace{\lambda\left(  a\rightharpoonup v\right)  \otimes w+\lambda
v\otimes\left(  a\rightharpoonup w\right)  }_{=\lambda\left(  \left(
a\rightharpoonup v\right)  \otimes w+v\otimes\left(  a\rightharpoonup
w\right)  \right)  }+\underbrace{\kappa\left(  b\rightharpoonup v\right)
\otimes w+\kappa v\otimes\left(  b\rightharpoonup w\right)  }_{=\kappa\left(
\left(  b\rightharpoonup v\right)  \otimes w+v\otimes\left(  b\rightharpoonup
w\right)  \right)  }\\
&  =\lambda\underbrace{\left(  \left(  a\rightharpoonup v\right)  \otimes
w+v\otimes\left(  a\rightharpoonup w\right)  \right)  }_{=\mathfrak{m}%
_{a}\left(  v\otimes w\right)  }+\kappa\underbrace{\left(  \left(
b\rightharpoonup v\right)  \otimes w+v\otimes\left(  b\rightharpoonup
w\right)  \right)  }_{=\mathfrak{m}_{b}\left(  v\otimes w\right)  }\\
&  =\lambda\mathfrak{m}_{a}\left(  v\otimes w\right)  +\kappa\mathfrak{m}%
_{b}\left(  v\otimes w\right)  =\left(  \lambda\mathfrak{m}_{a}+\kappa
\mathfrak{m}_{b}\right)  \left(  v\otimes w\right)  .
\end{align*}
Thus, the maps $\mathfrak{m}_{\lambda a+\kappa b}$ and $\lambda\mathfrak{m}%
_{a}+\kappa\mathfrak{m}_{b}$ are equal to each other on all pure tensors. But
since the pure tensors generate the tensor product $V\otimes W$, two arbitrary
$k$-linear maps defined on $V\otimes W$ must be equal if they are equal to
each other on all pure tensors. Thus, since the two $k$-linear maps
$\mathfrak{m}_{\lambda a+\kappa b}$ and $\lambda\mathfrak{m}_{a}%
+\kappa\mathfrak{m}_{b}$ are equal to each other on all pure tensors, we can
conclude that these $k$-linear maps $\mathfrak{m}_{\lambda a+\kappa b}$ and
$\lambda\mathfrak{m}_{a}+\kappa\mathfrak{m}_{b}$ are equal. That is,
$\mathfrak{m}_{\lambda a+\kappa b}=\lambda\mathfrak{m}_{a}+\kappa
\mathfrak{m}_{b}$, qed.}
\end{verlong}

\begin{verlong}
Now, let $M:\mathfrak{g}\times\left(  V\otimes W\right)  \rightarrow V\otimes
W$ be the map defined by%
\[
\left(  M\left(  a,T\right)  =\mathfrak{m}_{a}\left(  T\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{ and }T\in V\otimes
W\right)  .
\]
This map $M$ is $k$-bilinear (because $\mathfrak{m}_{a}\left(  T\right)  $
depends linearly on $a$ (since $\mathfrak{m}_{a}$ depends linearly on $a$, due
to (\ref{mu_ab})) and depends linearly on $T$ (because $\mathfrak{m}_{a}$ is
$k$-linear)) and satisfies%
\begin{align*}
M\left(  a,v\otimes w\right)   &  =\mathfrak{m}_{a}\left(  v\otimes w\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }M\right) \\
&  =\left(  a\rightharpoonup v\right)  \otimes w+v\otimes\left(
a\rightharpoonup w\right)
\end{align*}
for every $a\in\mathfrak{g}$, $v\in V$ and $w\in W$. In other words,
(\ref{prop.lie(X).cond}) is satisfied for $m=M$.
\end{verlong}

\begin{verlong}
Hence, we have shown that there exists a $k$-bilinear map $m:\mathfrak{g}%
\times\left(  V\otimes W\right)  \rightarrow V\otimes W$ which satisfies
(\ref{prop.lie(X).cond}) (namely, the map $M$). This proves Assertion $\alpha$.
\end{verlong}

\begin{verlong}
\textit{Proof of Assertion }$\beta$\textit{:} Let $m:\mathfrak{g}\times\left(
V\otimes W\right)  \rightarrow V\otimes W$ be a $k$-bilinear map which
satisfies (\ref{prop.lie(X).cond}). We are going to prove that $m=M$, where
$M$ is the map defined in the proof of Assertion $\alpha$ above.

Indeed, for every $a\in\mathfrak{g}$, let $\mathfrak{m}_{a}:V\otimes
W\rightarrow V\otimes W$ be the map $\mathfrak{m}_{a}$ defined in the proof of
Assertion $\alpha$ above, and let $m_{a}:V\otimes W\rightarrow V\otimes W$ be
the map defined by%
\[
\left(  m_{a}\left(  T\right)  =m\left(  a,T\right)  \text{ for all }%
a\in\mathfrak{g}\text{ and }T\in V\otimes W\right)  .
\]
This map $m_{a}$ is $k$-linear (since the map $m$ is $k$-bilinear).
Furthermore, every $v\in V$ and every $w\in W$ satisfy%
\begin{align*}
m_{a}\left(  v\otimes w\right)   &  =m\left(  a,v\otimes w\right)  =\left(
a\rightharpoonup v\right)  \otimes w+v\otimes\left(  a\rightharpoonup w\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since we assumed that }m\text{ satisfies
(\ref{prop.lie(X).cond})}\right) \\
&  =\mathfrak{m}_{a}\left(  v\otimes w\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{as proved in the proof of Assertion }\alpha\right)  .
\end{align*}
In other words, the maps $m_{a}$ and $\mathfrak{m}_{a}$ are equal to each
other on all pure tensors. But since the pure tensors generate the tensor
product $V\otimes W$, two arbitrary $k$-linear maps defined on $V\otimes W$
must be equal if they are equal to each other on all pure tensors. Thus, since
the two $k$-linear maps $m_{a}$ and $\mathfrak{m}_{a}$ are equal to each other
on all pure tensors, we can conclude that these $k$-linear maps $m_{a}$ and
$\mathfrak{m}_{a}$ are equal. That is, $m_{a}=\mathfrak{m}_{a}$.

Now, every $a\in\mathfrak{g}$ and $T\in V\otimes W$ satisfy%
\[
m\left(  a,T\right)  =\underbrace{m_{a}}_{=\mathfrak{m}_{a}}\left(  T\right)
=\mathfrak{m}_{a}\left(  T\right)  =M\left(  a,T\right)  .
\]
Thus, $m=M$.

So we have proved that every $k$-bilinear map $m:\mathfrak{g}\times\left(
V\otimes W\right)  \rightarrow V\otimes W$ which satisfies
(\ref{prop.lie(X).cond}) must satisfy $m=M$. Hence, there is \textit{at most
one} $k$-bilinear map $m:\mathfrak{g}\times\left(  V\otimes W\right)
\rightarrow V\otimes W$ which satisfies (\ref{prop.lie(X).cond}) (because
every such map $m$ must satisfy $m=M$, and hence be equal to $M$). Combined
with Assertion $\alpha$, this yields that there exists one and only one
$k$-bilinear map $m:\mathfrak{g}\times\left(  V\otimes W\right)  \rightarrow
V\otimes W$ which satisfies (\ref{prop.lie(X).cond}). This proves Assertion
$\beta$.
\end{verlong}

\begin{verlong}
\textit{Proof of Assertion }$\gamma$\textit{:} Let us denote by $\mu_{V\otimes
W}$ the $k$-bilinear map $m:\mathfrak{g}\times\left(  V\otimes W\right)
\rightarrow V\otimes W$ which satisfies (\ref{prop.lie(X).cond}). Let
$a\in\mathfrak{g}$ and $b\in\mathfrak{g}$. Define a map $P:V\otimes
W\rightarrow V\otimes W$ by%
\begin{equation}
\left(
\begin{array}
[c]{c}%
P\left(  T\right)  =\mu_{V\otimes W}\left(  a,\mu_{V\otimes W}\left(
b,T\right)  \right)  -\mu_{V\otimes W}\left(  b,\mu_{V\otimes W}\left(
a,T\right)  \right)  -\mu_{V\otimes W}\left(  \left[  a,b\right]  ,T\right) \\
\ \ \ \ \ \ \ \ \ \ \text{for every }T\in V\otimes W
\end{array}
\right)  . \label{pf.lie(X).def}%
\end{equation}
This map $P$ is $k$-linear (due to the $k$-bilinearity of $\mu_{V\otimes W}$),
and thus its kernel $\operatorname*{Ker}P$ is a $k$-submodule of $V\otimes W$.
\end{verlong}

\begin{verlong}
We have%
\begin{equation}
\mu_{V\otimes W}\left(  a,v\otimes w\right)  =\left(  a\rightharpoonup
v\right)  \otimes w+v\otimes\left(  a\rightharpoonup w\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{, }v\in V\text{ and
}w\in W \label{pf.lie(X).gamma.1}%
\end{equation}
(since $\mu_{V\otimes W}$ is the $k$-bilinear map $m:\mathfrak{g}\times\left(
V\otimes W\right)  \rightarrow V\otimes W$ which satisfies
(\ref{prop.lie(X).cond})).
\end{verlong}

\begin{verlong}
Every $a\in\mathfrak{g}$, $b\in\mathfrak{g}$, $v\in V$ and $w\in W$ satisfy%
\begin{align*}
\mu_{V\otimes W}\left(  b,v\otimes w\right)   &  =\left(  b\rightharpoonup
v\right)  \otimes w+v\otimes\left(  b\rightharpoonup w\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.lie(X).gamma.1}), applied to
}b\text{ instead of }a\right)
\end{align*}
and thus%
\begin{align}
&  \mu_{V\otimes W}\left(  a,\mu_{V\otimes W}\left(  b,v\otimes w\right)
\right) \nonumber\\
&  =\mu_{V\otimes W}\left(  a,\left(  b\rightharpoonup v\right)  \otimes
w+v\otimes\left(  b\rightharpoonup w\right)  \right) \nonumber\\
&  =\underbrace{\mu_{V\otimes W}\left(  a,\left(  b\rightharpoonup v\right)
\otimes w\right)  }_{\substack{=\left(  a\rightharpoonup\left(
b\rightharpoonup v\right)  \right)  \otimes w+\left(  b\rightharpoonup
v\right)  \otimes\left(  a\rightharpoonup w\right)  \\\text{(by
(\ref{pf.lie(X).gamma.1}), applied to}\\b\rightharpoonup v\text{ instead of
}v\text{)}}}+\underbrace{\mu_{V\otimes W}\left(  v\otimes\left(
b\rightharpoonup w\right)  \right)  }_{\substack{=\left(  a\rightharpoonup
v\right)  \otimes\left(  b\rightharpoonup w\right)  +v\otimes\left(
a\rightharpoonup\left(  b\rightharpoonup w\right)  \right)  \\\text{(by
(\ref{pf.lie(X).gamma.1}), applied to}\\b\rightharpoonup w\text{ instead of
}w\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu_{V\otimes W}\text{ is
}k\text{-bilinear}\right) \nonumber\\
&  =\left(  a\rightharpoonup\left(  b\rightharpoonup v\right)  \right)
\otimes w+\left(  b\rightharpoonup v\right)  \otimes\left(  a\rightharpoonup
w\right)  +\left(  a\rightharpoonup v\right)  \otimes\left(  b\rightharpoonup
w\right)  +v\otimes\left(  a\rightharpoonup\left(  b\rightharpoonup w\right)
\right)  . \label{pf.lie(X).gamma.2}%
\end{align}
The same argument, with $a$ and $b$ transposed, yields%
\begin{align*}
&  \mu_{V\otimes W}\left(  b,\mu_{V\otimes W}\left(  a,v\otimes w\right)
\right) \\
&  =\left(  b\rightharpoonup\left(  a\rightharpoonup v\right)  \right)
\otimes w+\left(  a\rightharpoonup v\right)  \otimes\left(  b\rightharpoonup
w\right)  +\left(  b\rightharpoonup v\right)  \otimes\left(  a\rightharpoonup
w\right)  +v\otimes\left(  b\rightharpoonup\left(  a\rightharpoonup w\right)
\right)  .
\end{align*}
Subtracting this equation from (\ref{pf.lie(X).gamma.2}), we obtain%
\begin{align*}
&  \mu_{V\otimes W}\left(  a,\mu_{V\otimes W}\left(  b,v\otimes w\right)
\right)  -\mu_{V\otimes W}\left(  b,\mu_{V\otimes W}\left(  a,v\otimes
w\right)  \right) \\
&  =\left(  \left(  a\rightharpoonup\left(  b\rightharpoonup v\right)
\right)  \otimes w+\left(  b\rightharpoonup v\right)  \otimes\left(
a\rightharpoonup w\right)  +\left(  a\rightharpoonup v\right)  \otimes\left(
b\rightharpoonup w\right)  +v\otimes\left(  a\rightharpoonup\left(
b\rightharpoonup w\right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  \left(  b\rightharpoonup\left(
a\rightharpoonup v\right)  \right)  \otimes w+\left(  a\rightharpoonup
v\right)  \otimes\left(  b\rightharpoonup w\right)  +\left(  b\rightharpoonup
v\right)  \otimes\left(  a\rightharpoonup w\right)  +v\otimes\left(
b\rightharpoonup\left(  a\rightharpoonup w\right)  \right)  \right) \\
&  =\underbrace{\left(  a\rightharpoonup\left(  b\rightharpoonup v\right)
\right)  \otimes w-\left(  b\rightharpoonup\left(  a\rightharpoonup v\right)
\right)  \otimes w}_{=\left(  a\rightharpoonup\left(  b\rightharpoonup
v\right)  -b\rightharpoonup\left(  a\rightharpoonup v\right)  \right)  \otimes
w}+\underbrace{v\otimes\left(  a\rightharpoonup\left(  b\rightharpoonup
w\right)  \right)  -v\otimes\left(  b\rightharpoonup\left(  a\rightharpoonup
w\right)  \right)  }_{=v\otimes\left(  a\rightharpoonup\left(
b\rightharpoonup w\right)  -b\rightharpoonup\left(  a\rightharpoonup w\right)
\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here we just cancelled some equal
terms}\right) \\
&  =\left(  a\rightharpoonup\left(  b\rightharpoonup v\right)
-b\rightharpoonup\left(  a\rightharpoonup v\right)  \right)  \otimes
w+v\otimes\left(  a\rightharpoonup\left(  b\rightharpoonup w\right)
-b\rightharpoonup\left(  a\rightharpoonup w\right)  \right)  .
\end{align*}
Comparing this to%
\begin{align*}
\mu_{V\otimes W}\left(  \left[  a,b\right]  ,v\otimes w\right)   &
=\underbrace{\left(  \left[  a,b\right]  \rightharpoonup v\right)
}_{\substack{=a\rightharpoonup\left(  b\rightharpoonup v\right)
-b\rightharpoonup\left(  a\rightharpoonup v\right)  \\\text{(by (\ref{[L-mod]}%
))}}}\otimes w+v\otimes\underbrace{\left(  \left[  a,b\right]  \rightharpoonup
w\right)  }_{\substack{=a\rightharpoonup\left(  b\rightharpoonup w\right)
-b\rightharpoonup\left(  a\rightharpoonup w\right)  \\\text{(by (\ref{[L-mod]}%
), applied}\\\text{to }w\text{ and }W\text{ instead of }v\text{ and }%
V\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.lie(X).gamma.1}), applied to
}\left[  a,b\right]  \text{ instead of }a\right) \\
&  =\left(  a\rightharpoonup\left(  b\rightharpoonup v\right)
-b\rightharpoonup\left(  a\rightharpoonup v\right)  \right)  \otimes
w+v\otimes\left(  a\rightharpoonup\left(  b\rightharpoonup w\right)
-b\rightharpoonup\left(  a\rightharpoonup w\right)  \right)  ,
\end{align*}
we conclude that%
\[
\mu_{V\otimes W}\left(  a,\mu_{V\otimes W}\left(  b,v\otimes w\right)
\right)  -\mu_{V\otimes W}\left(  b,\mu_{V\otimes W}\left(  a,v\otimes
w\right)  \right)  =\mu_{V\otimes W}\left(  \left[  a,b\right]  ,v\otimes
w\right)  .
\]
Now, (\ref{pf.lie(X).def}) (applied to $T=v\otimes w$) yields that%
\[
P\left(  v\otimes w\right)  =\underbrace{\mu_{V\otimes W}\left(
a,\mu_{V\otimes W}\left(  b,v\otimes w\right)  \right)  -\mu_{V\otimes
W}\left(  b,\mu_{V\otimes W}\left(  a,v\otimes w\right)  \right)  }%
_{=\mu_{V\otimes W}\left(  \left[  a,b\right]  ,v\otimes w\right)  }%
-\mu_{V\otimes W}\left(  \left[  a,b\right]  ,v\otimes w\right)  =0
\]
for every $v\in V$ and $w\in W$. In other words, the kernel
$\operatorname*{Ker}P$ contains $v\otimes w$ for every $v\in V$ and $w\in W$.
Since $\operatorname*{Ker}P$ is a $k$-submodule of $V\otimes W$, we can
conclude that%
\begin{align*}
&  \operatorname*{Ker}P\\
&  \supseteq\left(  \text{the smallest }k\text{-submodule of }V\otimes W\text{
which contains }v\otimes w\text{ for every }v\in V\text{ and }w\in W\right) \\
&  =\left\langle v\otimes w\ \mid\ \left(  v,w\right)  \in V\times
W\right\rangle =V\otimes W
\end{align*}
(because by the definition of the tensor product, $V\otimes W$ equals the span
$\left\langle v\otimes w\ \mid\ \left(  v,w\right)  \in V\times W\right\rangle
$). Thus, $P=0$. Hence, every $T\in V\otimes W$ satisfies $P\left(  T\right)
=0$. Due to (\ref{pf.lie(X).def}), this rewrites as follows: Every $T\in
V\otimes W$ satisfies%
\[
\mu_{V\otimes W}\left(  a,\mu_{V\otimes W}\left(  b,T\right)  \right)
-\mu_{V\otimes W}\left(  b,\mu_{V\otimes W}\left(  a,T\right)  \right)
-\mu_{V\otimes W}\left(  \left[  a,b\right]  ,T\right)  =0.
\]
In other words, every $T\in V\otimes W$ satisfies
\[
\mu_{V\otimes W}\left(  a,\mu_{V\otimes W}\left(  b,T\right)  \right)
-\mu_{V\otimes W}\left(  b,\mu_{V\otimes W}\left(  a,T\right)  \right)
=\mu_{V\otimes W}\left(  \left[  a,b\right]  ,T\right)  .
\]
Since this holds for all $a\in\mathfrak{g}$ and $b\in\mathfrak{g}$, we have
thus proven that (\ref{L-mod}) holds with $m$ and $V$ replaced by
$\mu_{V\otimes W}$ and $V\otimes W$. Since $\mu_{V\otimes W}$ is a
$k$-bilinear map, this yields (by Definition \ref{defs.liemod}) that $\left(
V\otimes W,\mu_{V\otimes W}\right)  $ is a $\mathfrak{g}$-module. Assertion
$\gamma$ is now proven.
\end{verlong}

\begin{verlong}
\textit{Proof of Assertion }$\delta$\textit{:} Consider the $\mathfrak{g}%
$-module $\left(  V\otimes W,\mu_{V\otimes W}\right)  $ defined in Assertion
$\gamma$. This $\mathfrak{g}$-module satisfies%
\begin{align*}
a\rightharpoonup\left(  v\otimes w\right)   &  =\mu_{V\otimes W}\left(
a,v\otimes w\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{because the Lie bracket
of this }\mathfrak{g}\text{-module is }\mu_{V\otimes W}\right) \\
&  =\left(  a\rightharpoonup v\right)  \otimes w+v\otimes\left(
a\rightharpoonup w\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lie(X).gamma.1})}\right)
\end{align*}
for every $a\in\mathfrak{g}$, $v\in V$ and $w\in W$. In other words, it
satisfies (\ref{prop.lie(X).form}). This proves Assertion $\delta$.
\end{verlong}

\begin{verlong}
Now that we have proven all four assertions $\alpha$, $\beta$, $\gamma$ and
$\delta$, Proposition \ref{prop.lie(X)} is proven.
\end{verlong}

\begin{verlong}
\begin{remark}
\label{rmk.lie(X)}In most advanced literature, proofs like the one we gave for
Proposition \ref{prop.lie(X)} are not detailed, and not even the necessity of
formulating such a proposition is mentioned. As a consequence, the definition
of the tensor product of two $\mathfrak{g}$-modules given in advanced texts
about Lie algebras usually does not look like the Definition \ref{defs.lie(X)}
that we gave, but instead roughly looks like follows:

\begin{definition}
\label{defs.lie(X).short}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a Lie algebra. Let $V$ and $W$ be two $\mathfrak{g}$-modules. Then, the
$k$-module $V\otimes W$ becomes a $\mathfrak{g}$-module by setting%
\begin{equation}
a\rightharpoonup\left(  v\otimes w\right)  =\left(  a\rightharpoonup v\right)
\otimes w+v\otimes\left(  a\rightharpoonup w\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{, }v\in V\text{ and
}w\in W. \label{rmk.lie(X).def}%
\end{equation}

\end{definition}

Such a definition can be confusing to an untrained reader: It is not
immediately clear that Definition \ref{defs.lie(X).short} really defines a
$\mathfrak{g}$-module structure on $V\otimes W$, or even just a well-defined
$k$-bilinear map $\mathfrak{g}\times\left(  V\otimes W\right)  \rightarrow
V\otimes W$. For example, if we would replace the equation
(\ref{rmk.lie(X).def}) by%
\[
a\rightharpoonup\left(  v\otimes w\right)  =\left(  a\rightharpoonup v\right)
\otimes w+v\otimes s\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}%
\text{, }v\in V\text{ and }w\in W,
\]
where $s$ is some fixed nonzero element of $W$, then it would \textit{not} be
well-defined, because one and the same pure tensor $T\in V\otimes W$ can be
written in the form $v\otimes w$ (with $v\in V$ and $w\in W$) in several
different ways, and they lead to different values of $\left(  a\rightharpoonup
v\right)  \otimes w+v\otimes s$. What saves us in the case of
(\ref{rmk.lie(X).def}) is that the right hand side of (\ref{rmk.lie(X).def})
is linear in $a$, linear in $v$ and linear in $w$, and thus
(\ref{rmk.lie(X).def}) can indeed be seen as the definition of a $k$-bilinear
map $\mathfrak{g}\times\left(  V\otimes W\right)  \rightarrow V\otimes
W$.\ \ \ \ {\footnotemark} The details of this argument we have given in our
proof of Proposition \ref{prop.lie(X)}.\newline A reader well-experienced with
bilinear maps and tensor products, of course, will \textit{not} have any
trouble with understanding a definition like Definition
\ref{defs.lie(X).short}. He will (in most cases) be able to check in his mind
that it is indeed a legitimate definition, except that sometimes he will need
to perform an explicit computation to show that the $k$-bilinear map indeed
satisfies the equation (\ref{L-mod}) (which is necessary to make sure that we
indeed get a $\mathfrak{g}$-module).
\end{remark}

\footnotetext{This still does not show that this $k$-bilinear map makes
$V\otimes W$ into a $\mathfrak{g}$-module (because to show this we still must
prove (\ref{L-mod})). But this can be shown separately by a direct
computation, which we did in the proof of Assertion $\gamma$ during the proof
of Proposition \ref{prop.lie(X)}.}
\end{verlong}

We now move on to showing properties of these tensor products:

\begin{proposition}
\label{prop.tensor-assoc}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a $k$-Lie algebra.\newline\textbf{(a)} Let $V$ be a $\mathfrak{g}$-module.
Then, the $k$-linear map%
\[
V\rightarrow k\otimes V,\ \ \ \ \ \ \ \ \ \ v\mapsto1\otimes v
\]
is a canonical isomorphism of $\mathfrak{g}$-modules. (Here, as usual, $k$
denotes the $\mathfrak{g}$-module $k$ defined in Definition \ref{defs.k-mod-g}%
.)\newline\textbf{(b)} Let $V$ be a $\mathfrak{g}$-module. Then, the
$k$-linear map%
\[
V\rightarrow V\otimes k,\ \ \ \ \ \ \ \ \ \ v\mapsto v\otimes1
\]
is a canonical isomorphism of $\mathfrak{g}$-modules. (Here, as usual, $k$
denotes the $\mathfrak{g}$-module $k$ defined in Definition \ref{defs.k-mod-g}%
.)\newline\textbf{(c)} Let $U$, $V$ and $W$ be $\mathfrak{g}$-modules. Then,
the $k$-linear map%
\[
\left(  U\otimes V\right)  \otimes W\rightarrow U\otimes\left(  V\otimes
W\right)  ,\ \ \ \ \ \ \ \ \ \ \left(  u\otimes v\right)  \otimes w\mapsto
u\otimes\left(  v\otimes w\right)
\]
is a canonical isomorphism of $\mathfrak{g}$-modules.
\end{proposition}

\begin{verlong}
\begin{remark}
Part \textbf{(c)} of this proposition suffers from the roughly the same flaw
as Definition \ref{defs.lie(X).short}: In Proposition \ref{prop.tensor-assoc}
\textbf{(c)}, it is not immediately clear why "the $k$-linear map%
\[
\left(  U\otimes V\right)  \otimes W\rightarrow U\otimes\left(  V\otimes
W\right)  ,\ \ \ \ \ \ \ \ \ \ \left(  u\otimes v\right)  \otimes w\mapsto
u\otimes\left(  v\otimes w\right)
\]
" is well-defined. We refer to standard algebra texts for a construction of
this map (it is similar to the construction of $M$ in the proof of Assertion
$\alpha$ during the proof of Proposition \ref{prop.lie(X)}); for us it is
enough to know that it is a $k$-linear map $\left(  U\otimes V\right)  \otimes
W\rightarrow U\otimes\left(  V\otimes W\right)  $ which maps $\left(  u\otimes
v\right)  \otimes w$ to $u\otimes\left(  v\otimes w\right)  $ for every $u\in
U$, $v\in V$ and $w\in W$ (and, in fact, the only such map).
\end{remark}
\end{verlong}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.tensor-assoc}.} \textbf{(a)} Let
$\delta_{k,V}$ denote the $k$-module homomorphism
\[
V\rightarrow k\otimes V,\ \ \ \ \ \ \ \ \ \ v\mapsto1\otimes v.
\]
Then, in order to prove Proposition \ref{prop.tensor-assoc} \textbf{(a)}, it
is enough to show that $\delta_{k,V}$ is a canonical isomorphism of
$\mathfrak{g}$-modules.
\end{verlong}

\begin{verlong}
It is clear that $\delta_{k,V}$ is canonical. Now let us prove that
$\delta_{k,V}$ is a homomorphism of $\mathfrak{g}$-modules:
\end{verlong}

\begin{verlong}
Every $a\in\mathfrak{g}$ and $v\in V$ satisfy $\delta_{k,V}\left(  v\right)
=1\otimes v$ (by the definition of $\delta_{k,V}$) and $\delta_{k,V}\left(
a\rightharpoonup v\right)  =1\otimes\left(  a\rightharpoonup v\right)  $ (by
the definition of $\delta_{k,V}$). Thus, every $a\in\mathfrak{g}$ and $v\in V$
satisfy%
\begin{align*}
a\rightharpoonup\underbrace{\delta_{k,V}\left(  v\right)  }_{=1\otimes v}  &
=a\rightharpoonup\left(  1\otimes v\right)  =\underbrace{\left(
a\rightharpoonup1\right)  }_{\substack{=0\text{ (since the Lie}\\\text{action
of }k\text{ is }0\text{)}}}\otimes v+1\otimes\left(  a\rightharpoonup v\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{prop.lie(X).form}), applied to
}k\text{, }V\text{, }1\text{ and }v\text{ instead of }V\text{, }W\text{,
}v\text{ and }w\right) \\
&  =\underbrace{0\otimes v}_{=0}+1\otimes\left(  a\rightharpoonup v\right)
=1\otimes\left(  a\rightharpoonup v\right)  =\delta_{k,V}\left(
a\rightharpoonup v\right)  .
\end{align*}
This proves that $\delta_{k,V}$ is a homomorphism of $\mathfrak{g}$-modules.
Since $\delta_{k,V}$ is an isomorphism of $k$-modules, this yields that
$\delta_{k,V}$ is an isomorphism of $\mathfrak{g}$-modules. Thus,
$\delta_{k,V}$ is a canonical isomorphism of $\mathfrak{g}$-modules. As we
said above, this proves Proposition \ref{prop.tensor-assoc} \textbf{(a)}.
\end{verlong}

\begin{verlong}
\textbf{(b)} The proof of Proposition \ref{prop.tensor-assoc} \textbf{(b)} is
completely analogous to the above-given proof of Proposition
\ref{prop.tensor-assoc} \textbf{(a)}, and thus will not be elaborated on.
\end{verlong}

\begin{verlong}
\textbf{(c)} Let $\alpha_{U,V,W}$ denote the $k$-linear map%
\[
\left(  U\otimes V\right)  \otimes W\rightarrow U\otimes\left(  V\otimes
W\right)  ,\ \ \ \ \ \ \ \ \ \ \left(  u\otimes v\right)  \otimes w\mapsto
u\otimes\left(  v\otimes w\right)  .
\]
We know from linear algebra that this map $\alpha_{U,V,W}$ is well-defined and
a canonical isomorphism of $k$-modules. Now, we are going to show that this
map $\alpha_{U,V,W}$ is an isomorphism of $\mathfrak{g}$-modules.
\end{verlong}

\begin{verlong}
The Lie action of a $\mathfrak{g}$-module is always $k$-bilinear. Therefore,
for every $a\in\mathfrak{g}$, the maps $\left(  U\otimes V\right)  \otimes
W\rightarrow\left(  U\otimes V\right)  \otimes W,$ $T\mapsto a\rightharpoonup
T$ and $U\otimes\left(  V\otimes W\right)  \rightarrow U\otimes\left(
V\otimes W\right)  ,$ $T\mapsto a\rightharpoonup T$ are $k$-linear.
\end{verlong}

\begin{verlong}
For every $a\in\mathfrak{g}$, let $\Phi_{a}:\left(  U\otimes V\right)  \otimes
W\rightarrow U\otimes\left(  V\otimes W\right)  $ be the $k$-linear map
defined by%
\[
\left(  \Phi_{a}\left(  T\right)  =\alpha_{U,V,W}\left(  a\rightharpoonup
T\right)  -a\rightharpoonup\left(  \alpha_{U,V,W}\left(  T\right)  \right)
\ \ \ \ \ \ \ \ \ \ \text{for every }T\in\left(  U\otimes V\right)  \otimes
W\right)  .
\]
This map $\Phi_{a}$ is $k$-linear (because the maps $\left(  U\otimes
V\right)  \otimes W\rightarrow\left(  U\otimes V\right)  \otimes W,$ $T\mapsto
a\rightharpoonup T$ and $U\otimes\left(  V\otimes W\right)  \rightarrow
U\otimes\left(  V\otimes W\right)  ,$ $T\mapsto a\rightharpoonup T$ are
$k$-linear, and so is the map $\alpha_{U,V,W}$). Hence, $\operatorname*{Ker}%
\Phi_{a}$ is a $k$-submodule of $\left(  U\otimes V\right)  \otimes W$.
\end{verlong}

\begin{verlong}
We will now show that $\left(  u\otimes v\right)  \otimes w\in
\operatorname*{Ker}\Phi_{a}$ for every $u\in U$, $v\in V$ and $w\in W$. In
fact, every $u\in U$, $v\in V$ and $w\in W$ satisfy%
\begin{align*}
&  \Phi_{a}\left(  \left(  u\otimes v\right)  \otimes w\right) \\
&  =\alpha_{U,V,W}\left(  \underbrace{a\rightharpoonup\left(  \left(  u\otimes
v\right)  \otimes w\right)  }_{\substack{=\left(  a\rightharpoonup\left(
u\otimes v\right)  \right)  \otimes w+\left(  u\otimes v\right)
\otimes\left(  a\rightharpoonup w\right)  \\\text{(by (\ref{prop.lie(X).form}%
), applied}\\\text{to }U\otimes V\text{ and }u\otimes v\text{ instead of
}V\text{ and }v\text{)}}}\right)  -a\rightharpoonup\left(  \underbrace{\alpha
_{U,V,W}\left(  \left(  u\otimes v\right)  \otimes w\right)  }%
_{\substack{=u\otimes\left(  v\otimes w\right)  \\\text{(by the definition of
}\alpha_{U,V,W}\text{)}}}\right) \\
&  =\alpha_{U,V,W}\left(  \underbrace{\left(  a\rightharpoonup\left(  u\otimes
v\right)  \right)  }_{\substack{=\left(  a\rightharpoonup u\right)  \otimes
v+u\otimes\left(  a\rightharpoonup v\right)  \\\text{(by
(\ref{prop.lie(X).form}), applied}\\\text{to }U\text{, }V\text{, }u\text{ and
}v\text{ instead of }V\text{, }W\text{, }v\text{ and }w\text{)}}}\otimes
w+\left(  u\otimes v\right)  \otimes\left(  a\rightharpoonup w\right)
\right)  -a\rightharpoonup\left(  u\otimes\left(  v\otimes w\right)  \right)
\\
&  =\alpha_{U,V,W}\left(  \underbrace{\left(  \left(  a\rightharpoonup
u\right)  \otimes v+u\otimes\left(  a\rightharpoonup v\right)  \right)
\otimes w}_{=\left(  \left(  a\rightharpoonup u\right)  \otimes v\right)
\otimes w+\left(  u\otimes\left(  a\rightharpoonup v\right)  \right)  \otimes
w}+\left(  u\otimes v\right)  \otimes\left(  a\rightharpoonup w\right)
\right)  -a\rightharpoonup\left(  u\otimes\left(  v\otimes w\right)  \right)
\\
&  =\underbrace{\alpha_{U,V,W}\left(  \left(  \left(  a\rightharpoonup
u\right)  \otimes v\right)  \otimes w+\left(  u\otimes\left(  a\rightharpoonup
v\right)  \right)  \otimes w+\left(  u\otimes v\right)  \otimes\left(
a\rightharpoonup w\right)  \right)  }_{\substack{=\alpha_{U,V,W}\left(
\left(  \left(  a\rightharpoonup u\right)  \otimes v\right)  \otimes w\right)
+\alpha_{U,V,W}\left(  \left(  u\otimes\left(  a\rightharpoonup v\right)
\right)  \otimes w\right)  +\alpha_{U,V,W}\left(  \left(  u\otimes v\right)
\otimes\left(  a\rightharpoonup w\right)  \right)  \\\text{(since }%
\alpha_{U,V,W}\text{ is }k\text{-linear)}}}-a\rightharpoonup\left(
u\otimes\left(  v\otimes w\right)  \right) \\
&  =\underbrace{\alpha_{U,V,W}\left(  \left(  \left(  a\rightharpoonup
u\right)  \otimes v\right)  \otimes w\right)  }_{\substack{=\left(
a\rightharpoonup u\right)  \otimes\left(  v\otimes w\right)  \\\text{(by the
definition of }\alpha_{U,V,W}\text{)}}}+\underbrace{\alpha_{U,V,W}\left(
\left(  u\otimes\left(  a\rightharpoonup v\right)  \right)  \otimes w\right)
}_{\substack{=u\otimes\left(  \left(  a\rightharpoonup v\right)  \otimes
w\right)  \\\text{(by the definition of }\alpha_{U,V,W}\text{)}}%
}+\underbrace{\alpha_{U,V,W}\left(  \left(  u\otimes v\right)  \otimes\left(
a\rightharpoonup w\right)  \right)  }_{\substack{=u\otimes\left(
v\otimes\left(  a\rightharpoonup w\right)  \right)  \\\text{(by the definition
of }\alpha_{U,V,W}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ -\underbrace{a\rightharpoonup\left(  u\otimes\left(
v\otimes w\right)  \right)  }_{\substack{=\left(  a\rightharpoonup u\right)
\otimes\left(  v\otimes w\right)  +u\otimes\left(  a\rightharpoonup\left(
v\otimes w\right)  \right)  \\\text{(by (\ref{prop.lie(X).form}),
applied}\\\text{to }U\text{, }V\otimes W\text{, }u\text{ and }v\otimes w\text{
instead of }V\text{, }W\text{, }v\text{ and }w\text{)}}}\\
&  =\left(  a\rightharpoonup u\right)  \otimes\left(  v\otimes w\right)
+u\otimes\left(  \left(  a\rightharpoonup v\right)  \otimes w\right)
+u\otimes\left(  v\otimes\left(  a\rightharpoonup w\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  \left(  a\rightharpoonup u\right)
\otimes\left(  v\otimes w\right)  +u\otimes\left(  a\rightharpoonup\left(
v\otimes w\right)  \right)  \right) \\
&  =u\otimes\left(  \left(  a\rightharpoonup v\right)  \otimes w\right)
+u\otimes\left(  v\otimes\left(  a\rightharpoonup w\right)  \right)
-u\otimes\underbrace{\left(  a\rightharpoonup\left(  v\otimes w\right)
\right)  }_{\substack{=\left(  a\rightharpoonup v\right)  \otimes
w+v\otimes\left(  a\rightharpoonup w\right)  \\\text{(by
(\ref{prop.lie(X).form}))}}}\\
&  =u\otimes\left(  \left(  a\rightharpoonup v\right)  \otimes w\right)
+u\otimes\left(  v\otimes\left(  a\rightharpoonup w\right)  \right)
-\underbrace{u\otimes\left(  \left(  a\rightharpoonup v\right)  \otimes
w+v\otimes\left(  a\rightharpoonup w\right)  \right)  }_{=u\otimes\left(
\left(  a\rightharpoonup v\right)  \otimes w\right)  +u\otimes\left(
v\otimes\left(  a\rightharpoonup w\right)  \right)  }\\
&  =u\otimes\left(  \left(  a\rightharpoonup v\right)  \otimes w\right)
+u\otimes\left(  v\otimes\left(  a\rightharpoonup w\right)  \right)  -\left(
u\otimes\left(  \left(  a\rightharpoonup v\right)  \otimes w\right)
+u\otimes\left(  v\otimes\left(  a\rightharpoonup w\right)  \right)  \right)
=0.
\end{align*}
Thus, every $u\in U$, $v\in V$ and $w\in W$ satisfy $\left(  u\otimes
v\right)  \otimes w\in\operatorname*{Ker}\Phi_{a}$. Hence,
$\operatorname*{Ker}\Phi_{a}$ contains $\left(  u\otimes v\right)  \otimes w$
for all $\left(  u,v,w\right)  \in U\times V\times W$. Since
$\operatorname*{Ker}\Phi_{a}$ is a $k$-submodule of $\left(  U\otimes
V\right)  \otimes W$, this yields that{\footnotemark}\footnotetext{Here, we
are using Convention \ref{conv.<>}.}
\begin{align*}
\operatorname*{Ker}\Phi_{a}  &  \supseteq\left(  \text{the smallest
}k\text{-submodule of }\left(  U\otimes V\right)  \otimes W\text{ which
contains}\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \left(  u\otimes v\right)  \otimes w\text{ for
all }\left(  u,v,w\right)  \in U\times V\times W\right) \\
&  =\left\langle \left(  u\otimes v\right)  \otimes w\ \mid\ \left(
u,v,w\right)  \in U\times V\times W\right\rangle .
\end{align*}
But since $\left\langle \left(  u\otimes v\right)  \otimes w\ \mid\ \left(
u,v,w\right)  \in U\times V\times W\right\rangle =\left(  U\otimes V\right)
\otimes W$ (this is easy to see{\footnotemark}\footnotetext{\textit{Proof.}
Let $T\in\left(  U\otimes V\right)  \otimes W$ be arbitrary. Then, $T$ is a
tensor in the tensor product $\left(  U\otimes V\right)  \otimes W$, and thus
can be written in the form $T=\sum\limits_{i=1}^{p}\lambda_{i}S_{i}\otimes
w_{i}$ for some $p\in\mathbb{N}$, some elements $\lambda_{1}$, $\lambda_{2}$,
$...$, $\lambda_{p}$ of $k$, some elements $S_{1}$, $S_{2}$, $...$, $S_{p}$ of
$U\otimes V$ and some elements $w_{1}$, $w_{2}$, $...$, $w_{p}$ of $W$
(because every tensor is a $k$-linear combination of pure tensors). For every
$i\in\left\{  1,2,...,p\right\}  $, we know that $S_{i}$ is a tensor in
$U\otimes V$, and thus can be written in the form $S_{i}=\sum\limits_{j=1}%
^{q_{i}}\mu_{i,j}u_{i,j}\otimes v_{i,j}$ for some $q_{i}\in\mathbb{N}$, some
elements $\mu_{i,1}$, $\mu_{i,2}$, $...$, $\mu_{i,q_{i}}$ of $k$, some
elements $u_{i,1}$, $u_{i,2}$, $...$, $u_{i,q_{i}}$ of $U$ and some elements
$v_{i,1}$, $v_{i,2}$, $...$, $v_{i,q_{i}}$ of $V$ (because every tensor is a
$k$-linear combination of pure tensors). Thus,%
\begin{align*}
T  &  =\sum\limits_{i=1}^{p}\lambda_{i}\underbrace{S_{i}}_{=\sum
\limits_{j=1}^{q_{i}}\mu_{i,j}u_{i,j}\otimes v_{i,j}}\otimes w_{i}%
=\sum\limits_{i=1}^{p}\lambda_{i}\left(  \sum\limits_{j=1}^{q_{i}}\mu
_{i,j}u_{i,j}\otimes v_{i,j}\right)  \otimes w_{i}=\sum\limits_{i=1}%
^{p}\lambda_{i}\sum\limits_{j=1}^{q_{i}}\mu_{i,j}\left(  u_{i,j}\otimes
v_{i,j}\right)  \otimes w_{i}\\
&  \in\left\langle \left(  u\otimes v\right)  \otimes w\ \mid\ \left(
u,v,w\right)  \in U\times V\times W\right\rangle
\end{align*}
(because for every $i\in\left\{  1,2,...,p\right\}  $ and every $j\in\left\{
1,2,...,q_{i}\right\}  $, the tensor $\left(  u_{i,j}\otimes v_{i,j}\right)
\otimes w_{i}$ has the form $\left(  u\otimes v\right)  \otimes w$ for some
$\left(  u,v,w\right)  \in U\times V\times W$ (namely, for $\left(
u,v,w\right)  =\left(  u_{i,j},v_{i,j},w_{i}\right)  $)). So we have proven
that every $T\in\left(  U\otimes V\right)  \otimes W$ satisfies $T\in
\left\langle \left(  u\otimes v\right)  \otimes w\ \mid\ \left(  u,v,w\right)
\in U\times V\times W\right\rangle $. In other words, $\left(  U\otimes
V\right)  \otimes W\subseteq\left\langle \left(  u\otimes v\right)  \otimes
w\ \mid\ \left(  u,v,w\right)  \in U\times V\times W\right\rangle $. Combined
with the obvious relation $\left\langle \left(  u\otimes v\right)  \otimes
w\ \mid\ \left(  u,v,w\right)  \in U\times V\times W\right\rangle
\subseteq\left(  U\otimes V\right)  \otimes W$, this yields $\left\langle
\left(  u\otimes v\right)  \otimes w\ \mid\ \left(  u,v,w\right)  \in U\times
V\times W\right\rangle =\left(  U\otimes V\right)  \otimes W$, qed.}), this
rewrites as $\operatorname*{Ker}\Phi_{a}\supseteq\left(  U\otimes V\right)
\otimes W$, so that $\Phi_{a}=0$. Hence, every $T\in\left(  U\otimes V\right)
\otimes W$ satisfies%
\[
0=\Phi_{a}\left(  T\right)  =\alpha_{U,V,W}\left(  a\rightharpoonup T\right)
-a\rightharpoonup\left(  \alpha_{U,V,W}\left(  T\right)  \right)  .
\]
In other words, every $T\in\left(  U\otimes V\right)  \otimes W$ satisfies
$\alpha_{U,V,W}\left(  a\rightharpoonup T\right)  =a\rightharpoonup\left(
\alpha_{U,V,W}\left(  T\right)  \right)  $. This shows that $\alpha_{U,V,W}$
is a $\mathfrak{g}$-module homomorphism. Since we also know that
$\alpha_{U,V,W}$ is a $k$-module isomorphism, we conclude that $\alpha
_{U,V,W}$ is a $\mathfrak{g}$-module isomorphism.
\end{verlong}

\begin{verlong}
Thus, $\alpha_{U,V,W}$ is a canonical isomorphism of $\mathfrak{g}$-modules.
Since $\alpha_{U,V,W}$ was defined as the $k$-linear map%
\[
\left(  U\otimes V\right)  \otimes W\rightarrow U\otimes\left(  V\otimes
W\right)  ,\ \ \ \ \ \ \ \ \ \ \left(  u\otimes v\right)  \otimes w\mapsto
u\otimes\left(  v\otimes w\right)  ,
\]
it thus follows that the $k$-linear map%
\[
\left(  U\otimes V\right)  \otimes W\rightarrow U\otimes\left(  V\otimes
W\right)  ,\ \ \ \ \ \ \ \ \ \ \left(  u\otimes v\right)  \otimes w\mapsto
u\otimes\left(  v\otimes w\right)
\]
is a canonical isomorphism of $\mathfrak{g}$-modules. This proves Proposition
\ref{prop.tensor-assoc} \textbf{(c)}.
\end{verlong}

\begin{verlong}
\begin{corollary}
\label{cor.k(X)V}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $V$ be a $\mathfrak{g}$-module. Then, the canonical
$k$-module isomorphism $k\otimes V\rightarrow V$ (this is the $k$-module
homomorphism that sends $\lambda\otimes v$ to $\lambda v$ for all $\lambda\in
k$ and $v\in V$) is a $\mathfrak{g}$-module isomorphism.
\end{corollary}

\textit{Proof of Corollary \ref{cor.k(X)V}.} According to Proposition
\ref{prop.tensor-assoc} \textbf{(a)}, the $k$-linear map $V\rightarrow
k\otimes V,$ $v\mapsto1\otimes v$ is an isomorphism of $\mathfrak{g}$-modules.
Thus, the inverse of this map $V\rightarrow k\otimes V,$ $v\mapsto1\otimes v$
is an isomorphism of $\mathfrak{g}$-modules as well (because the inverse of an
isomorphism of $\mathfrak{g}$-modules is an isomorphism of $\mathfrak{g}%
$-modules). But since the inverse of the map $V\rightarrow k\otimes V,$
$v\mapsto1\otimes v$ is the canonical $k$-module isomorphism $k\otimes
V\rightarrow V$ (this is easy to show), this yields that the canonical
$k$-module isomorphism $k\otimes V\rightarrow V$ is an isomorphism of
$\mathfrak{g}$-modules. In other words, Corollary \ref{cor.k(X)V} is proven.
\end{verlong}

\begin{condition}
\label{conv.V(X)k}Let $k$ be a commutative ring. Let $V$ be a $k$%
-module.\newline We are going to identify the three $k$-modules $V\otimes k$,
$k\otimes V$ and $V$ with each other (due to the canonical isomorphisms
$V\rightarrow V\otimes k$ and $V\rightarrow k\otimes V$).\newline If $V$ is a
$\mathfrak{g}$-module, where $\mathfrak{g}$ is some $k$-Lie algebra, then this
identification will not conflict with the $\mathfrak{g}$-module structures on
$V\otimes k$, $k\otimes V$ and $V$ (because the canonical isomorphisms
$V\rightarrow V\otimes k$ and $V\rightarrow k\otimes V$ are $\mathfrak{g}%
$-module isomorphisms (as Proposition \ref{prop.tensor-assoc} \textbf{(a)} and
\textbf{(b)} shows)).
\end{condition}

\begin{verlong}
Finally a straightforward fact:
\end{verlong}

\begin{proposition}
\label{prop.lie(X)f(X)g}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra.\newline Let $V$, $W$, $V^{\prime}$ and $W^{\prime}$ be four
$\mathfrak{g}$-modules, and let $f:V\rightarrow V^{\prime}$ and
$g:W\rightarrow W^{\prime}$ be two $\mathfrak{g}$-module homomorphisms. Then,
$f\otimes g:V\otimes W\rightarrow V^{\prime}\otimes W^{\prime}$ is a
$\mathfrak{g}$-module homomorphism.
\end{proposition}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.lie(X)f(X)g}.} Let $a\in\mathfrak{g}$
be arbitrary.

Clearly, the maps $V\otimes W\rightarrow V\otimes W,$ $T\mapsto
a\rightharpoonup T$ and $V^{\prime}\otimes W^{\prime}\rightarrow V^{\prime
}\otimes W^{\prime},$ $T\mapsto a\rightharpoonup T$ are $k$-linear (because
the Lie action of a $\mathfrak{g}$-module is $k$-bilinear).

Let $R_{a}:V\otimes W\rightarrow V^{\prime}\otimes W^{\prime}$ be the map
defined by%
\[
\left(  R_{a}\left(  T\right)  =a\rightharpoonup\left(  \left(  f\otimes
g\right)  \left(  T\right)  \right)  -\left(  f\otimes g\right)  \left(
a\rightharpoonup T\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }T\in V\otimes
W\right)  .
\]
This map $R_{a}$ is $k$-linear (since the maps $V\otimes W\rightarrow V\otimes
W,$ $T\mapsto a\rightharpoonup T$ and $V^{\prime}\otimes W^{\prime}\rightarrow
V^{\prime}\otimes W^{\prime},$ $T\mapsto a\rightharpoonup T$ are $k$-linear,
and so is the map $f\otimes g$). Thus, $\operatorname{Ker}R_{a}$ is a
$k$-submodule of $V\otimes W$.

Now let us prove that
\begin{equation}
\text{every }\left(  v,w\right)  \in V\times W\text{ satisfies }v\otimes
w\in\operatorname{Ker}R_{a}. \label{pf.lie(X)f(X)g.1}%
\end{equation}


\textit{Proof of (\ref{pf.lie(X)f(X)g.1}).} Let $\left(  v,w\right)  \in
V\times W$ be arbitrary. Then, $v\in V$ and $w\in W$. Now, the definition of
$R_{a}$ yields%
\begin{align*}
&  R_{a}\left(  v\otimes w\right) \\
&  =a\rightharpoonup\left(  \underbrace{\left(  f\otimes g\right)  \left(
v\otimes w\right)  }_{=f\left(  v\right)  \otimes g\left(  w\right)  }\right)
-\left(  f\otimes g\right)  \left(  \underbrace{a\rightharpoonup\left(
v\otimes w\right)  }_{\substack{=\left(  a\rightharpoonup v\right)  \otimes
w+v\otimes\left(  a\rightharpoonup w\right)  \\\text{(according to
(\ref{prop.lie(X).form}))}}}\right) \\
&  =\underbrace{a\rightharpoonup\left(  f\left(  v\right)  \otimes g\left(
w\right)  \right)  }_{\substack{=\left(  a\rightharpoonup\left(  f\left(
v\right)  \right)  \right)  \otimes g\left(  w\right)  +f\left(  v\right)
\otimes\left(  a\rightharpoonup\left(  g\left(  w\right)  \right)  \right)
\\\text{(according to (\ref{prop.lie(X).form}),}\\\text{applied to }V^{\prime
}\text{, }W^{\prime}\text{, }f\left(  v\right)  \text{ and }g\left(  w\right)
\text{ instead of }V\text{, }W\text{, }v\text{ and }w\text{)}}%
}-\underbrace{\left(  f\otimes g\right)  \left(  \left(  a\rightharpoonup
v\right)  \otimes w+v\otimes\left(  a\rightharpoonup w\right)  \right)
}_{\substack{=\left(  f\otimes g\right)  \left(  \left(  a\rightharpoonup
v\right)  \otimes w\right)  +\left(  f\otimes g\right)  \left(  v\otimes
\left(  a\rightharpoonup w\right)  \right)  \\\text{(since }f\otimes g\text{
is a }k\text{-linear map)}}}\\
&  =\left(  \left(  a\rightharpoonup\left(  f\left(  v\right)  \right)
\right)  \otimes g\left(  w\right)  +f\left(  v\right)  \otimes\left(
a\rightharpoonup\left(  g\left(  w\right)  \right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  \underbrace{\left(  f\otimes g\right)  \left(
\left(  a\rightharpoonup v\right)  \otimes w\right)  }_{=f\left(
a\rightharpoonup v\right)  \otimes g\left(  w\right)  }+\underbrace{\left(
f\otimes g\right)  \left(  v\otimes\left(  a\rightharpoonup w\right)  \right)
}_{=f\left(  v\right)  \otimes g\left(  a\rightharpoonup w\right)  }\right) \\
&  =\left(  \left(  a\rightharpoonup\left(  f\left(  v\right)  \right)
\right)  \otimes g\left(  w\right)  +f\left(  v\right)  \otimes\left(
a\rightharpoonup\left(  g\left(  w\right)  \right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  \underbrace{f\left(  a\rightharpoonup
v\right)  }_{\substack{=a\rightharpoonup\left(  f\left(  v\right)  \right)
\\\text{(since }f\text{ is a }\mathfrak{g}\text{-module map)}}}\otimes
g\left(  w\right)  +f\left(  v\right)  \otimes\underbrace{g\left(
a\rightharpoonup w\right)  }_{\substack{=a\rightharpoonup\left(  g\left(
w\right)  \right)  \\\text{(since }g\text{ is a }\mathfrak{g}\text{-module
map)}}}\right) \\
&  =\left(  \left(  a\rightharpoonup\left(  f\left(  v\right)  \right)
\right)  \otimes g\left(  w\right)  +f\left(  v\right)  \otimes\left(
a\rightharpoonup\left(  g\left(  w\right)  \right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  \left(  a\rightharpoonup\left(  f\left(
v\right)  \right)  \right)  \otimes g\left(  w\right)  +f\left(  v\right)
\otimes\left(  a\rightharpoonup\left(  g\left(  w\right)  \right)  \right)
\right) \\
&  =0.
\end{align*}
Thus, $v\otimes w\in\operatorname{Ker}R_{a}$. This proves
(\ref{pf.lie(X)f(X)g.1}).

The relation (\ref{pf.lie(X)f(X)g.1}) yields $\left\{  v\otimes w\ \mid
\ \left(  v,w\right)  \in V\times W\right\}  \subseteq\operatorname{Ker}R_{a}$.

Now, we have $V\otimes W=\left\langle v\otimes w\ \mid\ \left(  v,w\right)
\in V\times W\right\rangle $ (since the tensor product $V\otimes W$ is
generated by its pure tensors). In other words,%
\[
V\otimes W=\left\langle v\otimes w\ \mid\ \left(  v,w\right)  \in V\times
W\right\rangle =\left\langle \left\{  v\otimes w\ \mid\ \left(  v,w\right)
\in V\times W\right\}  \right\rangle \subseteq\operatorname{Ker}R_{a}%
\]
(due to Proposition \ref{prop.<>} \textbf{(a)} (applied to $V\otimes W$,
$\left\{  v\otimes w\ \mid\ \left(  v,w\right)  \in V\times W\right\}  $ and
$\operatorname{Ker}R_{a}$ instead of $M$, $S$ and $Q$), which can be applied
since $\left\{  v\otimes w\ \mid\ \left(  v,w\right)  \in V\times W\right\}
\subseteq\operatorname{Ker}R_{a}$ and since $\operatorname{Ker}R_{a}$ is a
$k$-module). Hence, $R_{a}=0$. Thus, every $T\in V\otimes W$ satisfies
$R_{a}\left(  T\right)  =0$. Therefore, every $T\in V\otimes W$ satisfies
$\left(  f\otimes g\right)  \left(  a\rightharpoonup T\right)
=a\rightharpoonup\left(  \left(  f\otimes g\right)  \left(  T\right)  \right)
$ (because $0=R_{a}\left(  T\right)  =a\rightharpoonup\left(  \left(  f\otimes
g\right)  \left(  T\right)  \right)  -\left(  f\otimes g\right)  \left(
a\rightharpoonup T\right)  $ and thus $a\rightharpoonup\left(  \left(
f\otimes g\right)  \left(  T\right)  \right)  =\left(  f\otimes g\right)
\left(  a\rightharpoonup T\right)  $). In other words, $f\otimes g$ is a
$\mathfrak{g}$-module homomorphism. This proves Proposition
\ref{prop.lie(X)f(X)g}.
\end{verlong}

We notice an analogue of Proposition \ref{prop.lie.(+)vsRes} for tensor products:

\begin{proposition}
\label{prop.lie.(X)vsRes}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$.\newline If $V$ and $W$ are two $\mathfrak{g}$-modules, then
$\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}\left(  V\otimes W\right)
=\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V\right)
\otimes\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}W\right)  $ as
$\mathfrak{h}$-modules. This allows us to speak of "the $\mathfrak{h}$-module
$V\otimes W$" without having to worry whether we mean $\operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}\left(  V\otimes W\right)  $ or $\left(
\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V\right)  \otimes\left(
\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}W\right)  $ (because it does
not matter, since $\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}\left(
V\otimes W\right)  =\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}%
}V\right)  \otimes\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}%
}W\right)  $).
\end{proposition}

This follows from the definitions.

\subsection{Tensor products of several $\mathfrak{g}$-modules}

We now define multi-factor tensor products of $\mathfrak{g}$-modules. First we
recall one of the possible definitions of the tensor product of several $k$-modules:

\begin{definition}
\label{defs.(X)mult}Let $k$ be a commutative ring. Let $n\in\mathbb{N}%
$.\newline Now, by induction over $n$, we are going to define a $k$-module
$V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ for any $n$ arbitrary $k$-modules
$V_{1}$, $V_{2}$, $...$, $V_{n}$:\newline\textit{Induction base:} For $n=0$,
we define $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ as the $k$-module
$k$.\newline\textit{Induction step:} Let $p\in\mathbb{N}$. Assuming that we
have defined a $k$-module $V_{1}\otimes V_{2}\otimes...\otimes V_{p}$ for any
$p$ arbitrary $k$-modules $V_{1}$, $V_{2}$, $...$, $V_{p}$, we now define a
$k$-module $V_{1}\otimes V_{2}\otimes...\otimes V_{p+1}$ for any $p+1$
arbitrary $k$-modules $V_{1}$, $V_{2}$, $...$, $V_{p+1}$ by the equation%
\begin{equation}
V_{1}\otimes V_{2}\otimes...\otimes V_{p+1}=V_{1}\otimes\left(  V_{2}\otimes
V_{3}\otimes...\otimes V_{p+1}\right)  . \label{defs.(X)mult.left}%
\end{equation}
Here, $V_{1}\otimes\left(  V_{2}\otimes V_{3}\otimes...\otimes V_{p+1}\right)
$ is to be understood as the tensor product of the $k$-module $V_{1}$ with the
$k$-module $V_{2}\otimes V_{3}\otimes...\otimes V_{p+1}$ (note that the
$k$-module $V_{2}\otimes V_{3}\otimes...\otimes V_{p+1}$ is already defined
because we assumed that we have defined a $k$-module $V_{1}\otimes
V_{2}\otimes...\otimes V_{p}$ for any $p$ arbitrary $k$-modules $V_{1}$,
$V_{2}$, $...$, $V_{p}$). This completes the inductive definition.\newline
Thus we have defined a $k$-module $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$
for any $n$ arbitrary $k$-modules $V_{1}$, $V_{2}$, $...$, $V_{n}$ for any
$n\in\mathbb{N}$. This $k$-module $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$
is called the \textit{tensor product} of the $k$-modules $V_{1}$, $V_{2}$,
$...$, $V_{n}$.
\end{definition}

\begin{remark}
\textbf{(a)} Definition \ref{defs.(X)mult} is not the only possible definition
of the tensor product of several $k$-modules. One could obtain a different
definition by replacing the equation (\ref{defs.(X)mult.left}) by%
\[
V_{1}\otimes V_{2}\otimes...\otimes V_{p+1}=\left(  V_{1}\otimes V_{2}%
\otimes...\otimes V_{p}\right)  \otimes V_{p+1}.
\]
This definition would have given us a \textit{different} $k$-module
$V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ for any $n$ arbitrary $k$-modules
$V_{1}$, $V_{2}$, $...$, $V_{n}$ for any $n\in\mathbb{N}$ than the one defined
in Definition \ref{defs.(X)mult}. However, this $k$-module would still be
\textit{canonically isomorphic} to the one defined in Definition
\ref{defs.(X)mult}, and thus it is commonly considered to be "more or less the
same $k$-module".\newline There is yet another definition of $V_{1}\otimes
V_{2}\otimes...\otimes V_{n}$, which proceeds by taking the free $k$-module on
the set $V_{1}\times V_{2}\times...\times V_{n}$ and factoring it modulo a
certain submodule. This definition gives yet \textit{another} $k$-module
$V_{1}\otimes V_{2}\otimes...\otimes V_{n}$, but this module is also
canonically isomorphic to the $k$-module $V_{1}\otimes V_{2}\otimes...\otimes
V_{n}$ defined in Definition \ref{defs.(X)mult}, and thus can be considered to
be "more or less the same $k$-module".\newline\textbf{(b)} Definition
\ref{defs.(X)mult}, applied to $n=1$, defines the tensor product of
\textit{one} $k$-module $V_{1}$ as $V_{1}\otimes k$. This takes some getting
used to, since it seems more natural to define the tensor product of one
$k$-module $V_{1}$ simply as $V_{1}$. But this isn't really different because
there is a canonical isomorphism of $k$-modules $V_{1}\cong V_{1}\otimes k$,
so most people consider $V_{1}$ to be "more or less the same $k$-module" as
$V_{1}\otimes k$.
\end{remark}

Now, by analogy, we define the tensor product of several $\mathfrak{g}$-modules.

\begin{definition}
\label{defs.lie(X)mult}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $n\in\mathbb{N}$.\newline Now, by induction over $n$, we
are going to define a $\mathfrak{g}$-module $V_{1}\otimes V_{2}\otimes
...\otimes V_{n}$ for any $n$ arbitrary $\mathfrak{g}$-modules $V_{1}$,
$V_{2}$, $...$, $V_{n}$:\newline\textit{Induction base:} For $n=0$, we define
$V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ as the $\mathfrak{g}$-module $k$
defined in Definition \ref{defs.k-mod-g}.\newline\textit{Induction step:} Let
$p\in\mathbb{N}$. Assuming that we have defined a $\mathfrak{g}$-module
$V_{1}\otimes V_{2}\otimes...\otimes V_{p}$ for any $p$ arbitrary
$\mathfrak{g}$-modules $V_{1}$, $V_{2}$, $...$, $V_{p}$, we now define a
$\mathfrak{g}$-module $V_{1}\otimes V_{2}\otimes...\otimes V_{p+1}$ for any
$p+1$ arbitrary $\mathfrak{g}$-modules $V_{1}$, $V_{2}$, $...$, $V_{p+1}$ by
the equation%
\begin{equation}
V_{1}\otimes V_{2}\otimes...\otimes V_{p+1}=V_{1}\otimes\left(  V_{2}\otimes
V_{3}\otimes...\otimes V_{p+1}\right)  . \label{defs.lie(X)mult.left}%
\end{equation}
Here, $V_{1}\otimes\left(  V_{2}\otimes V_{3}\otimes...\otimes V_{p+1}\right)
$ is to be understood as the tensor product of the $\mathfrak{g}$-module
$V_{1}$ with the $\mathfrak{g}$-module $V_{2}\otimes V_{3}\otimes...\otimes
V_{p+1}$ (note that the $\mathfrak{g}$-module $V_{2}\otimes V_{3}%
\otimes...\otimes V_{p+1}$ is already defined because we assumed that we have
defined a $\mathfrak{g}$-module $V_{1}\otimes V_{2}\otimes...\otimes V_{p}$
for any $p$ arbitrary $\mathfrak{g}$-modules $V_{1}$, $V_{2}$, $...$, $V_{p}%
$). This completes the inductive definition.\newline Thus we have defined a
$\mathfrak{g}$-module $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ for any $n$
arbitrary $\mathfrak{g}$-modules $V_{1}$, $V_{2}$, $...$, $V_{n}$ for any
$n\in\mathbb{N}$. This $\mathfrak{g}$-module $V_{1}\otimes V_{2}%
\otimes...\otimes V_{n}$ is called the \textit{tensor product} of the
$\mathfrak{g}$-modules $V_{1}$, $V_{2}$, $...$, $V_{n}$.
\end{definition}

\begin{remark}
\label{rmk.lie(X)mult}\textbf{(a)} In Definition \ref{defs.lie(X)mult}, we
could have replaced the equation (\ref{defs.lie(X)mult.left}) by%
\[
V_{1}\otimes V_{2}\otimes...\otimes V_{p+1}=\left(  V_{1}\otimes V_{2}%
\otimes...\otimes V_{p}\right)  \otimes V_{p+1}.
\]
This would have given us a \textit{different} $\mathfrak{g}$-module
$V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ for any $n$ arbitrary
$\mathfrak{g}$-modules $V_{1}$, $V_{2}$, $...$, $V_{n}$ for any $n\in
\mathbb{N}$ than the one defined in Definition \ref{defs.lie(X)mult}. However,
this $\mathfrak{g}$-module would still be \textit{canonically isomorphic} to
the one defined in Definition \ref{defs.lie(X)mult} (we will prove this and
actually something more general in Proposition \ref{prop.lie(X).bracketing}),
and thus it is commonly considered to be "more or less the same $\mathfrak{g}%
$-module".\newline\textbf{(b)} Definition \ref{defs.lie(X)mult}, applied to
$n=1$, defines the tensor product of \textit{one} $\mathfrak{g}$-module
$V_{1}$ as $V_{1}\otimes k$. This takes some getting used to, since it seems
more natural to define the tensor product of one $\mathfrak{g}$-module $V_{1}$
simply as $V_{1}$. But this isn't really different because Proposition
\ref{prop.tensor-assoc} \textbf{(b)} gives a canonical isomorphism of
$\mathfrak{g}$-modules $V_{1}\cong V_{1}\otimes k$, so most people consider
$V_{1}$ to be "more or less the same $\mathfrak{g}$-module" as $V_{1}\otimes
k$.\newline\textbf{(c)} Definition \ref{defs.lie(X)mult} does not conflict
with Definition \ref{defs.(X)mult}, because the underlying $k$-module of the
$\mathfrak{g}$-module $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ defined in
Definition \ref{defs.lie(X)mult} is indeed the $k$-module $V_{1}\otimes
V_{2}\otimes...\otimes V_{n}$ defined in Definition \ref{defs.(X)mult}. (This
is trivial by induction.)
\end{remark}

\begin{condition}
\label{conv.left-ind}A remark about notation is appropriate at this
point:\newline There are two different conflicting notions of a "pure tensor"
in a tensor product $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ of $n$
arbitrary $k$-modules $V_{1}$, $V_{2}$, $...$, $V_{n}$, where $n\geq1$. The
one notion defines a "pure tensor" as an element of the form $v\otimes T$ for
some $v\in V_{1}$ and some $T\in V_{2}\otimes V_{3}\otimes...\otimes V_{n}%
$\ \ \ \ {\footnotemark}. The other notion defines a "pure tensor" as an
element of the form $v_{1}\otimes v_{2}\otimes...\otimes v_{n}$ for some
$\left(  v_{1},v_{2},...,v_{n}\right)  \in V_{1}\times V_{2}\times...\times
V_{n}$. These two notions are not equivalent. In this note, we are going to
yield right of way to the second of these notions, i. e. we are going to
define a pure tensor in $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ as an
element of the form $v_{1}\otimes v_{2}\otimes...\otimes v_{n}$ for some
$\left(  v_{1},v_{2},...,v_{n}\right)  \in V_{1}\times V_{2}\times...\times
V_{n}$. The first notion, however, will also be used - but we will not call it
a "pure tensor" but rather a "left-induced tensor". Thus we define a
\textit{left-induced tensor} in $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ as
an element of the form $v\otimes T$ for some $v\in V_{1}$ and some $T\in
V_{2}\otimes V_{3}\otimes...\otimes V_{n}$.\newline We note that the
$k$-module $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ is generated by its
left-induced tensors, but also generated by its pure tensors.
\end{condition}

\footnotetext{In fact, if we look at Definition \ref{defs.(X)mult}, we see
that the $k$-module $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ was defined as
$V_{1}\otimes\left(  V_{2}\otimes V_{3}\otimes...\otimes V_{n}\right)  $, so
it is the $k$-module $A\otimes B$ where $A=V_{1}$ and $B=V_{2}\otimes
V_{3}\otimes...\otimes V_{n}$. Since the usual definition of a pure tensor in
$A\otimes B$ defines it as an element of the form $v\otimes T$ for some $v\in
A$ and $T\in B$, it thus is logical to say that a pure tensor in $V_{1}\otimes
V_{2}\otimes...\otimes V_{n}$ means an element of the form $v\otimes T$ for
$v\in V_{1}$ and $T\in V_{2}\otimes V_{3}\otimes...\otimes V_{n}$.}

Before we continue, we need a technical result:

\begin{proposition}
\label{prop.lie(X).bracketing}Let $k$ be a commutative ring. Let
$\mathfrak{g}$ be a $k$-Lie algebra. Let $n\in\mathbb{N}$.\newline Then, for
any $n$ arbitrary $\mathfrak{g}$-modules $V_{1}$, $V_{2}$, $...$, $V_{n}$ and
every $i\in\left\{  0,1,...,n\right\}  $, the canonical $k$-module isomorphism
$\left(  V_{1}\otimes V_{2}\otimes...\otimes V_{i}\right)  \otimes\left(
V_{i+1}\otimes V_{i+2}\otimes...\otimes V_{n}\right)  \rightarrow V_{1}\otimes
V_{2}\otimes...\otimes V_{n}$ is a $\mathfrak{g}$-module
isomorphism.{\footnotemark}
\end{proposition}

\footnotetext{Here and in the following, whenever I speak of "the canonical
$k$-module isomorphism $\left(  V_{1}\otimes V_{2}\otimes...\otimes
V_{i}\right)  \otimes\left(  V_{i+1}\otimes V_{i+2}\otimes...\otimes
V_{n}\right)  \rightarrow V_{1}\otimes V_{2}\otimes...\otimes V_{n}$", I mean
the $k$-module homomorphism $\left(  V_{1}\otimes V_{2}\otimes...\otimes
V_{i}\right)  \otimes\left(  V_{i+1}\otimes V_{i+2}\otimes...\otimes
V_{n}\right)  \rightarrow V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ which
sends $\left(  v_{1}\otimes v_{2}\otimes...\otimes v_{i}\right)
\otimes\left(  v_{i+1}\otimes v_{i+2}\otimes...\otimes v_{n}\right)  $ to
$v_{1}\otimes v_{2}\otimes...\otimes v_{n}$ for every $\left(  v_{1}%
,v_{2},...,v_{n}\right)  \in V_{1}\times V_{2}\times...\times V_{n}$. This
homomorphism is known (from linear algebra) to exist, be unique and be a
$k$-module isomorphism. (This is independent of the $\mathfrak{g}$-module
structures on $V_{1}$, $V_{2}$, $...$, $V_{n}$.)}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.lie(X).bracketing}.} Let $\mu
_{k}:k\otimes k\rightarrow k$ denote the multiplication map, i. e. the map
which sends $\lambda\left(  1\otimes1\right)  $ to $\lambda$ for every
$\lambda\in k$. (This map $\mu_{k}$ is well-defined because every element of
$k\otimes k$ can be written uniquely in the form $\lambda\left(
1\otimes1\right)  $ with $\lambda\in k$.)
\end{verlong}

\begin{verlong}
We are going to prove Proposition \ref{prop.lie(X).bracketing} by induction
over $n$:
\end{verlong}

\begin{verlong}
\textit{Induction base:} For $n=0$, Proposition \ref{prop.lie(X).bracketing}
states that for any $0$ arbitrary $\mathfrak{g}$-modules $V_{1}$, $V_{2}$,
$...$, $V_{0}$ and every $i\in\left\{  0\right\}  $, the canonical $k$-module
isomorphism $\left(  V_{1}\otimes V_{2}\otimes...\otimes V_{i}\right)
\otimes\left(  V_{i+1}\otimes V_{i+2}\otimes...\otimes V_{0}\right)
\rightarrow V_{1}\otimes V_{2}\otimes...\otimes V_{0}$ is a $\mathfrak{g}%
$-module isomorphism. But this is true because the $k$-module isomorphism
$\left(  V_{1}\otimes V_{2}\otimes...\otimes V_{i}\right)  \otimes\left(
V_{i+1}\otimes V_{i+2}\otimes...\otimes V_{0}\right)  \rightarrow V_{1}\otimes
V_{2}\otimes...\otimes V_{0}$ is simply the multiplication map $\mu
_{k}:k\otimes k\rightarrow k$ (because each of the tensor products
$V_{1}\otimes V_{2}\otimes...\otimes V_{i}$, $V_{i+1}\otimes V_{i+2}%
\otimes...\otimes V_{0}$ and $V_{1}\otimes V_{2}\otimes...\otimes V_{0}$ is an
empty tensor product (since $i=0$ due to $i\in\left\{  0\right\}  $), and
therefore equals $k$), and the multiplication map $\mu_{k}:k\otimes
k\rightarrow k$ is a $\mathfrak{g}$-module isomorphism (because every
$a\in\mathfrak{g}$ and $v\in k\otimes k$ satisfy $\mu_{k}\left(
a\rightharpoonup v\right)  =a\rightharpoonup\mu_{k}\left(  v\right)
$\ \ \ \ {\footnotemark}\footnotetext{\textit{Proof.} Let $a\in\mathfrak{g}$
and $v\in k\otimes k$. Then, there exists some $\lambda\in k$ such that
$v=\lambda\left(  1\otimes1\right)  $ (because $\left(  1\otimes1\right)  $ is
a basis of the $k$-module $k\otimes k$, since $\left(  1\right)  $ is a basis
of the $k$-module $k$). Thus,%
\begin{align*}
a\rightharpoonup v  &  =a\rightharpoonup\left(  \lambda\left(  1\otimes
1\right)  \right)  =\lambda\left(  a\rightharpoonup\left(  1\otimes1\right)
\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Lie action is
}k\text{-bilinear}\right) \\
&  =\lambda\left(  \underbrace{\left(  a\rightharpoonup1\right)
}_{\substack{=0\text{ (since the Lie}\\\text{action of }k\text{ is }0\text{)}%
}}\otimes1+1\otimes\underbrace{\left(  a\rightharpoonup1\right)
}_{\substack{=0\text{ (since the Lie}\\\text{action of }k\text{ is }0\text{)}%
}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{prop.lie(X).form}), applied to
}V=k\text{, }W=k\text{, }v=1\text{ and }w=1\right) \\
&  =\lambda\underbrace{\left(  0\otimes1+1\otimes0\right)  }_{=0}=0
\end{align*}
and thus $\mu_{k}\left(  a\rightharpoonup v\right)  =\mu_{k}\left(  0\right)
=0$. Combining this with $a\rightharpoonup\mu_{k}\left(  v\right)  =0$ (since
the Lie action of $k$ is $0$), we obtain $\mu_{k}\left(  a\rightharpoonup
v\right)  =a\rightharpoonup\mu_{k}\left(  v\right)  $, qed.}). Thus,
Proposition \ref{prop.lie(X).bracketing} is true for $n=0$. This completes the
induction base.
\end{verlong}

\begin{verlong}
\textit{Induction step:} Let $N\in\mathbb{N}_{+}$ be arbitrary. Assume that
Proposition \ref{prop.lie(X).bracketing} holds for $n=N-1$. We now must prove
that Proposition \ref{prop.lie(X).bracketing} holds for $n=N$. In other words,
we must prove that%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{for any }N\text{ arbitrary }\mathfrak{g}\text{-modules }V_{1}\text{,
}V_{2}\text{, }...\text{, }V_{N}\text{ and every}\\
i\in\left\{  0,1,...,N\right\}  \text{, the canonical }k\text{-module
isomorphism}\\
\left(  V_{1}\otimes V_{2}\otimes...\otimes V_{i}\right)  \otimes\left(
V_{i+1}\otimes V_{i+2}\otimes...\otimes V_{N}\right)  \rightarrow V_{1}\otimes
V_{2}\otimes...\otimes V_{N}\\
\text{is a }\mathfrak{g}\text{-module isomorphism}%
\end{array}
\right)  . \label{pf.lie(X).bracketing.claim}%
\end{equation}

\end{verlong}

\begin{verlong}
We have assume that Proposition \ref{prop.lie(X).bracketing} holds for $n=N$.
In other words, we have assumed that%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{for any }N-1\text{ arbitrary }\mathfrak{g}\text{-modules }V_{1}\text{,
}V_{2}\text{, }...\text{, }V_{N-1}\text{ and every}\\
i\in\left\{  0,1,...,N-1\right\}  \text{, the canonical }k\text{-module
isomorphism}\\
\left(  V_{1}\otimes V_{2}\otimes...\otimes V_{i}\right)  \otimes\left(
V_{i+1}\otimes V_{i+2}\otimes...\otimes V_{N-1}\right)  \rightarrow
V_{1}\otimes V_{2}\otimes...\otimes V_{N-1}\\
\text{is a }\mathfrak{g}\text{-module isomorphism}%
\end{array}
\right)  . \label{pf.lie(X).bracketing.ass}%
\end{equation}

\end{verlong}

\begin{verlong}
Now let us prove (\ref{pf.lie(X).bracketing.claim}). Let $V_{1}$, $V_{2}$,
$...$, $V_{N}$ be $N$ arbitrary $\mathfrak{g}$-modules. Let $i\in\left\{
0,1,...,N\right\}  $. Let $\rho$ be the canonical $k$-module isomorphism
$\left(  V_{1}\otimes V_{2}\otimes...\otimes V_{i}\right)  \otimes\left(
V_{i+1}\otimes V_{i+2}\otimes...\otimes V_{N}\right)  \rightarrow V_{1}\otimes
V_{2}\otimes...\otimes V_{N}$. We are now going to prove that $\rho$ is a
$\mathfrak{g}$-module isomorphism.
\end{verlong}

\begin{verlong}
We distinguish between two cases:
\end{verlong}

\begin{verlong}
\textit{Case 1:} We have $i=0$.
\end{verlong}

\begin{verlong}
\textit{Case 2:} We have $i>0$.
\end{verlong}

\begin{verlong}
Let us first consider Case 1: In this case, $i=0$, so that $V_{1}\otimes
V_{2}\otimes...\otimes V_{i}=k$ and $V_{i+1}\otimes V_{i+2}\otimes...\otimes
V_{N}=V_{1}\otimes V_{2}\otimes...\otimes V_{N}$. Hence, $\rho$ (being the
canonical $k$-module isomorphism $\left(  V_{1}\otimes V_{2}\otimes...\otimes
V_{i}\right)  \otimes\left(  V_{i+1}\otimes V_{i+2}\otimes...\otimes
V_{N}\right)  \rightarrow V_{1}\otimes V_{2}\otimes...\otimes V_{N}$) is the
canonical $k$-module isomorphism $k\otimes\left(  V_{1}\otimes V_{2}%
\otimes...\otimes V_{N}\right)  \rightarrow V_{1}\otimes V_{2}\otimes
...\otimes V_{N}$. But this canonical $k$-module isomorphism is a
$\mathfrak{g}$-module isomorphism (by Corollary \ref{cor.k(X)V}, applied to
$V=V_{1}\otimes V_{2}\otimes...\otimes V_{N}$). Thus, we have proven that
$\rho$ is a $\mathfrak{g}$-module isomorphism in Case 1.
\end{verlong}

\begin{verlong}
Let us now consider Case 2: In this case, $i>0$, so we have $V_{1}\otimes
V_{2}\otimes...\otimes V_{i}=V_{1}\otimes\left(  V_{2}\otimes V_{3}%
\otimes...\otimes V_{i}\right)  $ (as a $\mathfrak{g}$-module) by the
definition of $V_{1}\otimes V_{2}\otimes...\otimes V_{i}$. Also, $V_{1}\otimes
V_{2}\otimes...\otimes V_{N}=V_{1}\otimes\left(  V_{2}\otimes V_{3}%
\otimes...\otimes V_{N}\right)  $ (as a $\mathfrak{g}$-module) by the
definition of $V_{1}\otimes V_{2}\otimes...\otimes V_{N}$.
\end{verlong}

\begin{verlong}
We have denoted by $\rho$ the canonical $k$-module isomorphism $\left(
V_{1}\otimes V_{2}\otimes...\otimes V_{i}\right)  \otimes\left(
V_{i+1}\otimes V_{i+2}\otimes...\otimes V_{N}\right)  \rightarrow V_{1}\otimes
V_{2}\otimes...\otimes V_{N}$. Since $V_{1}\otimes V_{2}\otimes...\otimes
V_{i}=V_{1}\otimes\left(  V_{2}\otimes V_{3}\otimes...\otimes V_{i}\right)  $
and $V_{1}\otimes V_{2}\otimes...\otimes V_{N}=V_{1}\otimes\left(
V_{2}\otimes V_{3}\otimes...\otimes V_{N}\right)  $, this isomorphism $\rho$
thus is the canonical $k$-module isomorphism $\left(  V_{1}\otimes\left(
V_{2}\otimes V_{3}\otimes...\otimes V_{i}\right)  \right)  \otimes\left(
V_{i+1}\otimes V_{i+2}\otimes...\otimes V_{N}\right)  \rightarrow V_{1}%
\otimes\left(  V_{2}\otimes V_{3}\otimes...\otimes V_{N}\right)  $.

In the following, let $\operatorname*{id}$ denote the identity
$\operatorname*{id}\nolimits_{V_{1}}:V_{1}\rightarrow V_{1}$.

Let us denote by $\rho^{\prime}$ the canonical $k$-module isomorphism
\[
\left(  V_{2}\otimes V_{3}\otimes...\otimes V_{i}\right)  \otimes\left(
V_{i+1}\otimes V_{i+2}\otimes...\otimes V_{N}\right)  \rightarrow V_{2}\otimes
V_{3}\otimes...\otimes V_{N}.
\]
Then, $\operatorname*{id}\otimes\rho^{\prime}$ is the canonical $k$-module
isomorphism
\[
V_{1}\otimes\left(  \left(  V_{2}\otimes V_{3}\otimes...\otimes V_{i}\right)
\otimes\left(  V_{i+1}\otimes V_{i+2}\otimes...\otimes V_{N}\right)  \right)
\rightarrow V_{1}\otimes\left(  V_{2}\otimes V_{3}\otimes...\otimes
V_{N}\right)  .
\]

\end{verlong}

\begin{verlong}
We are going to denote the $\mathfrak{g}$-modules $V_{1}$, $V_{2}\otimes
V_{3}\otimes...\otimes V_{i}$ and $V_{i+1}\otimes V_{i+2}\otimes...\otimes
V_{N}$ by $U$, $V$ and $W$, respectively. Then,%
\begin{align*}
U\otimes V  &  =V_{1}\otimes\left(  V_{2}\otimes V_{3}\otimes...\otimes
V_{i}\right)  =V_{1}\otimes V_{2}\otimes...\otimes V_{i}%
\ \ \ \ \ \ \ \ \ \ \text{and thus}\\
\left(  U\otimes V\right)  \otimes W  &  =\left(  V_{1}\otimes V_{2}%
\otimes...\otimes V_{i}\right)  \otimes\left(  V_{i+1}\otimes V_{i+2}%
\otimes...\otimes V_{N}\right)  .
\end{align*}
Now, denote by $\alpha_{U,V,W}$ the $k$-linear map
\[
\left(  U\otimes V\right)  \otimes W\rightarrow U\otimes\left(  V\otimes
W\right)  ,\ \ \ \ \ \ \ \ \ \ \left(  u\otimes v\right)  \otimes w\mapsto
u\otimes\left(  v\otimes w\right)
\]
(this map is well-defined according to linear algebra). From Proposition
\ref{prop.tensor-assoc} \textbf{(c)}, we know that this map $\alpha_{U,V,W}$
is an isomorphism of $\mathfrak{g}$-modules. Moreover, this map $\alpha
_{U,V,W}$ turns out to be a map from $\left(  V_{1}\otimes\left(  V_{2}\otimes
V_{3}\otimes...\otimes V_{i}\right)  \right)  \otimes\left(  V_{i+1}\otimes
V_{i+2}\otimes...\otimes V_{N}\right)  $ to $V_{1}\otimes\left(  \left(
V_{2}\otimes V_{3}\otimes...\otimes V_{i}\right)  \otimes\left(
V_{i+1}\otimes V_{i+2}\otimes...\otimes V_{N}\right)  \right)  $ (because
$\alpha_{U,V,W}$ is a map from $\left(  U\otimes V\right)  \otimes W$ to
$U\otimes\left(  V\otimes W\right)  $, but we know that $U=V_{1}$,
$V=V_{2}\otimes V_{3}\otimes...\otimes V_{i}$ and $W=V_{i+1}\otimes
V_{i+2}\otimes...\otimes V_{N}$), and it is actually the canonical $k$-module
isomorphism
\begin{align*}
&  \left(  V_{1}\otimes\left(  V_{2}\otimes V_{3}\otimes...\otimes
V_{i}\right)  \right)  \otimes\left(  V_{i+1}\otimes V_{i+2}\otimes...\otimes
V_{N}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \rightarrow V_{1}\otimes\left(  \left(  V_{2}\otimes
V_{3}\otimes...\otimes V_{i}\right)  \otimes\left(  V_{i+1}\otimes
V_{i+2}\otimes...\otimes V_{N}\right)  \right)
\end{align*}
(because it maps
\[
\left(  v_{1}\otimes\left(  v_{2}\otimes v_{3}\otimes...\otimes v_{i}\right)
\right)  \otimes\left(  v_{i+1}\otimes v_{i+2}\otimes...\otimes v_{N}\right)
\]
to
\[
v_{1}\otimes\left(  \left(  v_{2}\otimes v_{3}\otimes...\otimes v_{i}\right)
\otimes\left(  v_{i+1}\otimes v_{i+2}\otimes...\otimes v_{N}\right)  \right)
\]
for every $\left(  v_{1},v_{2},...,v_{N}\right)  \in V_{1}\times V_{2}%
\times...\times V_{N}$, according to its definition).
\end{verlong}

\begin{verlong}
We now recollect our knowledge:

\begin{itemize}
\item The map $\rho$ is the canonical $k$-module isomorphism
\[
\left(  V_{1}\otimes\left(  V_{2}\otimes V_{3}\otimes...\otimes V_{i}\right)
\right)  \otimes\left(  V_{i+1}\otimes V_{i+2}\otimes...\otimes V_{N}\right)
\rightarrow V_{1}\otimes\left(  V_{2}\otimes V_{3}\otimes...\otimes
V_{N}\right)  .
\]


\item The map $\operatorname*{id}\otimes\rho^{\prime}$ is the canonical
$k$-module isomorphism
\[
V_{1}\otimes\left(  \left(  V_{2}\otimes V_{3}\otimes...\otimes V_{i}\right)
\otimes\left(  V_{i+1}\otimes V_{i+2}\otimes...\otimes V_{N}\right)  \right)
\rightarrow V_{1}\otimes\left(  V_{2}\otimes V_{3}\otimes...\otimes
V_{N}\right)  .
\]


\item The map $\alpha_{U,V,W}$ is the canonical $k$-module isomorphism
\begin{align*}
&  \left(  V_{1}\otimes\left(  V_{2}\otimes V_{3}\otimes...\otimes
V_{i}\right)  \right)  \otimes\left(  V_{i+1}\otimes V_{i+2}\otimes...\otimes
V_{N}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \rightarrow V_{1}\otimes\left(  \left(  V_{2}\otimes
V_{3}\otimes...\otimes V_{i}\right)  \otimes\left(  V_{i+1}\otimes
V_{i+2}\otimes...\otimes V_{N}\right)  \right)  .
\end{align*}

\end{itemize}

Hence, the diagram%
\[
\xymatrixcolsep{4pc}\xymatrix{
V_1 \otimes\left( \left( V_2 \otimes V_3 \otimes... \otimes V_i \right
) \otimes\left( V_{i+1} \otimes V_{i+2} \otimes... \otimes V_N \right
) \right)     \ar[r]^-{\operatorname*{id}\otimes\rho^{\prime}}     &    V_1 \otimes\left( V_2 \otimes V_3 \otimes
... \otimes V_N \right)    \\
\left( V_1 \otimes\left( V_2 \otimes V_3 \otimes... \otimes V_i \right
) \right) \otimes\left( V_{i+1} \otimes V_{i+2} \otimes... \otimes
V_N \right)     \ar[ru]_{\rho} \ar[u]_-{\alpha_{U,V,W}}
}
\]
is commutative (because each of its arrows is the canonical $k$-module
isomorphism between the respective objects, and because it is known from
linear algebra that canonical isomorphisms between various bracketings of a
tensor product form a commutative diagram). In other words, $\rho=\left(
\operatorname*{id}\otimes\rho^{\prime}\right)  \circ\alpha_{U,V,W}$.
\end{verlong}

\begin{verlong}
Now, we know that the canonical $k$-module isomorphism
\[
\left(  V_{2}\otimes V_{3}\otimes...\otimes V_{i}\right)  \otimes\left(
V_{i+1}\otimes V_{i+2}\otimes...\otimes V_{N}\right)  \rightarrow V_{2}\otimes
V_{3}\otimes...\otimes V_{N}%
\]
is a $\mathfrak{g}$-module isomorphism (by (\ref{pf.lie(X).bracketing.ass}),
applied to the $\mathfrak{g}$-modules $V_{2}$, $V_{3}$, $...$, $V_{N}$ and the
number $i-1$ instead of the $\mathfrak{g}$-modules $V_{1}$, $V_{2}$, $...$,
$V_{N-1}$ and the number $i$). Since the canonical $k$-module isomorphism
\[
\left(  V_{2}\otimes V_{3}\otimes...\otimes V_{i}\right)  \otimes\left(
V_{i+1}\otimes V_{i+2}\otimes...\otimes V_{N}\right)  \rightarrow V_{2}\otimes
V_{3}\otimes...\otimes V_{N}%
\]
is $\rho^{\prime}$, we thus have shown that $\rho^{\prime}$ is a
$\mathfrak{g}$-module isomorphism. Combined with the fact that
$\operatorname*{id}:V_{1}\rightarrow V_{1}$ is a $\mathfrak{g}$-module
isomorphism, this yields that $\operatorname*{id}\otimes\rho^{\prime}$ is a
$\mathfrak{g}$-module homomorphism (by Proposition \ref{prop.lie(X)f(X)g},
applied to $V_{1}$, $V_{1}$, $\left(  V_{2}\otimes V_{3}\otimes...\otimes
V_{i}\right)  \otimes\left(  V_{i+1}\otimes V_{i+2}\otimes...\otimes
V_{N}\right)  $, $V_{2}\otimes V_{3}\otimes...\otimes V_{N}$,
$\operatorname*{id}$ and $\rho^{\prime}$ instead of $V$, $V^{\prime}$, $W$,
$W^{\prime}$, $f$ and $g$). On the other hand, $\operatorname*{id}\otimes
\rho^{\prime}$ is a $k$-module isomorphism (since both $\operatorname*{id}$
and $\rho^{\prime}$ are $k$-module isomorphisms). Thus, $\operatorname*{id}%
\otimes\rho^{\prime}$ is a $\mathfrak{g}$-module homomorphism and a $k$-module
isomorphism at the same time. This yields that $\operatorname*{id}\otimes
\rho^{\prime}$ is a $\mathfrak{g}$-module isomorphism (by an application of
Proposition \ref{prop.liemodiso}). Combined with the fact that $\alpha
_{U,V,W}$ is a $\mathfrak{g}$-module isomorphism, this yields that $\left(
\operatorname*{id}\otimes\rho^{\prime}\right)  \circ\alpha_{U,V,W}$ is a
$\mathfrak{g}$-module isomorphism. In other words, $\rho$ is a $\mathfrak{g}%
$-module isomorphism (since $\rho=\left(  \operatorname*{id}\otimes
\rho^{\prime}\right)  \circ\alpha_{U,V,W}$).
\end{verlong}

\begin{verlong}
Hence, in both Cases 1 and 2, we have shown that $\rho$ is a canonical
$\mathfrak{g}$-module isomorphism. So we now know that $\rho$ is a canonical
$\mathfrak{g}$-module isomorphism in every case.
\end{verlong}

\begin{verlong}
Since $\rho$ is the canonical $k$-module isomorphism $\left(  V_{1}\otimes
V_{2}\otimes...\otimes V_{i}\right)  \otimes\left(  V_{i+1}\otimes
V_{i+2}\otimes...\otimes V_{N}\right)  \rightarrow V_{1}\otimes V_{2}%
\otimes...\otimes V_{N}$, this means that the canonical $k$-module isomorphism
$\left(  V_{1}\otimes V_{2}\otimes...\otimes V_{i}\right)  \otimes\left(
V_{i+1}\otimes V_{i+2}\otimes...\otimes V_{N}\right)  \rightarrow V_{1}\otimes
V_{2}\otimes...\otimes V_{N}$ is a $\mathfrak{g}$-module isomorphism. Since we
have proven this for any $N$ arbitrary $\mathfrak{g}$-modules $V_{1}$, $V_{2}%
$, $...$, $V_{N}$ and every $i\in\left\{  0,1,...,N\right\}  $, we have thus
verified (\ref{pf.lie(X).bracketing.claim}). This completes the induction
step, and thus the induction proof of Proposition \ref{prop.lie(X).bracketing}
is complete.
\end{verlong}

\begin{vershort}
\begin{remark}
\label{rmk.lie(X)mult.replace.short}In Definition \ref{defs.lie(X)mult}, we
defined the $\mathfrak{g}$-module $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$
by induction over $n$. We can also describe its Lie action explicitly: Let $k$
be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie algebra. Let
$n\in\mathbb{N}$. Let $V_{1}$, $V_{2}$, $...$, $V_{n}$ be $n$ arbitrary
$\mathfrak{g}$-modules. Then, the $\mathfrak{g}$-module $V_{1}\otimes
V_{2}\otimes...\otimes V_{n}$ defined in Definition \ref{defs.lie(X)mult}
satisfies%
\[
a\rightharpoonup\left(  v_{1}\otimes v_{2}\otimes...\otimes v_{n}\right)
=\sum\limits_{i=1}^{n}\underbrace{v_{1}\otimes v_{2}\otimes...\otimes v_{i-1}%
}_{\substack{\text{tensor product of the}\\\text{first }i-1\text{ vectors
}v_{\ell}}}\otimes\left(  a\rightharpoonup v_{i}\right)  \otimes
\underbrace{v_{i+1}\otimes v_{i+2}\otimes...\otimes v_{n}}%
_{\substack{\text{tensor product of the}\\\text{last }n-i\text{ vectors
}v_{\ell}}}
\]
for every $a\in\mathfrak{g}$ and every $\left(  v_{1},v_{2},...,v_{n}\right)
\in V_{1}\times V_{2}\times...\times V_{n}$.
\end{remark}
\end{vershort}

\begin{verlong}
In Definition \ref{defs.lie(X)mult}, we defined the $\mathfrak{g}$-module
$V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ by induction over $n$. It turns
out that we can also easily describe the $\mathfrak{g}$-module structure on
$V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ explicitly. First a convention:

\begin{condition}
\label{conv.replace}Let $k$ be a commutative ring. Let $n\in\mathbb{N}$. Let
$V_{1}$, $V_{2}$, $...$, $V_{n}$ be $n$ arbitrary $k$-modules. Let
$i\in\left\{  1,2,...,n\right\}  $, and let $W_{i}$ be a $k$-module. Let
$\left(  v_{1},v_{2},...,v_{n}\right)  \in V_{1}\times V_{2}\times...\times
V_{n}$ and $w_{i}\in W_{i}$ be arbitrary. Then, we are going to denote by
\[
v_{1}\otimes v_{2}\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$w_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
}\otimes...\otimes v_{n}%
\]
the result of replacing the $i$-th tensorand (this tensorand is equal to
$v_{i}$) of the tensor product $v_{1}\otimes v_{2}\otimes...\otimes v_{n}$ by
the vector $w_{i}$. In other words, $v_{1}\otimes v_{2}\otimes...\otimes\ $%
\begin{tabular}
[c]{|c|}\hline
$w_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
$\otimes...\otimes v_{n}$ is defined by%
\[
v_{1}\otimes v_{2}\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$w_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
}\otimes...\otimes v_{n}=\underbrace{v_{1}\otimes v_{2}\otimes...\otimes
v_{i-1}}_{\substack{\text{tensor product of the}\\\text{first }i-1\text{
vectors }v_{\ell}}}\otimes w_{i}\otimes\underbrace{v_{i+1}\otimes
v_{i+2}\otimes...\otimes v_{n}}_{\substack{\text{tensor product of
the}\\\text{last }n-i\text{ vectors }v_{\ell}}}.
\]

\end{condition}

\begin{proposition}
\label{prop.lie(X)mult.exp}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie algebra. Let $n\in\mathbb{N}$. Let $V_{1}$, $V_{2}$, $...$,
$V_{n}$ be $n$ arbitrary $\mathfrak{g}$-modules. Then, the $\mathfrak{g}%
$-module $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ defined in Definition
\ref{defs.lie(X)mult} satisfies%
\[
a\rightharpoonup\left(  v_{1}\otimes v_{2}\otimes...\otimes v_{n}\right)
=\sum\limits_{i=1}^{n}v_{1}\otimes v_{2}\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup v_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
}\otimes...\otimes v_{n}%
\]
for every $a\in\mathfrak{g}$ and every $\left(  v_{1},v_{2},...,v_{n}\right)
\in V_{1}\times V_{2}\times...\times V_{n}$. Here, we are using Convention
\ref{conv.replace}.
\end{proposition}

We are going to prove this later. First an obvious lemma:

\begin{proposition}
\label{prop.lie(X)mult.left}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie algebra. Let $n\in\mathbb{N}_{+}$. Let $V_{1}$, $V_{2}$, $...$,
$V_{n}$ be $n$ arbitrary $\mathfrak{g}$-modules. Then, the $\mathfrak{g}%
$-module $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ defined in Definition
\ref{defs.lie(X)mult} satisfies%
\begin{equation}
a\rightharpoonup\left(  v\otimes T\right)  =\left(  a\rightharpoonup v\right)
\otimes T+v\otimes\left(  a\rightharpoonup T\right)
\label{prop.lie(X)mult.left.form}%
\end{equation}
for every $a\in\mathfrak{g}$, $v\in V_{1}$ and $T\in V_{2}\otimes V_{3}%
\otimes...\otimes V_{n}$.
\end{proposition}

\textit{Proof of Proposition \ref{prop.lie(X)mult.left}.} According to
Definition \ref{defs.lie(X)mult}, the $\mathfrak{g}$-module $V_{1}\otimes
V_{2}\otimes...\otimes V_{n}$ is defined as the tensor product $V_{1}%
\otimes\left(  V_{2}\otimes V_{3}\otimes...\otimes V_{n}\right)  $. Hence,
applying (\ref{prop.lie(X).form}) to $V=V_{1}$ and $W=V_{2}\otimes
V_{3}\otimes...\otimes V_{n}$, we see that%
\begin{align*}
a\rightharpoonup\left(  v\otimes w\right)   &  =\left(  a\rightharpoonup
v\right)  \otimes w+v\otimes\left(  a\rightharpoonup w\right) \\
&  \ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{, }v\in
V_{1}\text{ and }w\in V_{2}\otimes V_{3}\otimes...\otimes V_{n}%
\end{align*}
(because the tensor product $V\otimes W$ of two $\mathfrak{g}$-modules $V$ and
$W$ always satisfies (\ref{prop.lie(X).form})). If we rename $w$ as $T$, this
rewrites as follows:%
\begin{align*}
a\rightharpoonup\left(  v\otimes T\right)   &  =\left(  a\rightharpoonup
v\right)  \otimes T+v\otimes\left(  a\rightharpoonup T\right) \\
&  \ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{, }v\in
V_{1}\text{ and }T\in V_{2}\otimes V_{3}\otimes...\otimes V_{n}.
\end{align*}
This proves Proposition \ref{prop.lie(X)mult.left}.

\textit{Proof of Proposition \ref{prop.lie(X)mult.exp}.} Let $a\in
\mathfrak{g}$ and $\left(  v_{1},v_{2},...,v_{n}\right)  \in V_{1}\times
V_{2}\times...\times V_{n}$ be arbitrary. We are now going to show that%
\begin{equation}
a\rightharpoonup\left(  v_{n-p+1}\otimes v_{n-p+2}\otimes...\otimes
v_{n}\right)  =\sum\limits_{i=n-p+1}^{n}v_{n-p+1}\otimes v_{n-p+2}%
\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup v_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
\ }\otimes...\otimes v_{n} \label{pf.lie(X)mult.exp.1}%
\end{equation}
for every $p\in\left\{  0,1,...,n\right\}  $. (The equation
(\ref{pf.lie(X)mult.exp.1}) is to be understood as an equation between two
elements of $V_{n-p+1}\otimes V_{n-p+2}\otimes...\otimes V_{n}$.)

\textit{Proof of (\ref{pf.lie(X)mult.exp.1}).} In fact, we are going to prove
(\ref{pf.lie(X)mult.exp.1}) by induction over $p$:

\textit{Induction base:} For $p=0$, the equation (\ref{pf.lie(X)mult.exp.1})
holds{\footnotemark}\footnotetext{This is because for $p=0$, we have
\[
a\rightharpoonup\underbrace{\left(  v_{n-p+1}\otimes v_{n-p+2}\otimes
...\otimes v_{n}\right)  }_{=\left(  \text{empty tensor product}\right)
=1}=a\rightharpoonup1=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since the Lie action
of }k\text{ is }0\right)
\]
and%
\[
\sum\limits_{i=n-p+1}^{n}v_{n-p+1}\otimes v_{n-p+2}\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup v_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
\ }\otimes...\otimes v_{n}=\left(  \text{empty sum}\right)  =0.
\]
}. This completes the induction base.

\textit{Induction step:} Let $q\in\left\{  0,1,...,n-1\right\}  $ be
arbitrary. Assume that (\ref{pf.lie(X)mult.exp.1}) holds for $p=q$. Now we
must show that (\ref{pf.lie(X)mult.exp.1}) holds for $p=q+1$.

We have assumed that (\ref{pf.lie(X)mult.exp.1}) holds for $p=q$; in other
words,%
\begin{equation}
a\rightharpoonup\left(  v_{n-q+1}\otimes v_{n-q+2}\otimes...\otimes
v_{n}\right)  =\sum\limits_{i=n-q+1}^{n}v_{n-q+1}\otimes v_{n-q+2}%
\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup v_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
\ }\otimes...\otimes v_{n}. \label{pf.lie(X)mult.exp.2}%
\end{equation}


Applying Proposition \ref{prop.lie(X)mult.left} to the number $q+1$ and the
$\mathfrak{g}$-modules $V_{n-q}$, $V_{n-q+1}$, $...$, $V_{n}$ instead of the
number $n$ and the $\mathfrak{g}$-modules $V_{1}$, $V_{2}$, $...$, $V_{n}$, we
obtain
\[
a\rightharpoonup\left(  v\otimes T\right)  =\left(  a\rightharpoonup v\right)
\otimes T+v\otimes\left(  a\rightharpoonup T\right)
\]
for every $a\in\mathfrak{g}$, $v\in V_{n-q}$ and $T\in V_{n-q+1}\otimes
V_{n-q+2}\otimes...\otimes V_{n}$. Applying this to $v=v_{n-q}$ and
$T=v_{n-q+1}\otimes v_{n-q+2}\otimes...\otimes v_{n}$, we obtain%
\begin{align*}
&  a\rightharpoonup\left(  v_{n-q}\otimes v_{n-q+1}\otimes v_{n-q+2}%
\otimes...\otimes v_{n}\right) \\
=  &  \underbrace{\left(  a\rightharpoonup v_{n-q}\right)  \otimes
v_{n-q+1}\otimes v_{n-q+2}\otimes...\otimes v_{n}}_{=v_{n-q}\otimes
v_{n-q+1}\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup v_{n-q}$\\\hline
$v_{n-q}$\\\hline
\end{tabular}
\ }\otimes...\otimes v_{n}}+v_{n-q}\otimes\underbrace{\left(  a\rightharpoonup
\left(  v_{n-q+1}\otimes v_{n-q+2}\otimes...\otimes v_{n}\right)  \right)
}_{\substack{=\sum\limits_{i=n-q+1}^{n}v_{n-q+1}\otimes v_{n-q+2}%
\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup v_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
\ }\otimes...\otimes v_{n}\\\text{(due to (\ref{pf.lie(X)mult.exp.2}))}}}\\
=  &  v_{n-q}\otimes v_{n-q+1}\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup v_{n-q}$\\\hline
$v_{n-q}$\\\hline
\end{tabular}
}\otimes...\otimes v_{n}\\
&  \ \ \ \ \ \ \ \ \ \ +\underbrace{v_{n-q}\otimes\sum\limits_{i=n-q+1}%
^{n}v_{n-q+1}\otimes v_{n-q+2}\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup v_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
}\otimes...\otimes v_{n}}_{=\sum\limits_{i=n-q+1}^{n}v_{n-q}\otimes\left(
v_{n-q+1}\otimes v_{n-q+2}\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup v_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
}\otimes...\otimes v_{n}\right)  }\\
=  &  \underbrace{v_{n-q}\otimes v_{n-q+1}\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup v_{n-q}$\\\hline
$v_{n-q}$\\\hline
\end{tabular}
}\otimes...\otimes v_{n}}_{=\sum\limits_{i=n-q}^{n-q}v_{n-q}\otimes
v_{n-q+1}\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup v_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
}\otimes...\otimes v_{n}}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{i=n-q+1}^{n}\underbrace{v_{n-q}%
\otimes\left(  v_{n-q+1}\otimes v_{n-q+2}\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup v_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
}\otimes...\otimes v_{n}\right)  }_{=v_{n-q}\otimes v_{n-q+1}\otimes
...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup v_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
}\otimes...\otimes v_{n}}\\
=  &  \sum\limits_{i=n-q}^{n-q}v_{n-q}\otimes v_{n-q+1}\otimes...\otimes
\ \text{%
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup v_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
}\otimes...\otimes v_{n}\\
&  \ \ \ \ \ \ \ \ \ \ +\sum\limits_{i=n-q+1}^{n}v_{n-q}\otimes v_{n-q+1}%
\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup v_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
\ }\otimes...\otimes v_{n}\\
=  &  \sum\limits_{i=n-q}^{n}v_{n-q}\otimes v_{n-q+1}\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup v_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
\ }\otimes...\otimes v_{n}\\
=  &  \sum\limits_{i=n-\left(  q+1\right)  +1}^{n}v_{n-\left(  q+1\right)
+1}\otimes v_{n-\left(  q+1\right)  +2}\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup v_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
\ }\otimes...\otimes v_{n}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }n-q=n-\left(  q+1\right)  +1\text{
and }n-q+1=n-\left(  q+1\right)  +2\right)  .
\end{align*}
In other words, (\ref{pf.lie(X)mult.exp.1}) holds for $p=q+1$. This completes
the induction step. Thus, the induction proof of (\ref{pf.lie(X)mult.exp.1})
is complete.

Now that we have proven (\ref{pf.lie(X)mult.exp.1}) for every $p\in\left\{
0,1,...,n\right\}  $, we can apply (\ref{pf.lie(X)mult.exp.1}) to $p=n$, and
obtain%
\[
a\rightharpoonup\left(  v_{n-n+1}\otimes v_{n-n+2}\otimes...\otimes
v_{n}\right)  =\sum\limits_{i=n-n+1}^{n}v_{n-n+1}\otimes v_{n-n+2}%
\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup v_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
\ }\otimes...\otimes v_{n}.
\]
Since $n-n+1=1$ and $n-n+2=2$, this simplifies to
\[
a\rightharpoonup\left(  v_{1}\otimes v_{2}\otimes...\otimes v_{n}\right)
=\sum\limits_{i=1}^{n}v_{1}\otimes v_{2}\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup v_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
\ }\otimes...\otimes v_{n}.
\]
This proves Proposition \ref{prop.lie(X)mult.exp}.
\end{verlong}

Finally we record the multi-factor version of Proposition
\ref{prop.lie(X)f(X)g}:

\begin{proposition}
\label{prop.lie(X)mor}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra.\newline Let $n\in\mathbb{N}$. Let $V_{1}$, $V_{2}$, $...$,
$V_{n}$ be $n$ arbitrary $\mathfrak{g}$-modules. Let $V_{1}^{\prime}$,
$V_{2}^{\prime}$, $...$, $V_{n}^{\prime}$ be $n$ arbitrary $\mathfrak{g}%
$-modules. Let $f_{i}:V_{i}\rightarrow V_{i}^{\prime}$ be a $\mathfrak{g}%
$-module homomorphism for every $i\in\left\{  1,2,...,n\right\}  $. Then,
$f_{1}\otimes f_{2}\otimes...\otimes f_{n}:V_{1}\otimes V_{2}\otimes...\otimes
V_{n}\rightarrow V_{1}^{\prime}\otimes V_{2}^{\prime}\otimes...\otimes
V_{n}^{\prime}$ is a $\mathfrak{g}$-module homomorphism.
\end{proposition}

This proposition follows from Proposition \ref{prop.lie(X)f(X)g} by induction
(the details being left to the reader).

We also notice that Proposition \ref{prop.lie.(X)vsRes} has a multi-factor version:

\begin{proposition}
\label{prop.lie.(X)multvsRes}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$.
Let $n\in\mathbb{N}$.\newline If $V_{1}$, $V_{2}$, $...$, $V_{n}$ are $n$
arbitrary $\mathfrak{g}$-modules, then $\operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}\left(  V_{1}\otimes V_{2}\otimes...\otimes V_{n}\right)
=\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V_{1}\right)
\otimes\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V_{2}\right)
\otimes...\otimes\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}%
V_{n}\right)  $ as $\mathfrak{h}$-modules. This allows us to speak of "the
$\mathfrak{h}$-module $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$" without
having to worry whether we mean $\operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}\left(  V_{1}\otimes V_{2}\otimes...\otimes V_{n}\right)  $ or
$\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V_{1}\right)
\otimes\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V_{2}\right)
\otimes...\otimes\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}%
V_{n}\right)  $ (because it does not matter, since $\operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}\left(  V_{1}\otimes V_{2}\otimes...\otimes
V_{n}\right)  =\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}%
V_{1}\right)  \otimes\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}%
}V_{2}\right)  \otimes...\otimes\left(  \operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}V_{n}\right)  $).
\end{proposition}

\subsection{Tensor powers of $\mathfrak{g}$-modules}

Next we define a particular case of tensor products of $\mathfrak{g}$-modules,
namely the tensor powers. First we recall the classical definition of the
tensor powers of a $k$-module:

\begin{definition}
\label{defs.(X)^n}Let $k$ be a commutative ring. Let $n\in\mathbb{N}$. For any
$k$-module $V$, we define a $k$-module $V^{\otimes n}$ by $V^{\otimes
n}=\underbrace{V\otimes V\otimes...\otimes V}_{n\text{ times}}$. This
$k$-module $V^{\otimes n}$ is called the $n$\textit{-th tensor power} of the
$k$-module $V$.
\end{definition}

\begin{remark}
\label{rmk.(X)^0}Let $k$ be a commutative ring, and let $V$ be a $k$-module.
Then, $V^{\otimes0}=k$ (because $V^{\otimes n}=\underbrace{V\otimes
V\otimes...\otimes V}_{0\text{ times}}=\left(  \text{tensor product of zero
}k\text{-modules}\right)  =k$ according to the induction base of Definition
\ref{defs.(X)mult}) and $V^{\otimes1}=V\otimes k$ (because $V^{\otimes
1}=\underbrace{V\otimes V\otimes...\otimes V}_{1\text{ times}}=V\otimes k$
according to the induction step of Definition \ref{defs.(X)mult}). Since we
identify $V\otimes k$ with $V$, we thus have $V^{\otimes1}=V$.
\end{remark}

Now, by analogy, we define the tensor powers of a $\mathfrak{g}$-module:

\begin{definition}
\label{defs.lie(X)^n}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $n\in\mathbb{N}$. For any $\mathfrak{g}$-module $V$, we
define a $\mathfrak{g}$-module $V^{\otimes n}$ by $V^{\otimes n}%
=\underbrace{V\otimes V\otimes...\otimes V}_{n\text{ times}}$. This
$\mathfrak{g}$-module $V^{\otimes n}$ is called the $n$\textit{-th tensor
power} of the $\mathfrak{g}$-module $V$.
\end{definition}

\begin{remark}
\label{rmk.lie(X)^0}Let $k$ be a commutative ring, let $\mathfrak{g}$ be a
$k$-Lie algebra, and let $V$ be a $\mathfrak{g}$-module. Then, $V^{\otimes
0}=k$ (as $\mathfrak{g}$-modules) and $V^{\otimes1}=V$ (as $\mathfrak{g}%
$-modules), where we identify the $\mathfrak{g}$-module $V\otimes k$ with $V$.
This is proven the same way as Remark \ref{rmk.(X)^0}.
\end{remark}

As a consequence of Proposition \ref{prop.lie(X)mor}, we now have:

\begin{proposition}
\label{prop.lie(X)^nmor}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $n\in\mathbb{N}$. Let $V$ and $V^{\prime}$ be
$\mathfrak{g}$-modules, and let $f:V\rightarrow V^{\prime}$ be a
$\mathfrak{g}$-module homomorphism. Then, $f^{\otimes n}:V^{\otimes
n}\rightarrow V^{\prime\otimes n}$ is a $\mathfrak{g}$-module homomorphism.
\end{proposition}

Here we are using the following convention:

\begin{condition}
\label{conv.f^(X)n}Let $k$ be a commutative ring. Let $n\in\mathbb{N}$. Let
$V$ and $V^{\prime}$ be $k$-modules, and let $f:V\rightarrow V^{\prime}$ be a
$k$-module homomorphism. Then, $f^{\otimes n}$ denotes the $k$-module
homomorphism $\underbrace{f\otimes f\otimes...\otimes f}_{n\text{ times}%
}:\underbrace{V\otimes V\otimes...\otimes V}_{n\text{ times}}\rightarrow
\underbrace{V^{\prime}\otimes V^{\prime}\otimes...\otimes V^{\prime}}_{n\text{
times}}$. Since $\underbrace{V\otimes V\otimes...\otimes V}_{n\text{ times}%
}=V^{\otimes n}$ and $\underbrace{V^{\prime}\otimes V^{\prime}\otimes
...\otimes V^{\prime}}_{n\text{ times}}=V^{\prime\otimes n}$, this $f^{\otimes
n}$ is thus a $k$-module homomorphism from $V^{\otimes n}$ to $V^{\prime
\otimes n}$.
\end{condition}

We notice the following:

\begin{proposition}
\label{prop.lie(X)^n.bracketing}Let $k$ be a commutative ring. Let
$\mathfrak{g}$ be a $k$-Lie algebra. Let $n\in\mathbb{N}$.\newline Then, for
any $\mathfrak{g}$-module $V$ and every $i\in\left\{  0,1,...,n\right\}  $,
the canonical $k$-module isomorphism $V^{\otimes i}\otimes V^{\otimes\left(
n-i\right)  }\rightarrow V^{\otimes n}$ is a $\mathfrak{g}$-module
isomorphism.{\footnotemark}
\end{proposition}

\footnotetext{Here and in the following, whenever I speak of "the canonical
$k$-module isomorphism $V^{\otimes i}\otimes V^{\otimes\left(  n-i\right)
}\rightarrow V^{\otimes n}$", I mean the $k$-module homomorphism $V^{\otimes
i}\otimes V^{\otimes\left(  n-i\right)  }\rightarrow V^{\otimes n}$ which
sends $\left(  v_{1}\otimes v_{2}\otimes...\otimes v_{i}\right)
\otimes\left(  v_{i+1}\otimes v_{i+2}\otimes...\otimes v_{n}\right)  $ to
$v_{1}\otimes v_{2}\otimes...\otimes v_{n}$ for every $\left(  v_{1}%
,v_{2},...,v_{n}\right)  \in V^{n}$. This homomorphism is known (from linear
algebra) to exist, be unique and be a $k$-module isomorphism. It is actually
the canonical $k$-module isomorphism $\underbrace{V\otimes V\otimes...\otimes
V}_{i\text{ times}}\otimes\underbrace{V\otimes V\otimes...\otimes
V}_{n-i\text{ times}}\rightarrow\underbrace{V\otimes V\otimes...\otimes
V}_{n\text{ times}}$.}

This proposition follows directly from applying Proposition
\ref{prop.lie(X).bracketing} to $V$, $V$, $...$, $V$ instead of $V_{1}$,
$V_{2}$, $...$, $V_{n}$.

\begin{condition}
\label{conv.lie(X)^n.ident}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie algebra. For every $\mathfrak{g}$-module $V$, every
$n\in\mathbb{N}$ and every $i\in\left\{  0,1,...,n\right\}  $, we are going to
identify the $\mathfrak{g}$-module $V^{\otimes i}\otimes V^{\otimes\left(
n-i\right)  }$ with the $\mathfrak{g}$-module $V^{\otimes n}$ (this is allowed
because of Proposition \ref{prop.lie(X)^n.bracketing}). In other words, for
every $\mathfrak{g}$-module $V$, every $a\in\mathbb{N}$ and every
$b\in\mathbb{N}$, we are going to identify the $\mathfrak{g}$-module
$V^{\otimes a}\otimes V^{\otimes b}$ with the $\mathfrak{g}$-module
$V^{\otimes\left(  a+b\right)  }$.
\end{condition}

Actually, there is no reason to restrict ourselves to $\mathfrak{g}$-modules
in this Convention \ref{conv.lie(X)^n.ident}; we can do the convention same
for $k$-modules:

\begin{condition}
\label{conv.(X)^n.ident}Let $k$ be a commutative ring. For every $k$-module
$V$, every $n\in\mathbb{N}$ and every $i\in\left\{  0,1,...,n\right\}  $, we
are going to identify the $k$-module $V^{\otimes i}\otimes V^{\otimes\left(
n-i\right)  }$ with the $k$-module $V^{\otimes n}$ (this is allowed because of
the canonical isomorphism $V^{\otimes i}\otimes V^{\otimes\left(  n-i\right)
}\cong V^{\otimes n}$). In other words, for every $k$-module $V$, every
$a\in\mathbb{N}$ and every $b\in\mathbb{N}$, we are going to identify the
$k$-module $V^{\otimes a}\otimes V^{\otimes b}$ with the $k$-module
$V^{\otimes\left(  a+b\right)  }$.
\end{condition}

Needless to say, Convention \ref{conv.(X)^n.ident} does not conflict with
Convention \ref{conv.lie(X)^n.ident}, because if $V$ is a $\mathfrak{g}%
$-module, then both of these conventions identify $V^{\otimes i}\otimes
V^{\otimes\left(  n-i\right)  }$ with $V^{\otimes n}$ by means of the same
isomorphism $V^{\otimes i}\otimes V^{\otimes\left(  n-i\right)  }\cong
V^{\otimes n}$.

Finally, Proposition \ref{prop.lie.(X)multvsRes} yields:

\begin{proposition}
\label{prop.lie.(X)nvsRes}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let
$n\in\mathbb{N}$.\newline If $V$ is any $\mathfrak{g}$-module, then
$\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}\left(  V^{\otimes n}\right)
=\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V\right)  ^{\otimes
n}$ as $\mathfrak{h}$-modules. This allows us to speak of "the $\mathfrak{h}%
$-module $V^{\otimes n}$" without having to worry whether we mean
$\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}\left(  V^{\otimes n}\right)
$ or $\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V\right)
^{\otimes n}$ (because it does not matter, since $\operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}\left(  V^{\otimes n}\right)  =\left(
\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V\right)  ^{\otimes n}$).
\end{proposition}

\subsection{Tensor algebra and universal enveloping algebra}

The tensor powers $V^{\otimes n}$ of a $k$-module $V$ can be combined to a
$k$-module $\otimes V$ which turns out to have an algebra structure: that of
the so-called tensor algebra. Let us recall its definition (which can easily
shown to be well-defined):

\begin{definition}
\label{defs.tensor}Let $k$ be a commutative ring.\newline\textbf{(a)} Let $V$
be a $k$-module. The \textit{tensor algebra} $\otimes V$ of $V$ over $k$ is
defined to be the $k$-algebra formed by the $k$-module $\bigoplus
\limits_{i\in\mathbb{N}}V^{\otimes i}=V^{\otimes0}\oplus V^{\otimes1}\oplus
V^{\otimes2}\oplus...$ equipped with a multiplication which is defined by%
\begin{equation}
\left(  a_{i}\right)  _{i\in\mathbb{N}}\cdot\left(  b_{i}\right)
_{i\in\mathbb{N}}=\left(  \sum_{i=0}^{n}a_{i}\otimes b_{n-i}\right)
_{n\in\mathbb{N}}\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  a_{i}\right)
_{i\in\mathbb{N}}\in\bigoplus\limits_{i\in\mathbb{N}}V^{\otimes i}\text{ and
}\left(  b_{i}\right)  _{i\in\mathbb{N}}\in\bigoplus\limits_{i\in\mathbb{N}%
}V^{\otimes i} \label{defs.tensor.mult}%
\end{equation}
(where for every $n\in\mathbb{N}$ and every $i\in\left\{  0,1,...,n\right\}
$, the tensor $a_{i}\otimes b_{n-i}\in V^{\otimes i}\otimes V^{\otimes\left(
n-i\right)  }$ is considered as an element of $V^{\otimes n}$ due to the
canonical identification $V^{\otimes i}\otimes V^{\otimes\left(  n-i\right)
}\cong V^{\otimes n}$ which was defined in Convention \ref{conv.(X)^n.ident}%
).\newline The $k$-module $\otimes V$ itself (without the $k$-algebra
structure) is called the \textit{tensor module} of $V$.\newline\textbf{(b)}
Let $V$ and $W$ be two $k$-modules, and let $f:V\rightarrow W$ be a $k$-module
homomorphism. The $k$-module homomorphisms $f^{\otimes i}:V^{\otimes
i}\rightarrow W^{\otimes i}$ for all $i\in\mathbb{N}$ can be combined together
to a $k$-module homomorphism from $V^{\otimes0}\oplus V^{\otimes1}\oplus
V^{\otimes2}\oplus...$ to $W^{\otimes0}\oplus W^{\otimes1}\oplus W^{\otimes
2}\oplus...$. This homomorphism is called $\otimes f$. Since $V^{\otimes
0}\oplus V^{\otimes1}\oplus V^{\otimes2}\oplus...=\otimes V$ and $W^{\otimes
0}\oplus W^{\otimes1}\oplus W^{\otimes2}\oplus...=\otimes W$, we see that this
homomorphism $\otimes f$ is a $k$-module homomorphism from $\otimes V$ to
$\otimes W$. Moreover, it follows easily from (\ref{defs.tensor.mult}) that
this $\otimes f$ is actually a $k$-algebra homomorphism from $\otimes V$ to
$\otimes W$.\newline\textbf{(c)} Let $V$ be a $k$-module. Then, according to
Convention \ref{conv.(++).inj}, we consider $V^{\otimes n}$ as a $k$-submodule
of the direct sum $\bigoplus\limits_{i\in\mathbb{N}}V^{\otimes i}=\otimes V$
for every $n\in\mathbb{N}$. In particular, every element of $k$ is considered
to be an element of $\otimes V$ by means of the canonical embedding
$k=V^{\otimes0}\subseteq\otimes V$, and every element of $V$ is considered to
be an element of $\otimes V$ by means of the canonical embedding
$V=V^{\otimes1}\subseteq\otimes V$. The element $1\in k\subseteq\otimes V$ is
easily seen to be the unity of the tensor algebra $\otimes V$.
\end{definition}

\begin{remark}
The formula (\ref{defs.tensor.mult}) (which defines the multiplication on the
tensor algebra $\otimes V$) is often put in words by saying that "the
multiplication in the tensor algebra $\otimes V$ is given by the tensor
product". This informal statement tempts many authors (including myself in
\cite{dg1}) to use the sign $\otimes$ for multiplication in the algebra
$\otimes V$, that is, to write $u\otimes v$ for the product of any two
elements $u$ and $v$ of the tensor algebra $\otimes V$. This notation,
however, can collide with the notation $u\otimes v$ for the tensor product of
two vectors $u$ and $v$ in a $k$-module.{\footnotemark} Due to this
possibility of collision, we are \textit{not} going to use the sign $\otimes$
for multiplication in the algebra $\otimes V$ in this paper. Instead we will
use the sign $\cdot$ for this multiplication. However, due to
(\ref{defs.tensor.mult}), we still have%
\begin{equation}
\left(  a\cdot b=a\otimes b\ \ \ \ \ \ \ \ \ \ \text{for any }n\in
\mathbb{N}\text{, any }m\in\mathbb{N}\text{, any }a\in V^{\otimes n}\text{ and
any }b\in V^{\otimes m}\right)  , \label{*=(X)}%
\end{equation}
where $a\otimes b$ is considered to be an element of $V^{\otimes\left(
n+m\right)  }$ by means of the identification of $V^{\otimes n}\otimes
V^{\otimes m}$ with $V^{\otimes\left(  n+m\right)  }$.
\end{remark}

\footnotetext{For example, if $z$ is a vector in the $k$-module $V$, then we
can define two elements $u$ and $v$ of $\otimes V$ by $u=1+z$ and $v=1-z$
(where $1$ and $z$ are considered to be elements of $\otimes V$ according to
Definition \ref{defs.tensor} \textbf{(c)}), and while the product of these
elements $u$ and $v$ in $\otimes V$ is the element $\left(  1+z\right)
\cdot\left(  1-z\right)  =1\cdot1-1\cdot z+1\cdot z-z\otimes z=1-z\otimes
z\in\otimes V$, the tensor product of these elements $u$ and $v$ is the
element $\left(  1+z\right)  \otimes\left(  1-z\right)  $ of $\left(  k\oplus
V\right)  \otimes\left(  k\oplus V\right)  \cong k\oplus V\oplus
V\oplus\left(  V\otimes V\right)  $, which is a different element of a totally
different $k$-module. So if we would use one and the same notation $u\otimes
v$ for both the product of $u$ and $v$ in $\otimes V$ and the tensor product
of $u$ and $v$ in $\left(  k\oplus V\right)  \otimes\left(  k\oplus V\right)
$, we would have ambiguous notations.}

Next we define the universal enveloping algebra of a Lie algebra
$\mathfrak{g}$. First, a convention about the notation we are using:

\begin{condition}
\label{conv.A*B}Let $k$ be a commutative ring. Let $U$ be a $k$%
-algebra.\newline\textbf{(a)} If $A$ and $B$ are two $k$-submodules of $U$,
then we denote by $A\cdot B$ the $k$-submodule $\left\langle ab\ \mid\ \left(
a,b\right)  \in A\times B\right\rangle $ of $U$. We also will sometimes
abbreviate $A\cdot B$ by $AB$.\newline\textbf{(b)} If $A$, $B$ and $C$ are
three $k$-submodules of $U$, then it is easily seen that $\left(  AB\right)
C=A\left(  BC\right)  $, so that we can denote each of the two equal
$k$-submodules $\left(  AB\right)  C$ and $A\left(  BC\right)  $ by $ABC$.
\end{condition}

\begin{definition}
\label{defs.U}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie
algebra. We define the \textit{universal enveloping algebra} $U\left(
\mathfrak{g}\right)  $ to be the factor algebra $\left(  \otimes
\mathfrak{g}\right)  \diagup I_{\mathfrak{g}}$, where $I_{\mathfrak{g}}$ is
the two-sided ideal%
\[
\left(  \otimes\mathfrak{g}\right)  \cdot\left\langle v\otimes w-w\otimes
v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}%
\times\mathfrak{g}\right\rangle \cdot\left(  \otimes\mathfrak{g}\right)
\]
of the algebra $\otimes\mathfrak{g}$.
\end{definition}

\begin{remark}
In Definition \ref{defs.U}, the term $\left\langle v\otimes w-w\otimes
v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}%
\times\mathfrak{g}\right\rangle $ is to be understood according to Convention
\ref{conv.<>}, and the multiplication sign $\cdot$ in
\[
\left(  \otimes\mathfrak{g}\right)  \cdot\left\langle v\otimes w-w\otimes
v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}%
\times\mathfrak{g}\right\rangle \cdot\left(  \otimes\mathfrak{g}\right)
\]
is to be understood according to Convention \ref{conv.A*B}. We note that,
although the multiplication in $\otimes\mathfrak{g}$ is related to the tensor
product by (\ref{*=(X)}), the product $\left(  \otimes\mathfrak{g}\right)
\cdot\left\langle v\otimes w-w\otimes v-\left[  v,w\right]  \ \mid\ \left(
v,w\right)  \in\mathfrak{g}\times\mathfrak{g}\right\rangle \cdot\left(
\otimes\mathfrak{g}\right)  $ has nothing to do with the tensor product
$\left(  \otimes\mathfrak{g}\right)  \otimes\left\langle v\otimes w-w\otimes
v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}%
\times\mathfrak{g}\right\rangle \otimes\left(  \otimes\mathfrak{g}\right)  $ !
\end{remark}

But we are not going to linger over universal enveloping algebras now. Let us
introduce the canonical $\mathfrak{g}$-module structure on $\otimes V$ for any
$\mathfrak{g}$-module $V$:

\begin{definition}
\label{defs.lietensor}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra.\newline Let $V$ be a $\mathfrak{g}$-module. Since $V^{\otimes
i}$ is a $\mathfrak{g}$-module for all $i\in\mathbb{N}$ (by Definition
\ref{defs.lie(X)^n}), the direct sum $\bigoplus\limits_{i\in\mathbb{N}%
}V^{\otimes i}$ is also a $\mathfrak{g}$-module (by Definition
\ref{defs.lie(++)}). In other words, the tensor algebra $\otimes V$ thus
becomes a $\mathfrak{g}$-module (since $\otimes V=\bigoplus\limits_{i\in
\mathbb{N}}V^{\otimes i}$). This $\mathfrak{g}$-module $\otimes V$ is called
the \textit{tensor }$\mathfrak{g}$\textit{-module} of the $\mathfrak{g}%
$-module $V$.\newline Whenever we will speak of the $\mathfrak{g}$-module
$\otimes V$, we will be meaning this $\mathfrak{g}$-module (although there
might be many different $\mathfrak{g}$-module structures on the $k$-module
$\otimes V$).
\end{definition}

\begin{vershort}
\begin{remark}
\label{rmk.g-mod-U(g).short}We will often consider the $\mathfrak{g}$-module
$\otimes\mathfrak{g}$. This is the tensor $\mathfrak{g}$-module of the
$\mathfrak{g}$-module $\mathfrak{g}$ itself. In other words, it is the result
of applying Definition \ref{defs.lietensor} to $V=\mathfrak{g}$. It is easily
seen to satisfy%
\begin{align*}
&  a\rightharpoonup\left(  v_{1}\otimes v_{2}\otimes...\otimes v_{n}\right)
=\sum\limits_{i=1}^{n}\underbrace{v_{1}\otimes v_{2}\otimes...\otimes v_{i-1}%
}_{\substack{\text{tensor product of the}\\\text{first }i-1\text{ vectors
}v_{\ell}}}\otimes\left[  a,v_{i}\right]  \otimes\underbrace{v_{i+1}\otimes
v_{i+2}\otimes...\otimes v_{n}}_{\substack{\text{tensor product of
the}\\\text{last }n-i\text{ vectors }v_{\ell}}}\\
&  \ \ \ \ \ \ \ \ \ \ \text{for all }n\in\mathbb{N}\text{, }a\in
\mathfrak{g}\text{ and }\left(  v_{1},v_{2},...,v_{n}\right)  \in
\mathfrak{g}^{n}.
\end{align*}
We can show that the ideal $I_{\mathfrak{g}}$ constructed in Definition
\ref{defs.U} is a $\mathfrak{g}$-submodule of this $\mathfrak{g}$-module
$\otimes\mathfrak{g}$. (In fact, this follows from Proposition \ref{prop.l25a}
\textbf{(a)}, applied to $\mathfrak{h}=\mathfrak{g}$.) Thus, $\left(
\otimes\mathfrak{g}\right)  \diagup I_{\mathfrak{g}}$ becomes a $\mathfrak{g}%
$-module. Since $\left(  \otimes\mathfrak{g}\right)  \diagup I_{\mathfrak{g}%
}=U\left(  \mathfrak{g}\right)  $, this yields that $U\left(  \mathfrak{g}%
\right)  $ becomes a $\mathfrak{g}$-module. This $\mathfrak{g}$-module
satisfies%
\begin{align*}
&  a\rightharpoonup\overline{\left(  v_{1}\otimes v_{2}\otimes...\otimes
v_{n}\right)  }=\sum\limits_{i=1}^{n}\overline{\underbrace{v_{1}\otimes
v_{2}\otimes...\otimes v_{i-1}}_{\substack{\text{tensor product of
the}\\\text{first }i-1\text{ vectors }v_{\ell}}}\otimes\left[  a,v_{i}\right]
\otimes\underbrace{v_{i+1}\otimes v_{i+2}\otimes...\otimes v_{n}%
}_{\substack{\text{tensor product of the}\\\text{last }n-i\text{ vectors
}v_{\ell}}}}\\
&  \ \ \ \ \ \ \ \ \ \ \text{for all }n\in\mathbb{N}\text{, }a\in
\mathfrak{g}\text{ and }\left(  v_{1},v_{2},...,v_{n}\right)  \in
\mathfrak{g}^{n}%
\end{align*}
(where $\overline{T}$ means the residue class of $T$ modulo $I_{\mathfrak{g}}$
for every $T\in\otimes\mathfrak{g}$). However, this $\mathfrak{g}$-module
structure is not the only interesting $\mathfrak{g}$-module structure on
$U\left(  \mathfrak{g}\right)  $. There is a different one, which satisfies%
\begin{align}
&  a\rightharpoonup\overline{\left(  v_{1}\otimes v_{2}\otimes...\otimes
v_{n}\right)  }=\overline{a\otimes v_{1}\otimes v_{2}\otimes...\otimes v_{n}%
}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \text{for all }n\in\mathbb{N}\text{, }a\in
\mathfrak{g}\text{ and }\left(  v_{1},v_{2},...,v_{n}\right)  \in
\mathfrak{g}^{n}.\label{rmk.g-mod-U(g).short.bad}%
\end{align}
It is important not to confuse these two $\mathfrak{g}$-module structures, as
they define two \textit{different} $\mathfrak{g}$-modules $U\left(
\mathfrak{g}\right)  $ (although, as $k$-modules, they are the same)!\newline%
\textbf{Whenever we will speak of the }$\mathfrak{g}$\textbf{-module
}$U\left(  \mathfrak{g}\right)  $ \textbf{in the following, we will mean the
}$\mathfrak{g}$\textbf{-module structure which is obtained by applying
Definition \ref{defs.lietensor} to }$V=\mathfrak{g}$ \textbf{and setting
}$U\left(  \mathfrak{g}\right)  =\left(  \otimes\mathfrak{g}\right)  \diagup
I_{\mathfrak{g}}$\textbf{, not the }$\mathfrak{g}$\textbf{-module structure
given by (\ref{rmk.g-mod-U(g).short.bad})!}
\end{remark}
\end{vershort}

\begin{verlong}
\begin{remark}
\label{rmk.g-mod-U(g).long}We will often consider the $\mathfrak{g}$-module
$\otimes\mathfrak{g}$. This is the tensor $\mathfrak{g}$-module of the
$\mathfrak{g}$-module $\mathfrak{g}$ itself. In other words, it is the result
of applying Definition \ref{defs.lietensor} to $V=\mathfrak{g}$. A look at
Proposition \ref{prop.lie(X)mult.exp} quickly yields that it satisfies%
\begin{align*}
&  a\rightharpoonup\left(  v_{1}\otimes v_{2}\otimes...\otimes v_{n}\right)
=\sum\limits_{i=1}^{n}v_{1}\otimes v_{2}\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$\left[  a,v_{i}\right]  $\\\hline
$v_{i}$\\\hline
\end{tabular}
\ }\otimes...\otimes v_{n}\\
&  \ \ \ \ \ \ \ \ \ \ \text{for all }n\in\mathbb{N}\text{, }a\in
\mathfrak{g}\text{ and }\left(  v_{1},v_{2},...,v_{n}\right)  \in
\mathfrak{g}^{n}.
\end{align*}
We can show that the ideal $I_{\mathfrak{g}}$ constructed in Definition
\ref{defs.U} is a $\mathfrak{g}$-submodule of this $\mathfrak{g}$-module
$\otimes\mathfrak{g}$. (In fact, this follows from Proposition \ref{prop.l25a}
\textbf{(a)}, applied to $\mathfrak{h}=\mathfrak{g}$.) Thus, $\left(
\otimes\mathfrak{g}\right)  \diagup I_{\mathfrak{g}}$ becomes a $\mathfrak{g}%
$-module. Since $\left(  \otimes\mathfrak{g}\right)  \diagup I_{\mathfrak{g}%
}=U\left(  \mathfrak{g}\right)  $, this yields that $U\left(  \mathfrak{g}%
\right)  $ becomes a $\mathfrak{g}$-module. This $\mathfrak{g}$-module
satisfies%
\begin{align*}
&  a\rightharpoonup\overline{\left(  v_{1}\otimes v_{2}\otimes...\otimes
v_{n}\right)  }=\sum\limits_{i=1}^{n}\overline{v_{1}\otimes v_{2}%
\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$\left[  a,v_{i}\right]  $\\\hline
$v_{i}$\\\hline
\end{tabular}
\ }\otimes...\otimes v_{n}}\\
&  \ \ \ \ \ \ \ \ \ \ \text{for all }n\in\mathbb{N}\text{, }a\in
\mathfrak{g}\text{ and }\left(  v_{1},v_{2},...,v_{n}\right)  \in
\mathfrak{g}^{n}%
\end{align*}
(where $\overline{T}$ means the residue class of $T$ modulo $I_{\mathfrak{g}}$
for every $T\in\otimes\mathfrak{g}$). However, this $\mathfrak{g}$-module
structure is not the only interesting $\mathfrak{g}$-module structure on
$U\left(  \mathfrak{g}\right)  $. There is a different one, which satisfies%
\begin{align}
&  a\rightharpoonup\overline{\left(  v_{1}\otimes v_{2}\otimes...\otimes
v_{n}\right)  }=\overline{a\otimes v_{1}\otimes v_{2}\otimes...\otimes v_{n}%
}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \text{for all }n\in\mathbb{N}\text{, }a\in
\mathfrak{g}\text{ and }\left(  v_{1},v_{2},...,v_{n}\right)  \in
\mathfrak{g}^{n}. \label{rmk.g-mod-U(g).long.bad}%
\end{align}
It is important not to confuse these two $\mathfrak{g}$-module structures, as
they define two \textit{different} $\mathfrak{g}$-modules $U\left(
\mathfrak{g}\right)  $ (although, as $k$-modules, they are the same)!\newline%
\textbf{Whenever we will speak of the }$\mathfrak{g}$\textbf{-module
}$U\left(  \mathfrak{g}\right)  $ \textbf{in the following, we will mean the
}$\mathfrak{g}$\textbf{-module structure which is obtained by applying
Definition \ref{defs.lietensor} to }$V=\mathfrak{g}$ \textbf{and setting
}$U\left(  \mathfrak{g}\right)  =\left(  \otimes\mathfrak{g}\right)  \diagup
I_{\mathfrak{g}}$\textbf{, not the }$\mathfrak{g}$\textbf{-module structure
given by (\ref{rmk.g-mod-U(g).long.bad})!}
\end{remark}
\end{verlong}

It is very easy to see that:

\begin{proposition}
\label{prop.(X)-hom-is-hom}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie algebra. Let $V$ and $W$ be two $\mathfrak{g}$-modules, and let
$f:V\rightarrow W$ be a $\mathfrak{g}$-module homomorphism. Then, $\otimes
f:\otimes V\rightarrow\otimes W$ is a $\mathfrak{g}$-module homomorphism.
\end{proposition}

\begin{definition}
\label{defs.lie.(X).inj}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $V$ be a $\mathfrak{g}$-module. Then, according to
Proposition \ref{prop.lie(++).inj}, we consider $V^{\otimes n}$ as a
$\mathfrak{g}$-submodule of the direct sum $\bigoplus\limits_{i\in\mathbb{N}%
}V^{\otimes i}=\otimes V$ for every $n\in\mathbb{N}$. In particular,
$k=V^{\otimes0}$ and $V=V^{\otimes1}$ become $\mathfrak{g}$-submodules of
$\otimes V$ this way.
\end{definition}

\subsection{$\mathfrak{g}$-algebras}

The next notion we introduce is that of a $\mathfrak{g}$\textit{-algebra}
(also called a $\mathfrak{g}$\textit{-module algebra} or $\mathfrak{g}%
$\textit{-Lie module algebra}):

\begin{definition}
\label{defs.g-alg}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. A $\mathfrak{g}$\textit{-algebra} will mean a $k$-algebra $A$
equipped with a $\mathfrak{g}$-module structure such that%
\begin{equation}
\left(  a\rightharpoonup\left(  uv\right)  =\left(  a\rightharpoonup u\right)
\cdot v+u\cdot\left(  a\rightharpoonup v\right)  \text{ for every }%
a\in\mathfrak{g}\text{, }u\in A\text{ and }v\in A\right)  .
\label{defs.g-alg.cond}%
\end{equation}

\end{definition}

\begin{remark}
In Definition \ref{defs.g-alg}, when we speak of "a $k$-algebra $A$ equipped
with a $\mathfrak{g}$-module structure", the words "a $\mathfrak{g}$-module
structure" mean "a $\mathfrak{g}$-module structure on the underlying
$k$-module of the $k$-algebra $A\ \ \ \ $". This $\mathfrak{g}$-module
structure must therefore be $k$-bilinear with respect to the underlying
$k$-module structure of the $k$-algebra $A$.
\end{remark}

\begin{remark}
Definition \ref{defs.g-alg} is often rewritten as follows:

\begin{definition}
\label{defs.g-alg.der}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. A $\mathfrak{g}$\textit{-algebra} will mean a $k$-algebra $A$
equipped with a $\mathfrak{g}$-module structure such that $\mathfrak{g}$ acts
on $A$ by means of derivations. Here, we say that "$\mathfrak{g}$ acts on $A$
by means of derivations" if and only if the map $A\rightarrow A,$
$u\mapsto\left(  a\rightharpoonup u\right)  $ is a derivation for every
$a\in\mathfrak{g}$.
\end{definition}

This Definition \ref{defs.g-alg.der} is indeed equivalent to Definition
\ref{defs.g-alg} because the condition that $\mathfrak{g}$ acts on $A$ by
means of derivations is equivalent to (\ref{defs.g-alg.cond}) (as can be
easily seen).
\end{remark}

It is easy to see that, when $\mathfrak{h}$ is a Lie subalgebra of a $k$-Lie
algebra $\mathfrak{g}$, every $\mathfrak{g}$-algebra canonically can be made
into an $\mathfrak{h}$-algebra:

\begin{definition}
\label{defs.g-as-h-algebra}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie algebra, and let $\mathfrak{h}$ be a Lie subalgebra of
$\mathfrak{g}$. Then, every $\mathfrak{g}$-algebra $A$ canonically becomes an
$\mathfrak{h}$-algebra. (In fact, the $\mathfrak{g}$-module $A$ canonically
becomes an $\mathfrak{h}$-module according to Definition
\ref{defs.g-as-h-module}, and thus $A$ is a $k$-algebra equipped with an
$\mathfrak{h}$-module structure which satisfies (\ref{defs.g-alg.cond}) with
$\mathfrak{g}$ replaced by $\mathfrak{h}$, so that $A$ thus is an
$\mathfrak{h}$-algebra.). This $\mathfrak{h}$-algebra is called the
\textit{restriction of }$A$ \textit{to }$\mathfrak{h}$, and denoted by
$\operatorname*{Res}\nolimits_{\mathfrak{h}}^{\mathfrak{g}}A$. However, when
there is no possibility of confusion, we will denote this $\mathfrak{h}%
$-algebra by $A$, and we will distinguish it from the original $\mathfrak{g}%
$-algebra $A$ by means of referring to the former one as "the $\mathfrak{h}%
$-algebra $A$" and referring to the latter one as "the $\mathfrak{g}$-algebra
$A$".
\end{definition}

A basic property of $\mathfrak{g}$-algebras:

\begin{proposition}
\label{prop.g-alg.PQ}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $A$ be a $\mathfrak{g}$-algebra. Let $P$ and $Q$ be two
$\mathfrak{g}$-submodules of $A$. Then, $P\cdot Q$ is a $\mathfrak{g}%
$-submodule of $A$.
\end{proposition}

\begin{remark}
Here, $P\cdot Q$ is to be understood as according to Convention \ref{conv.A*B}
\textbf{(a)}.
\end{remark}

\textit{Proof of Proposition \ref{prop.g-alg.PQ}.} According to Convention
\ref{conv.A*B} \textbf{(a)}, the $k$-submodule $P\cdot Q$ of $A$ is defined by
$P\cdot Q=\left\langle pq\ \mid\ \left(  p,q\right)  \in P\times
Q\right\rangle $.

Let $a\in\mathfrak{g}$ be arbitrary.

Now, let $\left(  u,v\right)  \in P\times Q$ be arbitrary. Then, $u\in P$ and
$v\in Q$. Since $P$ is a $\mathfrak{g}$-submodule of $A$, we have
$a\rightharpoonup u\in P$ (since $u\in P$). Since $Q$ is a $\mathfrak{g}%
$-submodule of $A$, we have $a\rightharpoonup v\in Q$ (since $v\in Q$). Now,
(\ref{defs.g-alg.cond}) shows us that
\begin{align*}
a\rightharpoonup\left(  uv\right)   &  =\underbrace{\left(  a\rightharpoonup
u\right)  }_{\in P}\cdot\underbrace{v}_{\in Q}+\underbrace{u}_{\in P}%
\cdot\underbrace{\left(  a\rightharpoonup v\right)  }_{\in Q}\in P\cdot
Q+P\cdot Q\\
&  \subseteq P\cdot Q\ \ \ \ \ \ \ \ \ \ \left(  \text{since }P\cdot Q\text{
is a }k\text{-module}\right)  .
\end{align*}


We have thus proven that
\begin{equation}
a\rightharpoonup\left(  uv\right)  \in P\cdot Q\text{ for every }\left(
u,v\right)  \in P\times Q. \label{pf.g-alg.PQ.1}%
\end{equation}


From this it is easily seen that $a\rightharpoonup t\in P\cdot Q$ for every
$t\in P\cdot Q$. \ \ \ \ {\footnotemark}\footnotetext{\textit{Proof.} In fact,
let $t\in P\cdot Q$ be arbitrary. Then, $t\in P\cdot Q=\left\langle
pq\ \mid\ \left(  p,q\right)  \in P\times Q\right\rangle $. Hence, $t$ is a
$k$-linear combination of terms of the form $pq$ for various $\left(
p,q\right)  \in P\times Q$. In other words, there exist some $n\in\mathbb{N}$,
some elements $\lambda_{1}$, $\lambda_{2}$, $...$, $\lambda_{n}$ of $k$, some
elements $p_{1}$, $p_{2}$, $...$, $p_{n}$ of $P$ and some elements $q_{1}$,
$q_{2}$, $...$, $q_{n}$ of $Q$ such that $t=\sum\limits_{i=1}^{n}\lambda
_{i}p_{i}q_{i}$. Hence,%
\begin{align*}
a\rightharpoonup t  &  =a\rightharpoonup\left(  \sum\limits_{i=1}^{n}%
\lambda_{i}p_{i}q_{i}\right)  =\sum\limits_{i=1}^{n}\lambda_{i}%
\underbrace{a\rightharpoonup\left(  p_{i}q_{i}\right)  }_{\substack{\in P\cdot
Q\\\text{(this follows from applying (\ref{pf.g-alg.PQ.1})}\\\text{to }%
p_{i}\text{ and }q_{i}\text{ instead of }u\text{ and }v\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Lie action of }A\text{ is
}k\text{-bilinear}\right) \\
&  \in\sum\limits_{i=1}^{n}\lambda_{i}P\cdot Q\subseteq P\cdot
Q\ \ \ \ \ \ \ \ \ \ \left(  \text{since }P\cdot Q\text{ is a }k\text{-module}%
\right)  ,
\end{align*}
qed.} Since this holds for every $a\in\mathfrak{g}$, we can conclude that
$P\cdot Q$ is a $\mathfrak{g}$-submodule of $A$. This proves Proposition
\ref{prop.g-alg.PQ}.

\subsection{$\otimes V$ is a $\mathfrak{g}$-algebra}

Now the prime example of a $\mathfrak{g}$-algebra is exactly what we would expect:

\begin{proposition}
\label{prop.lietensor}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra.\newline Let $V$ be a $\mathfrak{g}$-module. If we equip the
$k$-algebra $\otimes V$ (this $k$-algebra was defined in Definition
\ref{defs.tensor} \textbf{(a)}) with the $\mathfrak{g}$-module structure
defined in Definition \ref{defs.lietensor}, we obtain a $\mathfrak{g}$-algebra.
\end{proposition}

\begin{definition}
\label{defs.lietensor.g-alg}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie algebra.\newline Let $V$ be a $\mathfrak{g}$-module. The
$\mathfrak{g}$-algebra $\otimes V$ defined in Proposition \ref{prop.lietensor}
is called the \textit{tensor }$\mathfrak{g}$\textit{-algebra} of the
$\mathfrak{g}$-module $V$. Whenever we will speak of the $\mathfrak{g}%
$-algebra $\otimes V$, we will be meaning this tensor $\mathfrak{g}$-algebra
$\otimes V$ (unless we explicitly say that we are talking about a different
$\mathfrak{g}$-algebra structure on $\otimes V$).
\end{definition}

\begin{verlong}
\begin{remark}
By far the most important particular case of Proposition \ref{prop.lietensor}
is the following one: Whenever $k$ is a commutative ring and $\mathfrak{g}$ is
a $k$-Lie algebra, the tensor algebra $\otimes\mathfrak{g}$ of the
$\mathfrak{g}$-module $\mathfrak{g}$ is a $\mathfrak{g}$-algebra. This is a
consequence of Proposition \ref{prop.lietensor} (applied to $V=\mathfrak{g}$).
We are going to use this several times.
\end{remark}
\end{verlong}

\textit{Proof of Proposition \ref{prop.lietensor}.} In the following proof, we
are \textit{not} going to identify $V^{\otimes i}\otimes V^{\otimes\left(
n-i\right)  }$ with $V^{\otimes n}$ (for $n\in\mathbb{N}$ and $i\in\left\{
0,1,...,n\right\}  $) as we did in Definition \ref{defs.tensor}%
.{\footnotemark}\footnotetext{The purpose of this is to make the proof more
transparent. In fact, identifying $V^{\otimes i}\otimes V^{\otimes\left(
n-i\right)  }$ with $V^{\otimes n}$ would obscure the place where we actually
use Proposition \ref{prop.lie(X)^n.bracketing}.
\par
{}} As a consequence, the equation (\ref{defs.tensor.mult}) must be rewritten as%

\begin{equation}
\left(
\begin{array}
[c]{l}%
\left(  a_{i}\right)  _{i\in\mathbb{N}}\cdot\left(  b_{i}\right)
_{i\in\mathbb{N}}=\left(  \sum\limits_{i=0}^{n}\kappa_{i,n-i}\left(
a_{i}\otimes b_{n-i}\right)  \right)  _{n\in\mathbb{N}}\\
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  a_{i}\right)  _{i\in\mathbb{N}%
}\in\bigoplus\limits_{i\in\mathbb{N}}V^{\otimes i}\text{ and }\left(
b_{i}\right)  _{i\in\mathbb{N}}\in\bigoplus\limits_{i\in\mathbb{N}}V^{\otimes
i}%
\end{array}
\right)  , \label{defs.tensor.mult.lowlvl}%
\end{equation}
where $\kappa_{i,n-i}$ denotes the canonical $k$-module isomorphism
$V^{\otimes i}\otimes V^{\otimes\left(  n-i\right)  }\rightarrow V^{\otimes
n}$.

Proposition \ref{prop.lie(X)^n.bracketing} states that the canonical
$k$-module isomorphism $V^{\otimes i}\otimes V^{\otimes\left(  n-i\right)
}\rightarrow V^{\otimes n}$ is a $\mathfrak{g}$-module isomorphism. Since we
denoted this isomorphism by $\kappa_{i,n-i}$, we can thus conclude that
$\kappa_{i,n-i}$ is a $\mathfrak{g}$-module isomorphism.

Now, let $u\in\otimes V$ and $v\in\otimes V$ be any two elements. Let
$a\in\mathfrak{g}$ be arbitrary. Since $u\in\otimes V=\bigoplus\limits_{i\in
\mathbb{N}}V^{\otimes i}$, we can write $u$ as a family $\left(  a_{i}\right)
_{i\in\mathbb{N}}\in\bigoplus\limits_{i\in\mathbb{N}}V^{\otimes i}$. Since
$v\in\otimes V=\bigoplus\limits_{i\in\mathbb{N}}V^{\otimes i}$, we can write
$v$ as a family $\left(  b_{i}\right)  _{i\in\mathbb{N}}\in\bigoplus
\limits_{i\in\mathbb{N}}V^{\otimes i}$. Thus, $u=\left(  a_{i}\right)
_{i\in\mathbb{N}}$ and $v=\left(  b_{i}\right)  _{i\in\mathbb{N}}$, so that%
\[
uv=\left(  a_{i}\right)  _{i\in\mathbb{N}}\cdot\left(  b_{i}\right)
_{i\in\mathbb{N}}=\left(  \sum\limits_{i=0}^{n}\kappa_{i,n-i}\left(
a_{i}\otimes b_{n-i}\right)  \right)  _{n\in\mathbb{N}}%
\]
(by (\ref{defs.tensor.mult.lowlvl})), and therefore%
\[
a\rightharpoonup\left(  uv\right)  =a\rightharpoonup\left(  \sum
\limits_{i=0}^{n}\kappa_{i,n-i}\left(  a_{i}\otimes b_{n-i}\right)  \right)
_{n\in\mathbb{N}}=\left(  a\rightharpoonup\left(  \sum\limits_{i=0}^{n}%
\kappa_{i,n-i}\left(  a_{i}\otimes b_{n-i}\right)  \right)  \right)
_{n\in\mathbb{N}}%
\]
(by (\ref{prop.lie(++).form})). Since every $n\in\mathbb{N}$ satisfies%
\begin{align*}
a\rightharpoonup\left(  \sum\limits_{i=0}^{n}\kappa_{i,n-i}\left(
a_{i}\otimes b_{n-i}\right)  \right)   &  =\sum\limits_{i=0}^{n}%
\underbrace{a\rightharpoonup\left(  \kappa_{i,n-i}\left(  a_{i}\otimes
b_{n-i}\right)  \right)  }_{\substack{=\kappa_{i,n-i}\left(  a\rightharpoonup
\left(  a_{i}\otimes b_{n-i}\right)  \right)  \\\text{(since }\kappa
_{i,n-i}\text{ is a }\mathfrak{g}\text{-module}\\\text{isomorphism)}%
}}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the Lie action of}\\
\otimes V\text{ is }k\text{-bilinear}%
\end{array}
\right) \\
&  =\sum\limits_{i=0}^{n}\kappa_{i,n-i}\left(  \underbrace{a\rightharpoonup
\left(  a_{i}\otimes b_{n-i}\right)  }_{\substack{=\left(  a\rightharpoonup
a_{i}\right)  \otimes b_{n-i}+a_{i}\otimes\left(  a\rightharpoonup
b_{n-i}\right)  \\\text{(by (\ref{prop.lie(X).form}), applied to }V^{\otimes
i}\text{, }V^{\otimes\left(  n-i\right)  }\text{, }a_{i}\text{ and }%
b_{n-i}\\\text{instead of }V\text{, }W\text{, }v\text{ and }w\text{)}}}\right)
\\
&  =\sum\limits_{i=0}^{n}\underbrace{\kappa_{i,n-i}\left(  \left(
a\rightharpoonup a_{i}\right)  \otimes b_{n-i}+a_{i}\otimes\left(
a\rightharpoonup b_{n-i}\right)  \right)  }_{\substack{=\kappa_{i,n-i}\left(
\left(  a\rightharpoonup a_{i}\right)  \otimes b_{n-i}\right)  +\kappa
_{i,n-i}\left(  a_{i}\otimes\left(  a\rightharpoonup b_{n-i}\right)  \right)
\\\text{(since }\kappa_{i,n-i}\text{ is linear)}}}\\
&  =\sum\limits_{i=0}^{n}\left(  \kappa_{i,n-i}\left(  \left(
a\rightharpoonup a_{i}\right)  \otimes b_{n-i}\right)  +\kappa_{i,n-i}\left(
a_{i}\otimes\left(  a\rightharpoonup b_{n-i}\right)  \right)  \right) \\
&  =\sum\limits_{i=0}^{n}\kappa_{i,n-i}\left(  \left(  a\rightharpoonup
a_{i}\right)  \otimes b_{n-i}\right)  +\sum\limits_{i=0}^{n}\kappa
_{i,n-i}\left(  a_{i}\otimes\left(  a\rightharpoonup b_{n-i}\right)  \right)
,
\end{align*}
this rewrites as%
\begin{align}
a\rightharpoonup\left(  uv\right)   &  =\left(  \underbrace{a\rightharpoonup
\left(  \sum\limits_{i=0}^{n}\kappa_{i,n-i}\left(  a_{i}\otimes b_{n-i}%
\right)  \right)  }_{=\sum\limits_{i=0}^{n}\kappa_{i,n-i}\left(  \left(
a\rightharpoonup a_{i}\right)  \otimes b_{n-i}\right)  +\sum\limits_{i=0}%
^{n}\kappa_{i,n-i}\left(  a_{i}\otimes\left(  a\rightharpoonup b_{n-i}\right)
\right)  }\right)  _{n\in\mathbb{N}}\nonumber\\
&  =\left(  \sum\limits_{i=0}^{n}\kappa_{i,n-i}\left(  \left(
a\rightharpoonup a_{i}\right)  \otimes b_{n-i}\right)  +\sum\limits_{i=0}%
^{n}\kappa_{i,n-i}\left(  a_{i}\otimes\left(  a\rightharpoonup b_{n-i}\right)
\right)  \right)  _{n\in\mathbb{N}}\nonumber\\
&  =\left(  \sum\limits_{i=0}^{n}\kappa_{i,n-i}\left(  \left(
a\rightharpoonup a_{i}\right)  \otimes b_{n-i}\right)  \right)  _{n\in
\mathbb{N}}+\left(  \sum\limits_{i=0}^{n}\kappa_{i,n-i}\left(  a_{i}%
\otimes\left(  a\rightharpoonup b_{n-i}\right)  \right)  \right)
_{n\in\mathbb{N}}. \label{pf.lietensor.1}%
\end{align}


We have $u=\left(  a_{i}\right)  _{i\in\mathbb{N}}$ and thus $a\rightharpoonup
u=a\rightharpoonup\left(  a_{i}\right)  _{i\in\mathbb{N}}=\left(
a\rightharpoonup a_{i}\right)  _{i\in\mathbb{N}}$ (by (\ref{prop.lie(++).form}%
)). Together with $v=\left(  b_{i}\right)  _{i\in\mathbb{N}}$, this yields%
\begin{equation}
\left(  a\rightharpoonup u\right)  \cdot v=\left(  a\rightharpoonup
a_{i}\right)  _{i\in\mathbb{N}}\cdot\left(  b_{i}\right)  _{i\in\mathbb{N}%
}=\left(  \sum\limits_{i=0}^{n}\kappa_{i,n-i}\left(  \left(  a\rightharpoonup
a_{i}\right)  \otimes b_{n-i}\right)  \right)  _{n\in\mathbb{N}}
\label{pf.lietensor.2}%
\end{equation}
(by (\ref{defs.tensor.mult.lowlvl}), applied to $\left(  a\rightharpoonup
a_{i}\right)  _{i\in\mathbb{N}}$ instead of $\left(  a_{i}\right)
_{i\in\mathbb{N}}$).

On the other hand, $v=\left(  b_{i}\right)  _{i\in\mathbb{N}}$ and thus
$a\rightharpoonup v=a\rightharpoonup\left(  b_{i}\right)  _{i\in\mathbb{N}%
}=\left(  a\rightharpoonup b_{i}\right)  _{i\in\mathbb{N}}$ (by
(\ref{prop.lie(++).form})). Together with $u=\left(  a_{i}\right)
_{i\in\mathbb{N}}$, this yields%
\[
u\cdot\left(  a\rightharpoonup v\right)  =\left(  a_{i}\right)  _{i\in
\mathbb{N}}\cdot\left(  a\rightharpoonup b_{i}\right)  _{i\in\mathbb{N}%
}=\left(  \sum\limits_{i=0}^{n}\kappa_{i,n-i}\left(  a_{i}\otimes\left(
a\rightharpoonup b_{n-i}\right)  \right)  \right)  _{n\in\mathbb{N}}%
\]
(by (\ref{defs.tensor.mult.lowlvl}), applied to $\left(  a\rightharpoonup
b_{i}\right)  _{i\in\mathbb{N}}$ instead of $\left(  b_{i}\right)
_{i\in\mathbb{N}}$). Adding this equation to (\ref{pf.lietensor.2}), we obtain%
\[
\left(  a\rightharpoonup u\right)  \cdot v+u\cdot\left(  a\rightharpoonup
v\right)  =\left(  \sum\limits_{i=0}^{n}\kappa_{i,n-i}\left(  \left(
a\rightharpoonup a_{i}\right)  \otimes b_{n-i}\right)  \right)  _{n\in
\mathbb{N}}+\left(  \sum\limits_{i=0}^{n}\kappa_{i,n-i}\left(  a_{i}%
\otimes\left(  a\rightharpoonup b_{n-i}\right)  \right)  \right)
_{n\in\mathbb{N}}.
\]
Compared with (\ref{pf.lietensor.1}), this yields $a\rightharpoonup\left(
uv\right)  =\left(  a\rightharpoonup u\right)  \cdot v+u\cdot\left(
a\rightharpoonup v\right)  $. Since this holds for every $a\in\mathfrak{g}$,
$u\in\otimes V$ and $v\in\otimes V$, we can conclude that
(\ref{defs.g-alg.cond}) holds with $A$ replaced by $\otimes V$. Hence (by
Definition \ref{defs.g-alg}) we see that $\otimes V$ is a $\mathfrak{g}%
$-algebra. This proves Proposition \ref{prop.lietensor}.

We notice that the $\mathfrak{g}$-algebra $\otimes V$ behaves under
restriction as we would want it to:

\begin{proposition}
\label{prop.lie.(X)algvsRes}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$.\newline If $V$ is any $\mathfrak{g}$-module, then $\operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}\left(  \otimes V\right)  =\otimes\left(
\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V\right)  $ as $\mathfrak{h}%
$-algebras. This allows us to speak of "the $\mathfrak{h}$-algebra $\otimes
V$" without having to worry whether we mean $\operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}\left(  \otimes V\right)  $ or $\otimes\left(
\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V\right)  $ (because it does
not matter, since $\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}\left(
\otimes V\right)  =\otimes\left(  \operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}V\right)  $).
\end{proposition}

Proving this is a matter of applying the definitions.

\begin{verlong}
A consequence of Proposition \ref{prop.lietensor} that we are going to use:

\begin{corollary}
\label{cor.biideal}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$.\newline Let $V$ be a $\mathfrak{g}$-module. Then, $\otimes V$ is a
$\mathfrak{g}$-module (according to Definition \ref{defs.lietensor}). As we
know from Definition \ref{defs.g-as-h-module}, this $\otimes V$ therefore
becomes an $\mathfrak{h}$-module as well.\newline\textbf{(a)} Let $P$ and $Q$
be two $\mathfrak{h}$-submodules of $\otimes V$. Then, $P\cdot Q$ is an
$\mathfrak{h}$-submodule of $\otimes V$ as well.\newline\textbf{(b)} Let $R$
be an $\mathfrak{h}$-submodule of $\otimes V$. Then, $\left(  \otimes
V\right)  \cdot R\cdot\left(  \otimes V\right)  $ is an $\mathfrak{h}%
$-submodule of $\otimes V$ as well.\newline\textbf{(c)} Let $p\in\mathbb{N}$,
and let $R$ be an $\mathfrak{h}$-submodule of $\otimes V$. Then, $V^{\otimes
p}\cdot R$ is an $\mathfrak{h}$-submodule of $\otimes V$ as well.
\end{corollary}

Here, we are using Convention \ref{conv.A*B}.

\textit{Proof of Corollary \ref{cor.biideal}.} According to Definition
\ref{defs.g-as-h-algebra}, the $\mathfrak{h}$-module $\otimes V$, equipped
with its $k$-algebra structure, becomes an $\mathfrak{h}$-algebra (because it
is a $\mathfrak{g}$-algebra (according to Proposition \ref{prop.lietensor})).

\textbf{(a)} Since $\otimes V$ is an $\mathfrak{h}$-algebra, Proposition
\ref{prop.g-alg.PQ} (applied to $\otimes V$ and $\mathfrak{h}$ instead of $A$
and $\mathfrak{g}$) yields that $P\cdot Q$ is an $\mathfrak{h}$-submodule of
$\otimes V$. This proves Corollary \ref{cor.biideal} \textbf{(a)}.

\textbf{(b)} Corollary \ref{cor.biideal} \textbf{(a)} (applied to $\otimes V$
and $R$ instead of $P$ and $Q$) yields that $\left(  \otimes V\right)  \cdot
R$ is an $\mathfrak{h}$-submodule of $\otimes V$ (because both $\otimes V$ and
$R$ are $\mathfrak{h}$-submodules of $\otimes V$). Now, Corollary
\ref{cor.biideal} \textbf{(a)} (applied to $\left(  \otimes V\right)  \cdot R$
and $\otimes V$ instead of $P$ and $Q$) yields that $\left(  \left(  \otimes
V\right)  \cdot R\right)  \cdot\left(  \otimes V\right)  $ is an
$\mathfrak{h}$-submodule of $\otimes V$ (because both $\left(  \otimes
V\right)  \cdot R$ and $\otimes V$ are $\mathfrak{h}$-submodules of $\otimes
V$). Since $\left(  \left(  \otimes V\right)  \cdot R\right)  \cdot\left(
\otimes V\right)  =\left(  \otimes V\right)  \cdot R\cdot\left(  \otimes
V\right)  $, we can thus conclude that $\left(  \otimes V\right)  \cdot
R\cdot\left(  \otimes V\right)  $ is an $\mathfrak{h}$-submodule of $\otimes
V$. This proves Corollary \ref{cor.biideal} \textbf{(b)}.

\textbf{(c)} We know that $V^{\otimes p}$ is a $\mathfrak{g}$-submodule of
$\otimes V$, and thus an $\mathfrak{h}$-submodule of $\otimes V$ as well. Now,
Corollary \ref{cor.biideal} \textbf{(a)} (applied to $V^{\otimes p}$ and $R$
instead of $P$ and $Q$) yields that $V^{\otimes p}\cdot R$ is an
$\mathfrak{h}$-submodule of $\otimes V$ (because both $V^{\otimes p}$ and $R$
are $\mathfrak{h}$-submodules of $\otimes V$). This proves Corollary
\ref{cor.biideal} \textbf{(c)}.
\end{verlong}

\subsection{\label{subsect.U(g)}$\mathfrak{g}$-modules are $U\left(
\mathfrak{g}\right)  $-modules}

The present Subsection \ref{subsect.U(g)} is, apart from the (easy)
Proposition \ref{prop.Uhopf.easy}, of no importance for this paper, as the
results of this subsection will not be used anywhere further in this paper,
except in the (equally unimportant) Subsection \ref{subsect.U(g,h)}. However,
it provides some context for the theory of $\mathfrak{g}$-modules as well as
one motivation for considering the universal enveloping algebra $U\left(
\mathfrak{g}\right)  $.

Many results about $\mathfrak{g}$-modules (where $\mathfrak{g}$ is a Lie
algebra) are analogous to similar results about $A$-modules, where $A$ is an
associative algebra. For example, Proposition \ref{prop.liemodiso} is the
analogue of the following fact:

\begin{proposition}
\label{prop.algmodiso}Let $k$ be a commutative ring. Let $A$ be a $k$-algebra.
Let $V$ and $W$ be two $A$-modules. Let $f:V\rightarrow W$ be a $k$-linear
map. Then, $f$ is an $A$-module isomorphism if and only if $f$ is an
invertible $A$-module homomorphism. In other words, $f$ is an $A$-module
isomorphism if and only if $f$ is an $A$-module homomorphism and a $k$-module
isomorphism at the same time.
\end{proposition}

Now, of course, Proposition \ref{prop.liemodiso} and Proposition
\ref{prop.algmodiso} are both (more or less) trivial, so there is no wonder
that they are analogous. However, the analogy between $A$-modules and
$\mathfrak{g}$-modules goes much further, and there are much deeper theorems
about $A$-modules that have their counterparts for $\mathfrak{g}$-modules.
This has a reason: Namely, the category of $\mathfrak{g}$-modules is
isomorphic to the category of $A$-modules for a particular (associative)
algebra $A$ depending on $\mathfrak{g}$. Namely, this $A$ is the universal
enveloping algebra $U\left(  \mathfrak{g}\right)  $ defined in Definition
\ref{defs.U}. More precisely:

\begin{proposition}
\label{prop.U.equiv}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Consider the universal enveloping algebra $U\left(
\mathfrak{g}\right)  $ and the ideal $I_{\mathfrak{g}}$ defined in Definition
\ref{defs.U}.\newline\textbf{(a)} For every $\mathfrak{g}$-module $V$, there
is one and only one $U\left(  \mathfrak{g}\right)  $-module structure on $V$
satisfying%
\[
\left(  \overline{a}\cdot v=a\rightharpoonup v\ \ \ \ \ \ \ \ \ \ \text{for
every }a\in\mathfrak{g}\text{ and }v\in V\right)
\]
(where $\overline{a}$ denotes the projection of $a\in\mathfrak{g}%
\subseteq\otimes\mathfrak{g}$ on $\left(  \otimes\mathfrak{g}\right)  \diagup
I_{\mathfrak{g}}=U\left(  \mathfrak{g}\right)  $). This $U\left(
\mathfrak{g}\right)  $-module structure is canonical. Thus, every
$\mathfrak{g}$-module $V$ canonically becomes a $U\left(  \mathfrak{g}\right)
$-module.\newline\textbf{(b)} Conversely, for every $U\left(  \mathfrak{g}%
\right)  $-module $V$, we can define a $\mathfrak{g}$-module structure on $V$
by%
\[
\left(  a\rightharpoonup v=\overline{a}\cdot v\ \ \ \ \ \ \ \ \ \ \text{for
every }a\in\mathfrak{g}\text{ and }v\in V\right)
\]
(where $\overline{a}$ denotes the projection of $a\in\mathfrak{g}%
\subseteq\otimes\mathfrak{g}$ on $\left(  \otimes\mathfrak{g}\right)  \diagup
I_{\mathfrak{g}}=U\left(  \mathfrak{g}\right)  $). This $\mathfrak{g}$-module
structure is canonical. Thus, every $U\left(  \mathfrak{g}\right)  $-module
$V$ canonically becomes a $\mathfrak{g}$-module.\newline\textbf{(c)} Let $V$
and $W$ be two $\mathfrak{g}$-modules. Then, according to Proposition
\ref{prop.U.equiv} \textbf{(a)}, each of $V$ and $W$ canonically becomes a
$U\left(  \mathfrak{g}\right)  $-module. Let $f:V\rightarrow W$ be a map.
Then, $f$ is a homomorphism of $\mathfrak{g}$-modules if and only if $f$ is a
homomorphism of $U\left(  \mathfrak{g}\right)  $-modules.\newline\textbf{(d)}
Let $V$ and $W$ be two $U\left(  \mathfrak{g}\right)  $-modules. Then,
according to Proposition \ref{prop.U.equiv} \textbf{(b)}, each of $V$ and $W$
canonically becomes a $\mathfrak{g}$-module. Let $f:V\rightarrow W$ be a map.
Then, $f$ is a homomorphism of $\mathfrak{g}$-modules if and only if $f$ is a
homomorphism of $U\left(  \mathfrak{g}\right)  $-modules.\newline\textbf{(e)}
We can define a functor $U_{1}$ from the category of $\mathfrak{g}$-modules to
the category of $U\left(  \mathfrak{g}\right)  $-modules as follows: For every
$\mathfrak{g}$-module $V$, let $U_{1}\left(  V\right)  $ be the $U\left(
\mathfrak{g}\right)  $-module $V$ defined in Proposition \ref{prop.U.equiv}
\textbf{(a)}. For every homomorphism $f$ between $\mathfrak{g}$-modules, let
$U_{1}\left(  f\right)  $ be the same homomorphism $f$, but considered as a
homomorphism between $U\left(  \mathfrak{g}\right)  $-modules this time (this
is legitimate due to Proposition \ref{prop.U.equiv} \textbf{(c)}%
).\newline\textbf{(f)} We can define a functor $U_{2}$ from the category of
$U\left(  \mathfrak{g}\right)  $-modules to the category of $\mathfrak{g}%
$-modules as follows: For every $U\left(  \mathfrak{g}\right)  $-module $V$,
let $U_{2}\left(  V\right)  $ be the $\mathfrak{g}$-module $V$ defined in
Proposition \ref{prop.U.equiv} \textbf{(b)}. For every homomorphism $f$
between $U\left(  \mathfrak{g}\right)  $-modules, let $U_{2}\left(  f\right)
$ be the same homomorphism $f$, but considered as a homomorphism between
$\mathfrak{g}$-modules this time (this is legitimate due to Proposition
\ref{prop.U.equiv} \textbf{(d)}).\newline\textbf{(g)} The two functors $U_{1}$
and $U_{2}$ defined in Proposition \ref{prop.U.equiv} \textbf{(e)} and
\textbf{(f)} are mutually inverse.\newline\textbf{(h)} Both functors $U_{1}$
and $U_{2}$ are additive, exact and preserve kernels, cokernels and direct sums.
\end{proposition}

Note that the two functors $U_{1}$ and $U_{2}$ are mutually \textit{inverse}
(and not just quasi-inverse, as many pairs of functors in algebra tend to be)
in Proposition \ref{prop.U.equiv} \textbf{(g)}.

Proposition \ref{prop.U.equiv} explains why, for a number of results about
$A$-modules (with $A$ an associative algebra), there are analogous results for
$\mathfrak{g}$-modules (with $\mathfrak{g}$ a Lie algebra). However, it does
not explain everything; in particular, it does not explain anything about
tensor products of $\mathfrak{g}$-modules, because, in general, there is no
reasonable notion of a tensor product between two $A$-modules, where $A$ is an
associative algebra. However, there \textit{is} such a notion when $A$ is a
bialgebra. And there is a canonical way to turn $U\left(  \mathfrak{g}\right)
$ into a bialgebra, and even into a Hopf algebra (so that, in addition to
tensor products of modules, we also get $\operatorname*{Hom}$ modules). Before
I show this way, let us define a Hopf algebra structure on the tensor algebra
$\otimes\mathfrak{g}$, and, more generally, on $\otimes V$ for any $k$-module
$V$:

\begin{proposition}
\label{prop.tensorhopf}Let $k$ be a commutative ring. Let $V$ be a
$k$-module.\newline\textbf{(a)} Let $\varepsilon_{\otimes V}:\otimes
V\rightarrow k$ be the projection from the direct sum $\otimes V=\bigoplus
\limits_{i\in\mathbb{N}}V^{\otimes i}$ onto its addend $V^{\otimes0}=k$. Then,
$\varepsilon_{\otimes V}$ is a $k$-algebra homomorphism. It is given by%
\begin{align*}
&  \varepsilon_{\otimes V}\left(  v_{1}\otimes v_{2}\otimes...\otimes
v_{n}\right)  =\left\{
\begin{array}
[c]{c}%
1\text{, if }n=0\text{;}\\
0\text{, if }n>0
\end{array}
\right. \\
&  \ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}\text{ and every
}\left(  v_{1},v_{2},...,v_{n}\right)  \in V^{n}.
\end{align*}
\newline\textbf{(b)} There exists one and only one $k$-algebra homomorphism
$\mathbf{d}:\otimes V\rightarrow\left(  \otimes V\right)  \otimes\left(
\otimes V\right)  $ satisfying $\left(  \mathbf{d}\left(  v\right)  =1\otimes
v+v\otimes1\text{ for every }v\in V\right)  $. Denote this homomorphism
$\mathbf{d}$ by $\Delta_{\otimes V}$. Then, this homomorphism $\Delta_{\otimes
V}$ is given by%
\begin{align*}
&  \Delta_{\otimes V}\left(  v_{1}\otimes v_{2}\otimes...\otimes v_{n}\right)
\\
&  =\sum_{i=0}^{n}\sum_{\substack{\pi\in S_{n};\\\pi\left(  1\right)
<\pi\left(  2\right)  <...<\pi\left(  i\right)  ;\\\pi\left(  i+1\right)
<\pi\left(  i+2\right)  <...<\pi\left(  n\right)  }}\left(  v_{\pi\left(
1\right)  }\otimes v_{\pi\left(  2\right)  }\otimes...\otimes v_{\pi\left(
i\right)  }\right)  \otimes\left(  v_{\pi\left(  i+1\right)  }\otimes
v_{\pi\left(  i+2\right)  }\otimes...\otimes v_{\pi\left(  n\right)  }\right)
\\
&  \ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}\text{ and every
}\left(  v_{1},v_{2},...,v_{n}\right)  \in V^{n}.
\end{align*}
(Here, the two tensors $v_{\pi\left(  1\right)  }\otimes v_{\pi\left(
2\right)  }\otimes...\otimes v_{\pi\left(  i\right)  }$ and $v_{\pi\left(
i+1\right)  }\otimes v_{\pi\left(  i+2\right)  }\otimes...\otimes
v_{\pi\left(  n\right)  }$ are considered as elements of the tensor algebra
$\otimes V$, and their tensor product is a tensor in $\left(  \otimes
V\right)  \otimes\left(  \otimes V\right)  $.)\newline\textbf{(c)} There
exists one and only one $k$-algebra homomorphism $\mathbf{S}:\otimes
V\rightarrow\left(  \otimes V\right)  ^{\operatorname*{op}}$ satisfying
$\left(  \mathbf{S}\left(  v\right)  =-v\text{ for every }v\in V\right)  $.
Denote this homomorphism $\mathbf{S}$ by $S_{\otimes V}$. Concretely, this
$S_{\otimes V}$ is given by%
\begin{align*}
&  S_{\otimes V}\left(  v_{1}\otimes v_{2}\otimes...\otimes v_{n}\right)
=\left(  -1\right)  ^{n}v_{n}\otimes v_{n-1}\otimes...\otimes v_{1}\\
&  \ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}\text{ and every
}\left(  v_{1},v_{2},...,v_{n}\right)  \in V^{n}.
\end{align*}
\newline\textbf{(d)} The homomorphisms $\Delta_{\otimes V}$, $\varepsilon
_{\otimes V}$ and $S_{\otimes V}$ defined above make $\otimes V$ into a Hopf
algebra. This Hopf algebra is called the \textit{tensor Hopf algebra} of the
$k$-module $V$. This Hopf algebra $\otimes V$ is cocommutative.
\end{proposition}

\begin{proposition}
\label{prop.Uhopf}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. In the following, we will use the notations $U\left(
\mathfrak{g}\right)  $ and $I_{\mathfrak{g}}$ which were defined in Definition
\ref{defs.U}. Recall that $U\left(  \mathfrak{g}\right)  =\left(
\otimes\mathfrak{g}\right)  \diagup I_{\mathfrak{g}}$.\newline\textbf{(a)} The
ideal $I_{\mathfrak{g}}$ is a Hopf ideal of the tensor Hopf algebra
$\otimes\mathfrak{g}$ of the $k$-module $\mathfrak{g}$. (For the definition of
the tensor Hopf algebra $\otimes\mathfrak{g}$, see Proposition
\ref{prop.tensorhopf}.) Thus, $\left(  \otimes\mathfrak{g}\right)  \diagup
I_{\mathfrak{g}}$ becomes a Hopf algebra. Since $\left(  \otimes
\mathfrak{g}\right)  \diagup I_{\mathfrak{g}}=U\left(  \mathfrak{g}\right)  $,
this means that $U\left(  \mathfrak{g}\right)  $ becomes a Hopf algebra. This
Hopf algebra $U\left(  \mathfrak{g}\right)  $ is cocommutative.\newline%
\textbf{(b)} If the Hopf algebra $U\left(  \mathfrak{g}\right)  $ defined in
Proposition \ref{prop.Uhopf} \textbf{(a)} is used to define the tensor product
of two $U\left(  \mathfrak{g}\right)  $-modules, then the functors $U_{1}$ and
$U_{2}$ defined in Proposition \ref{prop.U.equiv} preserve tensor products.
These functors as well preserve internal $\operatorname*{Hom}$s and the
ground-field object, if these are also defined using the Hopf algebra
structure on $U\left(  \mathfrak{g}\right)  $.
\end{proposition}

A proof of Proposition \ref{prop.U.equiv} can be found in most standard
references about Lie algebras (such as \cite[Chapter III, Corollary
3.6]{knapp}), while a proof of Proposition \ref{prop.Uhopf} can be found in
references for Hopf algebras (such as \cite[\S 1.2.6, Example 6]%
{klimyk-schmudgen}; more precisely, \cite[\S 1.2.6, Example 6]%
{klimyk-schmudgen} proves what amounts to an equivalent form of Proposition
\ref{prop.Uhopf} independent of Proposition \ref{prop.tensorhopf}).
Proposition \ref{prop.tensorhopf} is proven in \cite[\S 1.2.6, Example
8]{klimyk-schmudgen}. All these proofs are rather standard, however. None of
these three propositions will be used in the following, except for the easiest
part of Proposition \ref{prop.Uhopf} \textbf{(a)} - namely, the one about the
counit of the Hopf algebra $U\left(  \mathfrak{g}\right)  $:

\begin{proposition}
\label{prop.Uhopf.easy}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. In the following, we will use the notations $U\left(
\mathfrak{g}\right)  $ and $I_{\mathfrak{g}}$ which were defined in Definition
\ref{defs.U}. Recall that $U\left(  \mathfrak{g}\right)  =\left(
\otimes\mathfrak{g}\right)  \diagup I_{\mathfrak{g}}$. Let $\psi$ be the
canonical projection from $\otimes\mathfrak{g}$ onto $U\left(  \mathfrak{g}%
\right)  $. Clearly, this projection $\psi$ is a surjective $k$-algebra
homomorphism.\newline Applying Proposition \ref{prop.tensorhopf} \textbf{(a)}
to $V=\mathfrak{g}$, we obtain a $k$-algebra homomorphism $\varepsilon
_{\otimes\mathfrak{g}}:\otimes\mathfrak{g}\rightarrow k$.\newline\textbf{(a)}
Then, $\varepsilon_{\otimes\mathfrak{g}}\left(  I_{\mathfrak{g}}\right)  =0$.
Thus, by the universal property of the factor algebra, there exists one and
only one $k$-algebra homomorphism $\overline{\varepsilon_{\otimes\mathfrak{g}%
}}:U\left(  \mathfrak{g}\right)  \rightarrow k$ such that $\varepsilon
_{\otimes\mathfrak{g}}=\overline{\varepsilon_{\otimes\mathfrak{g}}}\circ\psi$.
Denote this homomorphism $\overline{\varepsilon_{\otimes\mathfrak{g}}}$ by
$\varepsilon_{U\left(  \mathfrak{g}\right)  }$. Then, $\varepsilon
_{\otimes\mathfrak{g}}=\varepsilon_{U\left(  \mathfrak{g}\right)  }\circ\psi
$.\newline\textbf{(b)} This homomorphism $\varepsilon_{U\left(  \mathfrak{g}%
\right)  }$ satisfies
\begin{align*}
&  \varepsilon_{U\left(  \mathfrak{g}\right)  }\left(  \overline{v_{1}\otimes
v_{2}\otimes...\otimes v_{n}}\right)  =\left\{
\begin{array}
[c]{c}%
1\text{, if }n=0\text{;}\\
0\text{, if }n>0
\end{array}
\right. \\
&  \ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}\text{ and every
}\left(  v_{1},v_{2},...,v_{n}\right)  \in\mathfrak{g}^{n}.
\end{align*}

\end{proposition}

\begin{verlong}
For the sake of completeness, we record the proof of this proposition:
\end{verlong}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.Uhopf.easy}.} \textbf{(a)} In order to
prove Proposition \ref{prop.Uhopf.easy} \textbf{(a)}, we only need to show
that $\varepsilon_{\otimes\mathfrak{g}}\left(  I_{\mathfrak{g}}\right)  =0$,
because all the other assertions of Proposition \ref{prop.Uhopf.easy}
\textbf{(a)} follow trivially from it.

By the definition of $\varepsilon_{\otimes\mathfrak{g}}$, we know that
$\varepsilon_{\otimes\mathfrak{g}}$ is the projection from the direct sum
$\bigoplus\limits_{i\in\mathbb{N}}\mathfrak{g}^{\otimes i}$ onto its addend
$\mathfrak{g}^{\otimes0}=k$. Thus, $\varepsilon_{\otimes\mathfrak{g}}$ must
map every addend of the direct sum $\bigoplus\limits_{i\in\mathbb{N}%
}\mathfrak{g}^{\otimes i}$ different from $\mathfrak{g}^{\otimes0}$ to $0$. In
other words, every $n\in\mathbb{N}$ such that $n\neq0$ must satisfy
$\varepsilon_{\otimes\mathfrak{g}}\left(  \mathfrak{g}^{\otimes n}\right)  =0$
(because whenever $n\in\mathbb{N}$ satisfies $n\neq0$, the $k$-submodule
$\mathfrak{g}^{\otimes n}$ is an addend of the direct sum $\bigoplus
\limits_{i\in\mathbb{N}}\mathfrak{g}^{\otimes i}$ different from
$\mathfrak{g}^{\otimes0}$). Applying this to $n=1$ yields $\varepsilon
_{\otimes\mathfrak{g}}\left(  \mathfrak{g}^{\otimes1}\right)  =0$. On the
other hand, applying $\varepsilon_{\otimes\mathfrak{g}}\left(  \mathfrak{g}%
^{\otimes n}\right)  =0$ to $n=2$ yields $\varepsilon_{\otimes\mathfrak{g}%
}\left(  \mathfrak{g}^{\otimes2}\right)  =0$. Since $\varepsilon
_{\otimes\mathfrak{g}}$ is $k$-linear, we have $\varepsilon_{\otimes
\mathfrak{g}}\left(  \mathfrak{g}^{\otimes1}+\mathfrak{g}^{\otimes2}\right)
=\underbrace{\varepsilon_{\otimes\mathfrak{g}}\left(  \mathfrak{g}^{\otimes
1}\right)  }_{=0}+\underbrace{\varepsilon_{\otimes\mathfrak{g}}\left(
\mathfrak{g}^{\otimes2}\right)  }_{=0}=0+0=0$.

Now, every $\left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{g}$ satisfies%
\[
v\otimes w-w\otimes v-\left[  v,w\right]  =\underbrace{-\left[  v,w\right]
}_{\in\mathfrak{g}^{\otimes1}}+\underbrace{v\otimes w-w\otimes v}%
_{\in\mathfrak{g}^{\otimes2}}\in\mathfrak{g}^{\otimes1}+\mathfrak{g}%
^{\otimes2}.
\]
In other words, $\left\{  v\otimes w-w\otimes v-\left[  v,w\right]
\ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{g}\right\}
\subseteq\mathfrak{g}^{\otimes1}+\mathfrak{g}^{\otimes2}$. Thus, Proposition
\ref{prop.<>} \textbf{(a)} (applied to $\otimes\mathfrak{g}$, $\left\{
v\otimes w-w\otimes v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)
\in\mathfrak{g}\times\mathfrak{g}\right\}  $ and $\mathfrak{g}^{\otimes
1}+\mathfrak{g}^{\otimes2}$ instead of $M$, $S$ and $Q$) yields $\left\langle
\left\{  v\otimes w-w\otimes v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)
\in\mathfrak{g}\times\mathfrak{g}\right\}  \right\rangle \subseteq
\mathfrak{g}^{\otimes1}+\mathfrak{g}^{\otimes2}$.

Since $I_{\mathfrak{g}}=\left(  \otimes\mathfrak{g}\right)  \cdot\left\langle
v\otimes w-w\otimes v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)
\in\mathfrak{g}\times\mathfrak{g}\right\rangle \cdot\left(  \otimes
\mathfrak{g}\right)  $, we have%
\begin{align*}
\varepsilon_{\otimes\mathfrak{g}}\left(  I_{\mathfrak{g}}\right)   &
=\varepsilon_{\otimes\mathfrak{g}}\left(  \left(  \otimes\mathfrak{g}\right)
\cdot\left\langle v\otimes w-w\otimes v-\left[  v,w\right]  \ \mid\ \left(
v,w\right)  \in\mathfrak{g}\times\mathfrak{g}\right\rangle \cdot\left(
\otimes\mathfrak{g}\right)  \right) \\
&  =\underbrace{\varepsilon_{\otimes\mathfrak{g}}\left(  \otimes
\mathfrak{g}\right)  }_{\subseteq k}\cdot\varepsilon_{\otimes\mathfrak{g}%
}\left(  \underbrace{\left\langle v\otimes w-w\otimes v-\left[  v,w\right]
\ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{g}\right\rangle
}_{\substack{=\left\langle \left\{  v\otimes w-w\otimes v-\left[  v,w\right]
\ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{g}\right\}
\right\rangle \\\subseteq\mathfrak{g}^{\otimes1}+\mathfrak{g}^{\otimes2}%
}}\right)  \cdot\underbrace{\varepsilon_{\otimes\mathfrak{g}}\left(
\otimes\mathfrak{g}\right)  }_{\subseteq k}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{because }\varepsilon_{\otimes
\mathfrak{g}}\text{ is a }k\text{-algebra homomorphism}\right) \\
&  \subseteq k\cdot\underbrace{\varepsilon_{\otimes\mathfrak{g}}\left(
\mathfrak{g}^{\otimes1}+\mathfrak{g}^{\otimes2}\right)  }_{=0}\cdot
k=k\cdot0\cdot k=0.
\end{align*}
In other words, $\varepsilon_{\otimes\mathfrak{g}}\left(  I_{\mathfrak{g}%
}\right)  =0$. This completes the proof of Proposition \ref{prop.Uhopf.easy}
\textbf{(a)}.

\textbf{(b)} Every $n\in\mathbb{N}$ and every $\left(  v_{1},v_{2}%
,...,v_{n}\right)  \in\mathfrak{g}^{n}$ satisfy $\overline{v_{1}\otimes
v_{2}\otimes...\otimes v_{n}}=\psi\left(  v_{1}\otimes v_{2}\otimes...\otimes
v_{n}\right)  $ (since $\psi$ is the canonical projection from $\otimes
\mathfrak{g}$ onto $U\left(  \mathfrak{g}\right)  $) and thus%
\begin{align*}
\varepsilon_{U\left(  \mathfrak{g}\right)  }\left(  \overline{v_{1}\otimes
v_{2}\otimes...\otimes v_{n}}\right)   &  =\varepsilon_{U\left(
\mathfrak{g}\right)  }\left(  \psi\left(  v_{1}\otimes v_{2}\otimes...\otimes
v_{n}\right)  \right)  =\underbrace{\left(  \varepsilon_{U\left(
\mathfrak{g}\right)  }\circ\psi\right)  }_{=\varepsilon_{\otimes\mathfrak{g}}%
}\left(  v_{1}\otimes v_{2}\otimes...\otimes v_{n}\right) \\
&  =\varepsilon_{\otimes\mathfrak{g}}\left(  v_{1}\otimes v_{2}\otimes
...\otimes v_{n}\right)  =\left\{
\begin{array}
[c]{c}%
1\text{, if }n=0\text{;}\\
0\text{, if }n>0
\end{array}
\right. \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition \ref{prop.tensorhopf}
\textbf{(a)}, applied to }\mathfrak{g}\text{ instead of }V\right)  .
\end{align*}
This proves Proposition \ref{prop.Uhopf.easy} \textbf{(b)}.
\end{verlong}

Also, we note in passing that $\mathfrak{g}$-algebras (as introduced in
Definition \ref{defs.g-alg}) are left $U\left(  \mathfrak{g}\right)  $-module
algebras (where we are using the terminology of Hopf algebra theorists).

\subsection{Splitting of exact sequences of $\mathfrak{g}$-modules}

Next we are going to introduce the notion of \textit{splitting} for an exact
sequence of $\mathfrak{g}$-modules. This notion is based on the following
known result:

\begin{proposition}
\label{prop.g.split}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $A$, $B$ and $C$ be three $\mathfrak{g}$-modules. Let
$f:A\rightarrow B$ and $g:B\rightarrow C$ be two $\mathfrak{g}$-module
homomorphisms such that%
\[%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%0 \ar[r] & A \ar[r]^f & B \ar[r]^g & C \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrix{
0 \ar[r] & A \ar[r]^f & B \ar[r]^g & C \ar[r] & 0
}%
%EndExpansion
\]
is a short exact sequence. Then, the following three assertions $\mathbb{S}%
_{1}$, $\mathbb{S}_{2}$ and $\mathbb{S}_{3}$ are equivalent:\newline%
\textit{Assertion }$\mathbb{S}_{1}$\textit{:} There exists a $\mathfrak{g}%
$-module homomorphism $f^{\prime}:B\rightarrow A$ such that $f^{\prime}\circ
f=\operatorname*{id}\nolimits_{A}$.\newline\textit{Assertion }$\mathbb{S}_{2}%
$\textit{:} There exists a $\mathfrak{g}$-module homomorphism $g^{\prime
}:C\rightarrow B$ such that $g\circ g^{\prime}=\operatorname*{id}%
\nolimits_{C}$.\newline\textit{Assertion }$\mathbb{S}_{3}$\textit{:} There
exists a $\mathfrak{g}$-submodule $P$ of $B$ such that $B=f\left(  A\right)
\oplus P$.
\end{proposition}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.g.split}.} In order to prove
Proposition \ref{prop.g.split}, we must show the equivalence of the three
assertions $\mathbb{S}_{1}$, $\mathbb{S}_{2}$ and $\mathbb{S}_{3}$. In order
to achieve this, let us show that $\mathbb{S}_{1}\Longrightarrow\mathbb{S}%
_{3}$, $\mathbb{S}_{2}\Longrightarrow\mathbb{S}_{3}$, $\mathbb{S}%
_{3}\Longrightarrow\mathbb{S}_{1}$ and $\mathbb{S}_{3}\Longrightarrow
\mathbb{S}_{2}$:

\textit{Proof of the implication }$\mathbb{S}_{1}\Longrightarrow\mathbb{S}%
_{3}$\textit{:} Assume that Assertion $\mathbb{S}_{1}$ holds. Then, there
exists a $\mathfrak{g}$-module homomorphism $f^{\prime}:B\rightarrow A$ such
that $f^{\prime}\circ f=\operatorname*{id}\nolimits_{A}$. Consider this
homomorphism $f^{\prime}$. Now let $P=\operatorname*{Ker}f^{\prime}$. Then,
$P$ is a $\mathfrak{g}$-submodule of $B$ (since $f^{\prime}$ is a
$\mathfrak{g}$-module homomorphism, while the kernel of every $\mathfrak{g}%
$-module homomorphism is a $\mathfrak{g}$-submodule). Every $b\in B$ satisfies%
\begin{align*}
f^{\prime}\left(  b-f\left(  f^{\prime}\left(  b\right)  \right)  \right)   &
=f^{\prime}\left(  b\right)  -\underbrace{f^{\prime}\left(  f\left(
f^{\prime}\left(  b\right)  \right)  \right)  }_{=\left(  f^{\prime}\circ
f\circ f^{\prime}\right)  \left(  b\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }f^{\prime}\text{ is }k\text{-linear}\right) \\
&  =f^{\prime}\left(  b\right)  -\left(  \underbrace{f^{\prime}\circ
f}_{=\operatorname*{id}\nolimits_{A}}\circ f^{\prime}\right)  \left(
b\right)  =f^{\prime}\left(  b\right)  -\underbrace{\left(  \operatorname*{id}%
\nolimits_{A}\circ f^{\prime}\right)  }_{=f^{\prime}}\left(  b\right)
=f^{\prime}\left(  b\right)  -f^{\prime}\left(  b\right)  =0,
\end{align*}
so that $b-f\left(  f^{\prime}\left(  b\right)  \right)  \in
\operatorname*{Ker}f^{\prime}=P$ and thus%
\[
b=f\left(  \underbrace{f^{\prime}\left(  b\right)  }_{\in A}\right)
+\underbrace{\left(  b-f\left(  f^{\prime}\left(  b\right)  \right)  \right)
}_{\in P}\in f\left(  A\right)  +P.
\]
Thus, $B=f\left(  A\right)  +P$. Now let us show that $f\left(  A\right)  \cap
P=0$:

Let $x\in f\left(  A\right)  \cap P$ be arbitrary. Then, $x\in f\left(
A\right)  \cap P\subseteq f\left(  A\right)  $, so there exists some $y\in A$
such that $x=f\left(  y\right)  $. Consider this $y$. But on the other hand,
$x\in f\left(  A\right)  \cap P\subseteq P=\operatorname*{Ker}f^{\prime}$, so
that $f^{\prime}\left(  x\right)  =0$. Thus, $0=f^{\prime}\left(
\underbrace{x}_{=f\left(  y\right)  }\right)  =f^{\prime}\left(  f\left(
y\right)  \right)  =\underbrace{\left(  f^{\prime}\circ f\right)
}_{=\operatorname*{id}\nolimits_{A}}\left(  y\right)  =\operatorname*{id}%
\nolimits_{A}\left(  y\right)  =y$, so that $x=f\left(  \underbrace{y}%
_{=0}\right)  =f\left(  0\right)  =0$ (since $f$ is $k$-linear). We have thus
proven that every $x\in f\left(  A\right)  \cap P$ satisfies $x=0$. This shows
that $f\left(  A\right)  \cap P=0$.

Since $B=f\left(  A\right)  +P$ and $f\left(  A\right)  \cap P=0$, we have
$B=f\left(  A\right)  \oplus P$. Thus, we have shown that there exists a
$\mathfrak{g}$-submodule $P$ of $B$ such that $B=f\left(  A\right)  \oplus P$.
In other words, Assertion $\mathbb{S}_{3}$ holds. This proves the implication
$\mathbb{S}_{1}\Longrightarrow\mathbb{S}_{3}$.

\textit{Proof of the implication }$\mathbb{S}_{2}\Longrightarrow\mathbb{S}%
_{3}$\textit{:} Assume that Assertion $\mathbb{S}_{2}$ holds. Then, there
exists a $\mathfrak{g}$-module homomorphism $g^{\prime}:C\rightarrow B$ such
that $g\circ g^{\prime}=\operatorname*{id}\nolimits_{C}$. Consider this
homomorphism $g^{\prime}$. Now let $P=g^{\prime}\left(  C\right)  $. Then, $P$
is a $\mathfrak{g}$-submodule of $B$ (since $g^{\prime}$ is a $\mathfrak{g}%
$-module homomorphism, while the image of every $\mathfrak{g}$-module
homomorphism is a $\mathfrak{g}$-submodule). Every $b\in B$ satisfies%
\begin{align*}
g\left(  b-g^{\prime}\left(  g\left(  b\right)  \right)  \right)   &
=g\left(  b\right)  -\underbrace{g\left(  g^{\prime}\left(  g\left(  b\right)
\right)  \right)  }_{=\left(  g\circ g^{\prime}\circ g\right)  \left(
b\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }g\text{ is }%
k\text{-linear}\right) \\
&  =g\left(  b\right)  -\left(  \underbrace{g\circ g^{\prime}}%
_{=\operatorname*{id}\nolimits_{C}}\circ g\right)  \left(  b\right)  =g\left(
b\right)  -\underbrace{\left(  \operatorname*{id}\nolimits_{C}\circ g\right)
}_{=g}\left(  b\right)  =g\left(  b\right)  -g\left(  b\right)  =0,
\end{align*}
so that $b-g^{\prime}\left(  g\left(  b\right)  \right)  \in
\operatorname*{Ker}g=f\left(  A\right)  $ (since $%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%0 \ar[r] & A \ar[r]^f & B \ar[r]^g & C \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrix{
0 \ar[r] & A \ar[r]^f & B \ar[r]^g & C \ar[r] & 0
}%
%EndExpansion
$ is an exact sequence) and thus%
\[
b=\underbrace{\left(  b-g^{\prime}\left(  g\left(  b\right)  \right)  \right)
}_{\in f\left(  A\right)  }+g^{\prime}\left(  \underbrace{g\left(  b\right)
}_{\in C}\right)  \in f\left(  A\right)  +\underbrace{g^{\prime}\left(
C\right)  }_{=P}=f\left(  A\right)  +P.
\]
Thus, $B=f\left(  A\right)  +P$. Now let us show that $f\left(  A\right)  \cap
P=0$:

Let $x\in f\left(  A\right)  \cap P$ be arbitrary. Then, $x\in f\left(
A\right)  \cap P\subseteq f\left(  A\right)  =\operatorname*{Ker}g$. But on
the other hand, $x\in f\left(  A\right)  \cap P\subseteq P=g^{\prime}\left(
C\right)  $, so that there exists some $z\in C$ such that $x=g^{\prime}\left(
z\right)  $. Consider this $z$. Thus, $g\left(  \underbrace{g^{\prime}\left(
z\right)  }_{=x}\right)  =g\left(  x\right)  =0$ (since $x\in
\operatorname*{Ker}g$). In other words, $0=g\left(  g^{\prime}\left(
z\right)  \right)  =\underbrace{\left(  g\circ g^{\prime}\right)
}_{=\operatorname*{id}\nolimits_{C}}\left(  z\right)  =\operatorname*{id}%
\nolimits_{C}\left(  z\right)  =z$, so that $x=g^{\prime}\left(
\underbrace{z}_{=0}\right)  =g^{\prime}\left(  0\right)  =0$ (since
$g^{\prime}$ is $k$-linear). We have thus proven that every $x\in f\left(
A\right)  \cap P$ satisfies $x=0$. This shows that $f\left(  A\right)  \cap
P=0$.

Since $B=f\left(  A\right)  +P$ and $f\left(  A\right)  \cap P=0$, we have
$B=f\left(  A\right)  \oplus P$. Thus, we have shown that there exists a
$\mathfrak{g}$-submodule $P$ of $B$ such that $B=f\left(  A\right)  \oplus P$.
In other words, Assertion $\mathbb{S}_{3}$ holds. This proves the implication
$\mathbb{S}_{2}\Longrightarrow\mathbb{S}_{3}$.

\textit{Proof of the implication }$\mathbb{S}_{3}\Longrightarrow\mathbb{S}%
_{1}$\textit{:} Assume that Assertion $\mathbb{S}_{3}$ holds. Then, there
exists a $\mathfrak{g}$-submodule $P$ of $B$ such that $B=f\left(  A\right)
\oplus P$. Consider this $\mathfrak{g}$-submodule $P$.

The $\mathfrak{g}$-module homomorphism $f:A\rightarrow B$ induces a
$\mathfrak{g}$-module homomorphism $f_{1}:A\rightarrow f\left(  A\right)  $
defined by $\left(  f_{1}\left(  a\right)  =f\left(  a\right)  \text{ for
every }a\in A\right)  $. Since $f$ is injective (because $%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%0 \ar[r] & A \ar[r]^f & B \ar[r]^g & C \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrix{
0 \ar[r] & A \ar[r]^f & B \ar[r]^g & C \ar[r] & 0
}%
%EndExpansion
$ is an exact sequence), this homomorphism $f_{1}$ is bijective. That is,
$f_{1}$ is a $k$-module isomorphism. Thus $f_{1}$ is a $\mathfrak{g}$-module
homomorphism and a $k$-module isomorphism at the same time. By an application
of Proposition \ref{prop.liemodiso}, this yields that $f_{1}$ is a
$\mathfrak{g}$-module isomorphism. Thus, its inverse $f_{1}^{-1}$ also is a
$\mathfrak{g}$-module isomorphism.

On the other hand, let $\pi:f\left(  A\right)  \oplus P\rightarrow f\left(
A\right)  $ be the projection from the direct sum $f\left(  A\right)  \oplus
P$ to its addend $f\left(  A\right)  $. Since $f\left(  A\right)  $ and $P$
are $\mathfrak{g}$-modules, this projection $\pi$ is a surjective
$\mathfrak{g}$-module homomorphism from $f\left(  A\right)  \oplus P$ onto
$f\left(  A\right)  $. Since $f\left(  A\right)  \oplus P=B$, this $\pi$ is
thus a surjective $\mathfrak{g}$-module homomorphism from $B$ onto $f\left(
A\right)  $.

Let $f^{\prime}=f_{1}^{-1}\circ\pi$. Then, $f^{\prime}$ is a $\mathfrak{g}%
$-module homomorphism (since it is the composition of the $\mathfrak{g}%
$-module homomorphisms $\pi$ and $f_{1}^{-1}$). Moreover, every $a\in A$
satisfies%
\[
\pi\left(  f\left(  a\right)  \right)  =f\left(  a\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\left(  a\right)  \in f\left(
A\right)  \text{, while }\pi\text{ is a projection onto }f\left(  A\right)
\right)
\]
and thus%
\[
\left(  \underbrace{f^{\prime}}_{=f_{1}^{-1}\circ\pi}\circ f\right)  \left(
a\right)  =\left(  f_{1}^{-1}\circ\pi\circ f\right)  \left(  a\right)
=f_{1}^{-1}\left(  \underbrace{\pi\left(  f\left(  a\right)  \right)
}_{=f\left(  a\right)  =f_{1}\left(  a\right)  }\right)  =f_{1}^{-1}\left(
f_{1}\left(  a\right)  \right)  =a=\operatorname*{id}\nolimits_{A}\left(
a\right)  .
\]
Thus, $f^{\prime}\circ f=\operatorname*{id}\nolimits_{A}$.

Thus, we have shown that there exists a $\mathfrak{g}$-module homomorphism
$f^{\prime}:B\rightarrow A$ such that $f^{\prime}\circ f=\operatorname*{id}%
\nolimits_{A}$. In other words, Assertion $\mathbb{S}_{1}$ holds. This proves
the implication $\mathbb{S}_{3}\Longrightarrow\mathbb{S}_{1}$.

\textit{Proof of the implication }$\mathbb{S}_{3}\Longrightarrow\mathbb{S}%
_{2}$\textit{:} Assume that Assertion $\mathbb{S}_{3}$ holds. Then, there
exists a $\mathfrak{g}$-submodule $P$ of $B$ such that $B=f\left(  A\right)
\oplus P$. Consider this $\mathfrak{g}$-submodule $P$.

Since $%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%0 \ar[r] & A \ar[r]^f & B \ar[r]^g & C \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrix{
0 \ar[r] & A \ar[r]^f & B \ar[r]^g & C \ar[r] & 0
}%
%EndExpansion
$ is an exact sequence, we have $g\left(  B\right)  =C$ and $f\left(
A\right)  =\operatorname*{Ker}g$.

Define a map $g_{1}:P\rightarrow C$ through $g_{1}=g\mid_{P}$. Then,%
\[
\operatorname*{Ker}g_{1}=\operatorname*{Ker}\left(  g\mid_{P}\right)
=\underbrace{\operatorname*{Ker}g}_{=f\left(  A\right)  }\cap P=f\left(
A\right)  \cap P=0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }B=f\left(
A\right)  \oplus P\right)  ,
\]
so that the map $g_{1}$ is injective.

On the other hand, $B=f\left(  A\right)  \oplus P=f\left(  A\right)  +P$, so
that%
\begin{align*}
g\left(  B\right)   &  =g\left(  f\left(  A\right)  +P\right)  =g\left(
\underbrace{f\left(  A\right)  }_{=\operatorname*{Ker}g}\right)
+\underbrace{g\left(  P\right)  }_{=\left(  g\mid_{P}\right)  \left(
P\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }g\text{ is }%
k\text{-linear}\right) \\
&  =\underbrace{g\left(  \operatorname*{Ker}g\right)  }_{=0}%
+\underbrace{\left(  g\mid_{P}\right)  }_{=g_{1}}\left(  P\right)
=0+g_{1}\left(  P\right)  =g_{1}\left(  P\right)  .
\end{align*}
Since $g\left(  B\right)  =C$, this becomes $C=g_{1}\left(  P\right)  $. Thus,
$g_{1}$ is surjective.

The map $g_{1}$ is bijective (since it is injective and surjective), and thus
it is a $k$-module isomorphism. An application of Proposition
\ref{prop.liemodiso} now shows that $g_{1}$ is a $\mathfrak{g}$-module
isomorphism (since it is a $\mathfrak{g}$-module homomorphism and a $k$-module
isomorphism). Thus, its inverse $g_{1}^{-1}$ also is a $\mathfrak{g}$-module isomorphism.

Let $\iota$ be the canonical inclusion of $P$ in $B$. Then, $\iota$ is a
$\mathfrak{g}$-module homomorphism.

Let $g^{\prime}=\iota\circ g_{1}^{-1}$. Then, $g^{\prime}$ is a $\mathfrak{g}%
$-module homomorphism (since it is the composition of the $\mathfrak{g}%
$-module homomorphisms $g_{1}^{-1}$ and $\iota$). Moreover, every $c\in C$
satisfies%
\begin{align*}
\left(  g\circ\underbrace{g^{\prime}}_{=\iota\circ g_{1}^{-1}}\right)  \left(
c\right)   &  =\left(  g\circ\iota\circ g_{1}^{-1}\right)  \left(  c\right)
=\underbrace{g\left(  \left(  \iota\circ g_{1}^{-1}\right)  \left(  c\right)
\right)  }_{\substack{=\left(  g\mid_{P}\right)  \left(  \left(  \iota\circ
g_{1}^{-1}\right)  \left(  c\right)  \right)  \\\text{(since }\left(
\iota\circ g_{1}^{-1}\right)  \left(  c\right)  \in P\text{)}}%
}=\underbrace{\left(  g\mid_{P}\right)  }_{=g_{1}}\left(  \left(  \iota\circ
g_{1}^{-1}\right)  \left(  c\right)  \right) \\
&  =g_{1}\left(  \underbrace{\left(  \iota\circ g_{1}^{-1}\right)  \left(
c\right)  }_{=\iota\left(  g_{1}^{-1}\left(  c\right)  \right)  }\right)
=g_{1}\left(  \underbrace{\iota\left(  g_{1}^{-1}\left(  c\right)  \right)
}_{\substack{=g_{1}^{-1}\left(  c\right)  \\\text{(since }\iota\text{ is just
an}\\\text{inclusion map)}}}\right)  =g_{1}\left(  g_{1}^{-1}\left(  c\right)
\right)  =c=\operatorname*{id}\nolimits_{C}\left(  c\right)  .
\end{align*}
Thus, $g\circ g^{\prime}=\operatorname*{id}\nolimits_{C}$.

Thus, we have shown that there exists a $\mathfrak{g}$-module homomorphism
$g^{\prime}:C\rightarrow B$ such that $g\circ g^{\prime}=\operatorname*{id}%
\nolimits_{C}$. In other words, Assertion $\mathbb{S}_{2}$ holds. This proves
the implication $\mathbb{S}_{3}\Longrightarrow\mathbb{S}_{2}$.

We have thus shown the four implications $\mathbb{S}_{1}\Longrightarrow
\mathbb{S}_{3}$, $\mathbb{S}_{2}\Longrightarrow\mathbb{S}_{3}$, $\mathbb{S}%
_{3}\Longrightarrow\mathbb{S}_{1}$ and $\mathbb{S}_{3}\Longrightarrow
\mathbb{S}_{2}$. These implications, when combined, yield the equivalence
$\mathbb{S}_{1}\Longleftrightarrow\mathbb{S}_{2}\Longleftrightarrow
\mathbb{S}_{3}$. This proves Proposition \ref{prop.g.split}.
\end{verlong}

Proposition \ref{prop.g.split} is in no way peculiar to $\mathfrak{g}%
$-modules. A completely analogous proposition (with exactly the same proof)
would hold if we replace $\mathfrak{g}$ by $A$ and "Lie algebra" by "algebra"
throughout Proposition \ref{prop.g.split}. More generally, an analogue of
Proposition \ref{prop.g.split} holds in every abelian category.

\begin{definition}
\label{defs.g.split}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $A$, $B$ and $C$ be three $\mathfrak{g}$-modules. Let
$f:A\rightarrow B$ and $g:B\rightarrow C$ be two $\mathfrak{g}$-module
homomorphisms such that%
\[%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%0 \ar[r] & A \ar[r]^f & B \ar[r]^g & C \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrix{
0 \ar[r] & A \ar[r]^f & B \ar[r]^g & C \ar[r] & 0
}%
%EndExpansion
\]
is a short exact sequence. Then, we say that \textit{the exact sequence }$%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%0 \ar[r] & A \ar[r]^f & B \ar[r]^g & C \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrix{
0 \ar[r] & A \ar[r]^f & B \ar[r]^g & C \ar[r] & 0
}%
%EndExpansion
$\textit{ of }$\mathfrak{g}$\textit{-modules is }$\mathfrak{g}$\textit{-split}
if and only if the three equivalent assertions $\mathbb{S}_{1}$,
$\mathbb{S}_{2}$ and $\mathbb{S}_{3}$ from Proposition \ref{prop.g.split} are satisfied.
\end{definition}

\begin{verlong}
\begin{corollary}
\label{cor.g.split}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $A$, $B$ and $C$ be three $\mathfrak{g}$-modules. Let
$f:A\rightarrow B$ and $g:B\rightarrow C$ be two $\mathfrak{g}$-module
homomorphisms such that%
\[%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%0 \ar[r] & A \ar[r]^f & B \ar[r]^g & C \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrix{
0 \ar[r] & A \ar[r]^f & B \ar[r]^g & C \ar[r] & 0
}%
%EndExpansion
\]
is a short exact sequence. Then, the exact sequence $%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%0 \ar[r] & A \ar[r]^f & B \ar[r]^g & C \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrix{
0 \ar[r] & A \ar[r]^f & B \ar[r]^g & C \ar[r] & 0
}%
%EndExpansion
$ of $\mathfrak{g}$-modules is $\mathfrak{g}$-split if and only if there
exists a $\mathfrak{g}$-submodule $P$ of $B$ such that $B=f\left(  A\right)
\oplus P$.
\end{corollary}

\textit{Proof of Corollary \ref{cor.g.split}.} We have the following
equivalence of assertions:%
\begin{align*}
&  \ \left(  \text{the exact sequence }%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%0 \ar[r] & A \ar[r]^f & B \ar[r]^g & C \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrix{
0 \ar[r] & A \ar[r]^f & B \ar[r]^g & C \ar[r] & 0
}%
%EndExpansion
\text{ of }\mathfrak{g}\text{-modules is }\mathfrak{g}\text{-split}\right) \\
\Longleftrightarrow &  \ \left(  \text{the three equivalent assertions
}\mathbb{S}_{1}\text{, }\mathbb{S}_{2}\text{ and }\mathbb{S}_{3}\text{ from
Proposition \ref{prop.g.split} are satisfied}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Definition \ref{defs.g.split}}\right)
\\
\Longleftrightarrow &  \ \left(  \text{the assertion }\mathbb{S}_{3}\text{
from Proposition \ref{prop.g.split} is satisfied}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because the assertions }\mathbb{S}_{1}\text{, }\mathbb{S}_{2}\text{ and
}\mathbb{S}_{3}\text{ from Proposition \ref{prop.g.split} are}\\
\text{equivalent, and thus they are all satisfied if any of them is satisfied}%
\end{array}
\right) \\
\Longleftrightarrow &  \ \left(  \text{there exists a }\mathfrak{g}%
\text{-submodule }P\text{ of }B\text{ such that }B=f\left(  A\right)  \oplus
P\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because Assertion }\mathbb{S}_{3}\text{ says that there exists a}\\
\mathfrak{g}\text{-submodule }P\text{ of }B\text{ such that }B=f\left(
A\right)  \oplus P
\end{array}
\right)  .
\end{align*}
This proves Corollary \ref{cor.g.split}.
\end{verlong}

\subsection{Filtrations of $\mathfrak{g}$-modules}

Now we are going to define the notion of a \textit{filtration} of a
$k$-module. Not that this is an unknown notion, but it is one of the most
overloaded notions in algebra (there are at least four different meanings of a
"filtration", and every author defines it to mean the one he wants), so let us
settle what \textit{we} are going to call a filtration:

\begin{definition}
\label{defs.filtr-k}Let $k$ be a commutative ring. Let $V$ be a $k$%
-module.\newline A $k$\textit{-module filtration} of $V$ will mean a sequence
$\left(  V_{n}\right)  _{n\geq0}$ of $k$-submodules of $V$ such that
$\bigcup\limits_{n\geq0}V_{n}=V$ and $V_{0}\subseteq V_{1}\subseteq
V_{2}\subseteq...$.
\end{definition}

Similarly we define the notion of a $\mathfrak{g}$-module filtration:

\begin{definition}
\label{defs.filtr-g}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $V$ be a $\mathfrak{g}$-module.\newline A $\mathfrak{g}%
$\textit{-module filtration} of $V$ will mean a sequence $\left(
V_{n}\right)  _{n\geq0}$ of $\mathfrak{g}$-submodules of $V$ such that
$\bigcup\limits_{n\geq0}V_{n}=V$ and $V_{0}\subseteq V_{1}\subseteq
V_{2}\subseteq...$.
\end{definition}

The following is obvious:

\begin{proposition}
\label{prop.filtr-k-g}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $V$ be a $\mathfrak{g}$-module.\newline\textbf{(a)} Every
$\mathfrak{g}$-module filtration of $V$ is a $k$-module filtration of
$V$.\newline\textbf{(b)} If $\left(  V_{n}\right)  _{n\geq0}$ is a $k$-module
filtration of $V$ such that $\left(  V_{n}\text{ is a }\mathfrak{g}%
\text{-submodule of }V\text{ for every }n\in\mathbb{N}\right)  $, then
$\left(  V_{n}\right)  _{n\geq0}$ is a $\mathfrak{g}$-module filtration of $V$.
\end{proposition}

We also record a definition:

\begin{definition}
\label{defs.lietensor.g-alg.filt}Let $k$ be a commutative ring. Let
$\mathfrak{g}$ be a $k$-Lie algebra.\newline Let $V$ be a $\mathfrak{g}%
$-module. Consider the tensor $\mathfrak{g}$-module $\otimes V$ of the
$\mathfrak{g}$-module $V$.\newline For every $n\in\mathbb{Z}$, let
$V^{\otimes\leq n}$ denote the $\mathfrak{g}$-submodule $\bigoplus
\limits_{i=0}^{n}V^{\otimes i}$ of the $\mathfrak{g}$-module $\otimes V$. (In
fact, $\bigoplus\limits_{i=0}^{n}V^{\otimes i}$ is a $\mathfrak{g}$-submodule
of $\otimes V$, because $\otimes V=\bigoplus\limits_{i\in\mathbb{N}}V^{\otimes
i}$ as a $\mathfrak{g}$-module.) Note that this definition yields
$V^{\otimes\leq n}=0$ for every integer $n<0$.\newline It is clear that
$\left(  V^{\otimes\leq n}\right)  _{n\geq0}$ is a $\mathfrak{g}$-module
filtration of $\otimes V$. This filtration is called the \textit{degree
filtration} of $\otimes V$.
\end{definition}

We will also use the notation $V^{\otimes\leq n}$ in a slightly more general context:

\begin{definition}
\label{defs.tensor.filt}Let $k$ be a commutative ring. Let $V$ be a
$k$-module. Consider the tensor $k$-module $\otimes V$ of the $k$-module
$V$.\newline For every $n\in\mathbb{N}$, let $V^{\otimes\leq n}$ denote the
$k$-submodule $\bigoplus\limits_{i=0}^{n}V^{\otimes i}$ of the $k$-module
$\otimes V$. It is clear that $\left(  V^{\otimes\leq n}\right)  _{n\geq0}$ is
a $k$-module filtration of $\otimes V$. This filtration is called the
\textit{degree filtration} of $\otimes V$.
\end{definition}

We notice a known fact:

\begin{proposition}
\label{prop.tensor.filt-is-alg-filt}Let $k$ be a commutative ring. Let $V$ be
a $k$-module. Then, considering $V^{\otimes n}$ as a $k$-submodule of $\otimes
V$ for every $n\in\mathbb{N}$, we have:\newline\textbf{(a)} Every
$i\in\mathbb{N}$ and $j\in\mathbb{N}$ satisfy $V^{\otimes i}\cdot V^{\otimes
j}=V^{\otimes\left(  i+j\right)  }$.\newline\textbf{(b)} Every $n\in
\mathbb{N}$ and $m\in\mathbb{N}$ satisfy $V^{\otimes\leq n}\cdot
V^{\otimes\leq m}\subseteq V^{\otimes\leq\left(  n+m\right)  }$.
\end{proposition}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.tensor.filt-is-alg-filt}.} \textbf{(a)}
Every $i\in\left\{  0,1,...,n\right\}  $ and $j\in\left\{  0,1,...,m\right\}
$ satisfy%
\begin{align*}
&  V^{\otimes i}\cdot V^{\otimes j}\\
&  =\left\langle ab\ \mid\ \left(  a,b\right)  \in V^{\otimes i}\times
V^{\otimes j}\right\rangle =\left\langle a\otimes b\ \mid\ \left(  a,b\right)
\in V^{\otimes i}\times V^{\otimes j}\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since every }\left(  a,b\right)  \in V^{\otimes i}\times V^{\otimes
j}\text{ satisfy }ab=a\cdot b=a\otimes b\\
\text{(according to (\ref{*=(X)}) (applied to }i\text{ and }j\text{ instead of
}n\text{ and }m\text{),}\\
\text{because }\left(  a,b\right)  \in V^{\otimes i}\times V^{\otimes j}\text{
yields }a\in V^{\otimes i}\text{ and }b\in V^{\otimes j}\text{)}%
\end{array}
\right) \\
&  =V^{\otimes i}\otimes V^{\otimes j}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since the tensor product }V^{\otimes i}\otimes V^{\otimes j}\text{ is
spanned by pure tensors,}\\
\text{i. e., by tensors of the form }a\otimes b\text{ for }\left(  a,b\right)
\in V^{\otimes i}\times V^{\otimes j}%
\end{array}
\right) \\
&  =V^{\otimes\left(  i+j\right)  }.
\end{align*}
This proves Proposition \ref{prop.tensor.filt-is-alg-filt} \textbf{(a)}.

\textbf{(b)} Definition \ref{defs.tensor.filt} yields%
\[
V^{\otimes\leq\left(  n+m\right)  }=\bigoplus\limits_{i=0}^{n+m}V^{\otimes
i}=\bigoplus\limits_{r=0}^{n+m}V^{\otimes r}\ \ \ \ \ \ \ \ \ \ \left(
\text{here we renamed }i\text{ as }r\text{ in the sum}\right)  .
\]


On the other hand, Proposition \ref{prop.tensor.filt-is-alg-filt} \textbf{(a)}
yields that every $i\in\left\{  0,1,...,n\right\}  $ and $j\in\left\{
0,1,...,m\right\}  $ satisfy $V^{\otimes i}\cdot V^{\otimes j}=V^{\otimes
\left(  i+j\right)  }\subseteq\bigoplus\limits_{r=0}^{n+m}V^{\otimes r}$
(because $i\in\left\{  0,1,...,n\right\}  $ and $j\in\left\{
0,1,...,m\right\}  $ yield $i+j\in\left\{  0,1,...,n+m\right\}  $, so that
$V^{\otimes\left(  i+j\right)  }$ is one of the addends of the direct sum
$\bigoplus\limits_{r=0}^{n+m}V^{\otimes r}$). Since $\bigoplus\limits_{r=0}%
^{n+m}V^{\otimes r}=V^{\otimes\leq\left(  n+m\right)  }$, we have thus proven
that every $i\in\left\{  0,1,...,n\right\}  $ and $j\in\left\{
0,1,...,m\right\}  $ satisfy $V^{\otimes i}\cdot V^{\otimes j}\subseteq
V^{\otimes\leq\left(  n+m\right)  }$.

Now, Definition \ref{defs.tensor.filt} yields $V^{\otimes\leq n}%
=\bigoplus\limits_{i=0}^{n}V^{\otimes i}$, $V^{\otimes\leq m}=\bigoplus
\limits_{i=0}^{m}V^{\otimes i}$ and $V^{\otimes\leq\left(  n+m\right)
}=\bigoplus\limits_{i=0}^{n+m}V^{\otimes i}$.

Since $V^{\otimes\leq n}=\bigoplus\limits_{i=0}^{n}V^{\otimes i}%
=\sum\limits_{i=0}^{n}V^{\otimes i}$ (since direct sums are sums) and
\begin{align*}
V^{\otimes\leq m}  &  =\bigoplus\limits_{i=0}^{m}V^{\otimes i}=\bigoplus
\limits_{j=0}^{m}V^{\otimes j}\ \ \ \ \ \ \ \ \ \ \left(  \text{here we
renamed }i\text{ as }j\text{ in the sum}\right) \\
&  =\sum\limits_{j=0}^{m}V^{\otimes j}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
direct sums are sums}\right)  ,
\end{align*}
we have%
\begin{align*}
\underbrace{V^{\otimes\leq n}}_{=\sum\limits_{i=0}^{n}V^{\otimes i}}%
\cdot\underbrace{V^{\otimes\leq m}}_{=\sum\limits_{j=0}^{m}V^{\otimes j}}  &
=\left(  \sum\limits_{i=0}^{n}V^{\otimes i}\right)  \cdot\left(
\sum\limits_{j=0}^{m}V^{\otimes j}\right)  =\sum\limits_{i=0}^{n}%
\sum\limits_{j=0}^{m}\underbrace{V^{\otimes i}\cdot V^{\otimes j}}_{\subseteq
V^{\otimes\leq\left(  n+m\right)  }}\subseteq\sum\limits_{i=0}^{n}%
\sum\limits_{j=0}^{m}V^{\otimes\leq\left(  n+m\right)  }\\
&  \subseteq V^{\otimes\leq\left(  n+m\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }V^{\otimes\leq\left(  n+m\right)  }\text{ is a }k\text{-module}%
\right)  .
\end{align*}
This proves Proposition \ref{prop.tensor.filt-is-alg-filt} \textbf{(b)}.
\end{verlong}

Our next definition is concerned with $k$-module homomorphisms:

\begin{definition}
\label{defs.respect}Let $k$ be a commutative ring. Let $V$ and $W$ be two
$k$-modules. Let $f:V\rightarrow W$ be a $k$-module homomorphism. Let $\left(
V_{n}\right)  _{n\geq0}$ be a $k$-module filtration of $V$, and let $\left(
W_{n}\right)  _{n\geq0}$ be a $k$-module filtration of $W$.\newline We say
that \textit{the map }$f$ \textit{respects the filtrations }$\left(
V_{n}\right)  _{n\geq0}$ \textit{and }$\left(  W_{n}\right)  _{n\geq0}$ if it
satisfies $\left(  f\left(  V_{n}\right)  \subseteq W_{n}\text{ for every
}n\in\mathbb{N}\right)  $. Sometimes we abbreviate "the map $f$ respects the
filtrations $\left(  V_{n}\right)  _{n\geq0}$ and $\left(  W_{n}\right)
_{n\geq0}$\ \ \ \ " to "\textit{the map }$f$\textit{ respects the
filtration}", as long as the filtrations $\left(  V_{n}\right)  _{n\geq0}$ and
$\left(  W_{n}\right)  _{n\geq0}$ can be inferred from the context.
\end{definition}

Filtrations of $k$-modules and homomorphisms respecting them lead to new
modules rsp. homomorphisms:

\begin{definition}
\label{defs.gr}Let $k$ be a commutative ring.\newline\textbf{(a)} Let $V$ be a
$k$-module. Let $\left(  V_{n}\right)  _{n\geq0}$ be a $k$-module filtration
of $V$. Then, for every $p\in\mathbb{N}$, we denote the $k$-module
$V_{p}\diagup V_{p-1}$ (where $V_{-1}$ means $0$) by $\operatorname*{gr}%
\nolimits_{p}\left(  V,\left(  V_{n}\right)  _{n\geq0}\right)  $. When the
filtration $\left(  V_{n}\right)  _{n\geq0}$ is clear from the context, we
will abbreviate $\operatorname*{gr}\nolimits_{p}\left(  V,\left(
V_{n}\right)  _{n\geq0}\right)  $ by $\operatorname*{gr}\nolimits_{p}%
V$.\newline\textbf{(b)} Let $V$ and $W$ be two $k$-modules. Let $\left(
V_{n}\right)  _{n\geq0}$ be a $k$-module filtration of $V$, and let $\left(
W_{n}\right)  _{n\geq0}$ be a $k$-module filtration of $W$. Let
$f:V\rightarrow W$ be a $k$-module homomorphism respecting the
filtration.{\footnotemark} Then, for every $p\in\mathbb{N}$, we can define a
$k$-module homomorphism $\operatorname*{gr}\nolimits_{p}f:\operatorname*{gr}%
\nolimits_{p}V\rightarrow\operatorname*{gr}\nolimits_{p}W$ as
follows{\footnotemark}: Since $f$ respects the filtration, we have $f\left(
V_{p}\right)  \subseteq W_{p}$. Thus, $f$ induces a $k$-module homomorphism
$f_{p}:V_{p}\rightarrow W_{p}$ defined by $\left(  f_{p}\left(  v\right)
=f\left(  v\right)  \text{ for every }v\in V_{p}\right)  $. This homomorphism
$f_{p}$ sends $V_{p-1}$ to $W_{p-1}$ (since $f\left(  V_{p-1}\right)
\subseteq W_{p-1}$, which is because $f$ respects the filtration), and thus
gives rise to a $k$-module homomorphism $f_{p}^{\prime}:V_{p}\diagup
V_{p-1}\rightarrow W_{p}\diagup W_{p-1}$ which satisfies $\left(
f_{p}^{\prime}\left(  \overline{v}\right)  =\overline{f_{p}\left(  v\right)
}\text{ for every }v\in V_{p}\right)  $ (where $\overline{v}$ denotes the
residue class of $v$ modulo $V_{p-1}$, and $\overline{f_{p}\left(  v\right)
}$ denotes the residue class of $f_{p}\left(  v\right)  $ modulo $W_{p-1}$).
Since $V_{p}\diagup V_{p-1}=\operatorname*{gr}\nolimits_{p}V$ and
$W_{p}\diagup W_{p-1}=\operatorname*{gr}\nolimits_{p}W$, this $k$-module
homomorphism $f_{p}^{\prime}:V_{p}\diagup V_{p-1}\rightarrow W_{p}\diagup
W_{p-1}$ is a $k$-module homomorphism $f_{p}^{\prime}:\operatorname*{gr}%
\nolimits_{p}V\rightarrow\operatorname*{gr}\nolimits_{p}W$. We will denote
this homomorphism $f_{p}^{\prime}$ by $\operatorname*{gr}\nolimits_{p}f$.
(Strictly speaking, the notation $\operatorname*{gr}\nolimits_{p}f$ is
ambiguous, because the homomorphism $\operatorname*{gr}\nolimits_{p}f$ depends
not only on $p$ and $f$, but also on the filtrations $\left(  V_{n}\right)
_{n\geq0}$ and $\left(  W_{n}\right)  _{n\geq0}$. But we will never run into
ambiguities with this notation, because in our cases the filtrations $\left(
V_{n}\right)  _{n\geq0}$ and $\left(  W_{n}\right)  _{n\geq0}$ will always be
clear form the context.)
\end{definition}

\addtocounter{footnote}{-1} \addtocounter{Hfootnote}{-1} \footnotetext{Of
course, "respecting the filtration" means "respecting the filtrations $\left(
V_{n}\right)  _{n\geq0}$ and $\left(  W_{n}\right)  _{n\geq0}$\ \ \ \ " here,
because the only filtrations of $V$ and $W$ inferrable from the context are
$\left(  V_{n}\right)  _{n\geq0}$ and $\left(  W_{n}\right)  _{n\geq0}$.}
\addtocounter{footnote}{1} \addtocounter{Hfootnote}{1} \footnotetext{Here,
$\operatorname*{gr}\nolimits_{p}V$ means $\operatorname*{gr}\nolimits_{p}%
\left(  V,\left(  V_{n}\right)  _{n\geq0}\right)  $, and $\operatorname*{gr}%
\nolimits_{p}W$ means $\operatorname*{gr}\nolimits_{p}\left(  W,\left(
W_{n}\right)  _{n\geq0}\right)  $, because the only filtrations of $V$ and $W$
inferrable from the context are $\left(  V_{n}\right)  _{n\geq0}$ and $\left(
W_{n}\right)  _{n\geq0}$.}

As one could expect, we can get additional structure if we start at
$\mathfrak{g}$-modules and $\mathfrak{g}$-module homomorphisms:

\begin{proposition}
\label{prop.gr.lie}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra.\newline\textbf{(a)} Let $V$ be a $\mathfrak{g}$-module. Let
$\left(  V_{n}\right)  _{n\geq0}$ be a $\mathfrak{g}$-module filtration of
$V$. Then, for every $p\in\mathbb{N}$, the $k$-module $\operatorname*{gr}%
\nolimits_{p}V$ canonically becomes a $\mathfrak{g}$-module, since it is the
quotient module $V_{p}\diagup V_{p-1}$ and since both $V_{p}$ and $V_{p-1}$
are $\mathfrak{g}$-modules.\newline\textbf{(b)} Let $V$ and $W$ be two
$\mathfrak{g}$-modules. Let $\left(  V_{n}\right)  _{n\geq0}$ be a
$\mathfrak{g}$-module filtration of $V$, and let $\left(  W_{n}\right)
_{n\geq0}$ be a $\mathfrak{g}$-module filtration of $W$. Let $f:V\rightarrow
W$ be a $\mathfrak{g}$-module homomorphism respecting the filtration. Then,
for every $p\in\mathbb{N}$, the $k$-module homomorphism $\operatorname*{gr}%
\nolimits_{p}f:\operatorname*{gr}\nolimits_{p}V\rightarrow\operatorname*{gr}%
\nolimits_{p}W$ is a $\mathfrak{g}$-module homomorphism.
\end{proposition}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.gr.lie}.} Part \textbf{(a)} of this
proposition is obviously true (there is nothing to prove here). Part
\textbf{(b)} follows from the definitions (the details are left to the reader).
\end{verlong}

The following fact is also easy to see:

\begin{proposition}
\label{prop.gr.functor}Let $k$ be a commutative ring.\newline\textbf{(a)} Let
$V$ be a $k$-module. Let $\left(  V_{n}\right)  _{n\geq0}$ be a $k$-module
filtration of $V$. Then, the map $\operatorname*{id}:V\rightarrow V$ respects
the filtration, and satisfies $\operatorname*{gr}\nolimits_{p}%
\operatorname*{id}=\operatorname*{id}$ for every $p\in\mathbb{N}$%
.\newline\textbf{(b)} Let $U$, $V$ and $W$ be three $k$-modules. Let $\left(
U_{n}\right)  _{n\geq0}$ be a $k$-module filtration of $U$. Let $\left(
V_{n}\right)  _{n\geq0}$ be a $k$-module filtration of $V$. Let $\left(
W_{n}\right)  _{n\geq0}$ be a $k$-module filtration of $W$. Let
$f:U\rightarrow V$ and $g:V\rightarrow W$ be two $k$-module homomorphisms
respecting the filtration. Then, the homomorphism $g\circ f:U\rightarrow W$
also respects the filtration and satisfies $\operatorname*{gr}\nolimits_{p}%
g\circ\operatorname*{gr}\nolimits_{p}f=\operatorname*{gr}\nolimits_{p}\left(
g\circ f\right)  $ for every $p\in\mathbb{N}$.\newline\textbf{(c)} Let $V$ and
$W$ be two $k$-modules. Let $\left(  V_{n}\right)  _{n\geq0}$ be a $k$-module
filtration of $V$. Let $\left(  W_{n}\right)  _{n\geq0}$ be a $k$-module
filtration of $W$. Let $f:V\rightarrow W$ and $g:V\rightarrow W$ be two
$k$-module homomorphisms respecting the filtration. Then, $f-g:V\rightarrow W$
also respects the filtration, and satisfies $\operatorname*{gr}\nolimits_{p}%
\left(  f-g\right)  =\operatorname*{gr}\nolimits_{p}f-\operatorname*{gr}%
\nolimits_{p}g$ for every $p\in\mathbb{N}$.\newline\textbf{(d)} Let $V$ and
$W$ be two $k$-modules. Let $\left(  V_{n}\right)  _{n\geq0}$ be a $k$-module
filtration of $V$. Let $\left(  W_{n}\right)  _{n\geq0}$ be a $k$-module
filtration of $W$. Then, the $k$-module homomorphism $0:V\rightarrow W$ (which
maps everything to $0$) respects the filtration and satisfies
$\operatorname*{gr}\nolimits_{p}0=0$ for every $p\in\mathbb{N}$.
\end{proposition}

In the language of category theory, Proposition \ref{prop.gr.functor} says
that for each $p\in\mathbb{N}$, Definition \ref{defs.gr} defines an additive
functor $\operatorname*{gr}\nolimits_{p}$ from the category of $k$-modules
with filtration (where morphisms are $k$-module homomorphisms respecting the
filtration) to the category of $k$-modules.

\begin{conclusion}
\label{warn.gr.iso}Filtrations of $k$-modules have one somewhat dangerous
property: If we have two $k$-modules $V$ and $W$ with filtrations $\left(
V_{n}\right)  _{n\geq0}$ and $\left(  W_{n}\right)  _{n\geq0}$, respectively,
and an isomorphism $f:V\rightarrow W$ of $k$-modules which respects the
filtration, then we \textit{cannot} (in general) be sure that
$\operatorname*{gr}\nolimits_{p}f:\operatorname*{gr}\nolimits_{p}%
V\rightarrow\operatorname*{gr}\nolimits_{p}W$ is an isomorphism for every
$p\in\mathbb{N}$. In order to be able to tell that $\operatorname*{gr}%
\nolimits_{p}f$ is an isomorphism, we need to require that $f^{-1}$ also
respect the filtration. This is enough due to the following fact:
\end{conclusion}

\begin{proposition}
\label{prop.gr.iso}Let $k$ be a commutative ring. Let $V$ and $W$ be two
$k$-modules. Let $\left(  V_{n}\right)  _{n\geq0}$ be a $k$-module filtration
of $V$, and let $\left(  W_{n}\right)  _{n\geq0}$ be a $k$-module filtration
of $W$. Let $f:V\rightarrow W$ be a $k$-module isomorphism. If each of the
maps $f$ and $f^{-1}$ respects the filtration, then $\operatorname*{gr}%
\nolimits_{p}f:\operatorname*{gr}\nolimits_{p}V\rightarrow\operatorname*{gr}%
\nolimits_{p}W$ is a $k$-module isomorphism for every $p\in\mathbb{N}$.
\end{proposition}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.gr.iso}.} Assume that each of the maps
$f$ and $f^{-1}$ respects the filtration. Then, according to Definition
\ref{defs.gr} \textbf{(b)}, both maps $\operatorname*{gr}\nolimits_{p}%
f:\operatorname*{gr}\nolimits_{p}V\rightarrow\operatorname*{gr}\nolimits_{p}W$
and $\operatorname*{gr}\nolimits_{p}\left(  f^{-1}\right)  :\operatorname*{gr}%
\nolimits_{p}W\rightarrow\operatorname*{gr}\nolimits_{p}V$ are well-defined.
Proposition \ref{prop.gr.functor} \textbf{(b)} (applied to $W$, $\left(
W_{n}\right)  _{n\geq0}$, $f^{-1}$ and $f$ instead of $U$, $\left(
U_{n}\right)  _{n\geq0}$, $f$ and $g$) now yields that $\operatorname*{gr}%
\nolimits_{p}f\circ\operatorname*{gr}\nolimits_{p}\left(  f^{-1}\right)
=\operatorname*{gr}\nolimits_{p}\left(  f\circ f^{-1}\right)  $ for every
$p\in\mathbb{N}$. On the other hand, Proposition \ref{prop.gr.functor}
\textbf{(b)} (applied to $V$, $\left(  V_{n}\right)  _{n\geq0}$, $W$, $\left(
W_{n}\right)  _{n\geq0}$, $V$, $\left(  V_{n}\right)  _{n\geq0}$ and $f^{-1}$
instead of $U$, $\left(  U_{n}\right)  _{n\geq0}$, $V$, $\left(  V_{n}\right)
_{n\geq0}$, $W$, $\left(  W_{n}\right)  _{n\geq0}$ and $g$) now yields that
$\operatorname*{gr}\nolimits_{p}\left(  f^{-1}\right)  \circ\operatorname*{gr}%
\nolimits_{p}f=\operatorname*{gr}\nolimits_{p}\left(  f^{-1}\circ f\right)  $
for every $p\in\mathbb{N}$. Thus, every $p\in\mathbb{N}$ satisfies
$\operatorname*{gr}\nolimits_{p}f\circ\operatorname*{gr}\nolimits_{p}\left(
f^{-1}\right)  =\operatorname*{gr}\nolimits_{p}\underbrace{\left(  f\circ
f^{-1}\right)  }_{=\operatorname*{id}}=\operatorname*{gr}\nolimits_{p}%
\operatorname*{id}=\operatorname*{id}$ (according to Proposition
\ref{prop.gr.functor} \textbf{(a)}) and $\operatorname*{gr}\nolimits_{p}%
\left(  f^{-1}\right)  \circ\operatorname*{gr}\nolimits_{p}%
f=\operatorname*{gr}\nolimits_{p}\underbrace{\left(  f^{-1}\circ f\right)
}_{=\operatorname*{id}}=\operatorname*{gr}\nolimits_{p}\operatorname*{id}%
=\operatorname*{id}$ (according to Proposition \ref{prop.gr.functor}
\textbf{(a)}). This means that for every $p\in\mathbb{N}$, the two $k$-module
homomorphisms $\operatorname*{gr}\nolimits_{p}f$ and $\operatorname*{gr}%
\nolimits_{p}\left(  f^{-1}\right)  $ are inverse to each other. Thus,
$\operatorname*{gr}\nolimits_{p}f$ is a $k$-module isomorphism. This proves
Proposition \ref{prop.gr.iso}.
\end{verlong}

Another easy fact:

\begin{proposition}
\label{prop.gr.surj}Let $k$ be a commutative ring. Let $V$ and $W$ be two
$k$-modules. Let $\left(  V_{n}\right)  _{n\geq0}$ be a $k$-module filtration
of $V$, and let $\left(  W_{n}\right)  _{n\geq0}$ be a $k$-module filtration
of $W$. Let $f:V\rightarrow W$ be a $k$-module homomorphism which respects the
filtration. Let $p\in\mathbb{N}$. If $f\left(  V_{p}\right)  =W_{p}$, then
$\operatorname*{gr}\nolimits_{p}f:\operatorname*{gr}\nolimits_{p}%
V\rightarrow\operatorname*{gr}\nolimits_{p}W$ is surjective.
\end{proposition}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.gr.surj}.} Assume that $f\left(
V_{p}\right)  =W_{p}$.

We are going to use the notations introduced in Definition \ref{defs.gr}
\textbf{(b)}. With these notations, $\operatorname*{gr}\nolimits_{p}%
f=f_{p}^{\prime}$ (because this is how $\operatorname*{gr}\nolimits_{p}f$ was
defined), $\left(  f_{p}^{\prime}\left(  \overline{v}\right)  =\overline
{f_{p}\left(  v\right)  }\text{ for every }v\in V_{p}\right)  $ and $\left(
f_{p}\left(  v\right)  =f\left(  v\right)  \text{ for every }v\in
V_{p}\right)  $.

Now let $w\in\operatorname*{gr}\nolimits_{p}W$ be arbitrary. Then,
$w\in\operatorname*{gr}\nolimits_{p}W=W_{p}\diagup W_{p-1}$, so there exists
some $u\in W_{p}$ such that $w=\overline{u}$ (where $\overline{u}$ denotes the
residue class of $w$ modulo $W_{p-1}$). Since $u\in W_{p}=f\left(
V_{p}\right)  $, there exists some $v\in V_{p}$ such that $u=f\left(
v\right)  $. Thus, $\underbrace{\left(  \operatorname*{gr}\nolimits_{p}%
f\right)  }_{=f_{p}^{\prime}}\left(  \overline{v}\right)  =f_{p}^{\prime
}\left(  \overline{v}\right)  =\overline{f_{p}\left(  v\right)  }=\overline
{u}$ (since $f_{p}\left(  v\right)  =f\left(  v\right)  =u$) leads to $\left(
\operatorname*{gr}\nolimits_{p}f\right)  \left(  \overline{v}\right)
=\overline{u}=w$. Thus, $w\in\left(  \operatorname*{gr}\nolimits_{p}f\right)
\left(  \operatorname*{gr}\nolimits_{p}U\right)  $. Hence, we have shown that
every $w\in\operatorname*{gr}\nolimits_{p}W$ satisfies $w\in\left(
\operatorname*{gr}\nolimits_{p}f\right)  \left(  \operatorname*{gr}%
\nolimits_{p}U\right)  $. In other words, $\operatorname*{gr}\nolimits_{p}%
W\subseteq\left(  \operatorname*{gr}\nolimits_{p}f\right)  \left(
\operatorname*{gr}\nolimits_{p}U\right)  $. Thus, $\operatorname*{gr}%
\nolimits_{p}f:\operatorname*{gr}\nolimits_{p}V\rightarrow\operatorname*{gr}%
\nolimits_{p}W$ is surjective. Proposition \ref{prop.gr.surj} is now proven.
\end{verlong}

\begin{proposition}
\label{prop.gr.0}Let $k$ be a commutative ring. Let $V$ and $W$ be two
$k$-modules. Let $\left(  V_{n}\right)  _{n\geq0}$ be a $k$-module filtration
of $V$, and let $\left(  W_{n}\right)  _{n\geq0}$ be a $k$-module filtration
of $W$. Let $f:V\rightarrow W$ be a $k$-module homomorphism which respects the
filtration. Let $p\in\mathbb{N}$. If $f\left(  V_{p}\right)  \subseteq
W_{p-1}$, then $\operatorname*{gr}\nolimits_{p}f=0$.
\end{proposition}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.gr.0}.} Assume that $f\left(
V_{p}\right)  \subseteq W_{p-1}$.

We are going to use the notations introduced in Definition \ref{defs.gr}
\textbf{(b)}. With these notations, $\operatorname*{gr}\nolimits_{p}%
f=f_{p}^{\prime}$ (because this is how $\operatorname*{gr}\nolimits_{p}f$ was
defined), $\left(  f_{p}^{\prime}\left(  \overline{v}\right)  =\overline
{f_{p}\left(  v\right)  }\text{ for every }v\in V_{p}\right)  $ and $\left(
f_{p}\left(  v\right)  =f\left(  v\right)  \text{ for every }v\in
V_{p}\right)  $. This means that every $v\in V_{p}$ satisfies $f_{p}^{\prime
}\left(  \overline{v}\right)  =\overline{f_{p}\left(  v\right)  }=0$ (because
$v\in V_{p}$ yields $f\left(  v\right)  \in f\left(  V_{p}\right)  \subseteq
W_{p-1}$ and thus $f_{p}\left(  v\right)  =f\left(  v\right)  \in W_{p-1}$).
Since every element $x\in V_{p}\diagup V_{p-1}$ can be written in the form
$x=\overline{v}$ for some $v\in V_{p}$, this yields that $f_{p}^{\prime
}\left(  x\right)  =0$ for every $x\in V_{p}\diagup V_{p-1}$ (because writing
$x$ in the form $x=\overline{v}$ for some $v\in V_{p}$, we get $f_{p}^{\prime
}\left(  x\right)  =f_{p}^{\prime}\left(  \overline{v}\right)  =0$). In other
words, $f_{p}^{\prime}=0$. Thus, $\operatorname*{gr}\nolimits_{p}%
f=f_{p}^{\prime}=0$. Proposition \ref{prop.gr.0} is now proven.
\end{verlong}

One more triviality:

\begin{proposition}
\label{prop.tensor.filt.(X)f}Let $k$ be a commutative ring. Let $V$ and $W$ be
two $k$-modules. Let $f:V\rightarrow W$ be a $k$-module homomorphism. Then,
the $k$-module homomorphism $\otimes f:\otimes V\rightarrow\otimes W$ respects
the filtration, where the filtrations on $\otimes V$ and $\otimes W$ are the
degree filtrations (i. e., the filtration on $\otimes V$ is $\left(
V^{\otimes\leq n}\right)  _{n\geq0}$, and the filtration on $\otimes W$ is
$\left(  W^{\otimes\leq n}\right)  _{n\geq0}$).
\end{proposition}

Another definition related to degree filtrations:

\begin{definition}
\label{defs.filtr.grad}Let $k$ be a commutative ring. Let $V$ be a $k$-module.
Let $p\in\mathbb{N}$. Let $\operatorname*{grad}\nolimits_{V,p}$ denote the map
$V^{\otimes p}\rightarrow V^{\otimes\leq p}\diagup V^{\otimes\leq\left(
p-1\right)  }$ which sends every $T\in V^{\otimes p}$ to the equivalence class
of $T\in V^{\otimes p}\subseteq V^{\otimes\leq p}$ modulo $V^{\otimes
\leq\left(  p-1\right)  }$.\newline Since $V^{\otimes\leq p}\diagup
V^{\otimes\leq\left(  p-1\right)  }=\operatorname*{gr}\nolimits_{p}\left(
\otimes V\right)  $ (because the filtration on $\otimes V$ is the degree
filtration $\left(  V^{\otimes\leq n}\right)  _{n\geq0}$, and thus
$\operatorname*{gr}\nolimits_{p}\left(  \otimes V\right)  $ is defined as
$V^{\otimes\leq p}\diagup V^{\otimes\leq\left(  p-1\right)  }$), we see that
this map $\operatorname*{grad}\nolimits_{V,p}:V^{\otimes p}\rightarrow
V^{\otimes\leq p}\diagup V^{\otimes\leq\left(  p-1\right)  }$ is a map
$V^{\otimes p}\rightarrow\operatorname*{gr}\nolimits_{p}\left(  \otimes
V\right)  $.
\end{definition}

\begin{proposition}
\label{prop.filtr.grad}Let $k$ be a commutative ring. Let $V$ be a $k$-module.
Let $p\in\mathbb{N}$. Then, the map $\operatorname*{grad}\nolimits_{V,p}%
:V^{\otimes p}\rightarrow\operatorname*{gr}\nolimits_{p}\left(  \otimes
V\right)  $ defined in Definition \ref{defs.filtr.grad} is a canonical
$k$-module isomorphism.
\end{proposition}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.filtr.grad}.} We have $V^{\otimes
\leq\left(  p-1\right)  }=\bigoplus\limits_{i=0}^{p-1}V^{\otimes i}$ (by the
definition of $V^{\otimes\leq\left(  p-1\right)  }$) and
\begin{align*}
V^{\otimes\leq p}  &  =\bigoplus\limits_{i=0}^{p}V^{\otimes i}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by the definition of }V^{\otimes\leq
p}\right) \\
&  =\underbrace{\left(  \bigoplus\limits_{i=0}^{p-1}V^{\otimes i}\right)
}_{=V^{\otimes\leq\left(  p-1\right)  }}\oplus V^{\otimes p}=V^{\otimes
\leq\left(  p-1\right)  }\oplus V^{\otimes p}.
\end{align*}


Now, we need a known lemma from algebra:

\begin{lemma}
\label{lem.filtr.grad}Let $A$ and $B$ be two $k$-modules. Consider $A$ as a
$k$-submodule of $A\oplus B$. Then, the map $B\rightarrow\left(  A\oplus
B\right)  \diagup A$ which sends every $T\in B$ to the equivalence class of
$T\in B\subseteq A\oplus B$ modulo $A$ is a $k$-module isomorphism.
\end{lemma}

Lemma \ref{lem.filtr.grad} (applied to $A=V^{\otimes\leq\left(  p-1\right)  }$
and $B=V^{\otimes p}$) yields that the map $V^{\otimes p}\rightarrow\left(
V^{\otimes\leq\left(  p-1\right)  }\oplus V^{\otimes p}\right)  \diagup
V^{\otimes\leq\left(  p-1\right)  }$ which sends every $T\in V^{\otimes p}$ to
the equivalence class of $T\in V^{\otimes p}\subseteq V^{\otimes\leq\left(
p-1\right)  }\oplus V^{\otimes p}$ modulo $V^{\otimes\leq\left(  p-1\right)
}$ is a $k$-module isomorphism. Since $V^{\otimes\leq\left(  p-1\right)
}\oplus V^{\otimes p}=V^{\otimes\leq p}$, this rewrites as follows: The map
$V^{\otimes p}\rightarrow V^{\otimes\leq p}\diagup V^{\otimes\leq\left(
p-1\right)  }$ which sends every $T\in V^{\otimes p}$ to the equivalence class
of $T\in V^{\otimes p}\subseteq V^{\otimes\leq\left(  p-1\right)  }\oplus
V^{\otimes p}$ modulo $V^{\otimes\leq\left(  p-1\right)  }$ is a $k$-module
isomorphism. Since the map $V^{\otimes p}\rightarrow V^{\otimes\leq p}\diagup
V^{\otimes\leq\left(  p-1\right)  }$ which sends every $T\in V^{\otimes p}$ to
the equivalence class of $T\in V^{\otimes p}\subseteq V^{\otimes\leq\left(
p-1\right)  }\oplus V^{\otimes p}$ modulo $V^{\otimes\leq\left(  p-1\right)
}$ is simply the map $\operatorname*{grad}\nolimits_{V,p}$ (by the definition
of $\operatorname*{grad}\nolimits_{V,p}$), this yields that the map
$\operatorname*{grad}\nolimits_{V,p}$ is a $k$-module isomorphism. Since this
map $\operatorname*{grad}\nolimits_{V,p}$ is clearly canonical, we have thus
proven Proposition \ref{prop.filtr.grad}.
\end{verlong}

Also, we can easily see that:

\begin{proposition}
\label{prop.filtr.grad.g}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a $k$-Lie algebra. Let $V$ be a $\mathfrak{g}$-module. Let $p\in\mathbb{N}$.
Then, the map $\operatorname*{grad}\nolimits_{V,p}:V^{\otimes p}%
\rightarrow\operatorname*{gr}\nolimits_{p}\left(  \otimes V\right)  $ defined
in Definition \ref{defs.filtr.grad} is a canonical $\mathfrak{g}$-module isomorphism.
\end{proposition}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.filtr.grad.g}.} In order to obtain a
proof of Proposition \ref{prop.filtr.grad.g}, it is enough to apply the
following changes to the proof of Proposition \ref{prop.filtr.grad}:

\begin{itemize}
\item Replace "Proposition \ref{prop.filtr.grad}" by "Proposition
\ref{prop.filtr.grad.g}".

\item Replace the words "$k$-module" by "$\mathfrak{g}$-module".
\end{itemize}

This completes the proof of Proposition \ref{prop.filtr.grad.g}.
\end{verlong}

\subsection{Filtrations and isomorphisms}

The next proposition will not be used until Section \ref{sect.split}, but is
still very elementary:

\begin{proposition}
\label{prop.filtr.iso}Let $k$ be a commutative ring. Let $V$ be a $k$-module.
Let $\left(  V_{n}\right)  _{n\geq0}$ be a $k$-module filtration of $V$. Let
$f:V\rightarrow V$ be a $k$-module homomorphism which satisfies%
\[
\left(  f\left(  V_{n}\right)  \subseteq V_{n-1}\ \ \ \ \ \ \ \ \ \ \text{for
every }n\in\mathbb{N}\right)  ,
\]
where $V_{-1}$ denotes the $k$-submodule $0$ of $V$. Then:\newline\textbf{(a)}
The $k$-module homomorphism $\operatorname{id}-f$ is an isomorphism.\newline%
\textbf{(b)} Each of the maps $\operatorname{id}-f$ and $\left(
\operatorname{id}-f\right)  ^{-1}$ respects the filtration.
\end{proposition}

Note that this proposition is a generalization of the following well-known fact:

\begin{corollary}
\label{cor.filtr.iso.triangle}Let $k$ be a commutative ring. Let
$m\in\mathbb{N}$. Let $A\in\operatorname{M}_{m}\left(  k\right)  $ be a
strictly upper triangular $m\times m$ matrix over $k$. Then, $I_{m}-A$ is an
invertible matrix, and its inverse is an upper triangular matrix.
\end{corollary}

The following quick proof of this fact by means of applying Proposition
\ref{prop.filtr.iso} is not here because we are going to use it, but rather
because it provides a kind of intuition for the meaning of Proposition
\ref{prop.filtr.iso} (even if only in a particular case).

\textit{Proof of Corollary \ref{cor.filtr.iso.triangle} (sketched).} Define a
filtration $\left(  V_{n}\right)  _{n\geq0}$ of the $k$-module $k^{m}$ by
$\left(  V_{n}=\left\{
\begin{array}
[c]{c}%
\left\langle e_{1},e_{2},...,e_{n}\right\rangle \text{, if }n\leq m;\\
k^{m}\text{, if }n>m
\end{array}
\right.  \text{ for every }n\in\mathbb{N}\right)  $, where $\left(
e_{1},e_{2},...,e_{m}\right)  $ is the standard basis of the $k$-module
$k^{m}$. Let $f$ be the $k$-linear map $k^{m}\rightarrow k^{m}$ represented by
the matrix $A$. Then, since $A$ is a strictly upper triangular matrix, it is
easy to see that $f\left(  V_{n}\right)  \subseteq V_{n-1}$ for every
$n\in\mathbb{N}$ (where $V_{-1}$ denotes the $k$-submodule $0$ of $k^{m}$).
Thus, Proposition \ref{prop.filtr.iso} \textbf{(a)} yields that
$\operatorname{id}-f$ is an isomorphism, and Proposition \ref{prop.filtr.iso}
\textbf{(b)} yields that its inverse $\left(  \operatorname{id}-f\right)
^{-1}$ respects the filtration. Translating this back into the language of
matrices, we obtain that $I_{m}-A$ is an invertible matrix, and its inverse is
an upper triangular matrix. This proves Corollary \ref{cor.filtr.iso.triangle}.

\begin{vershort}
\textit{Proof of Proposition \ref{prop.filtr.iso}.} First it is easy to see
that $f^{\kappa}\left(  V_{n}\right)  =0$ for every $\kappa\in\mathbb{N}$ and
$n\in\mathbb{N}$ satisfying $\kappa>n$. (This follows easily by induction.)

Now we want to explicitly construct the inverse of the map $\operatorname*{id}%
-f$. We are going to construct it as the infinite sum $\sum\limits_{\kappa
=0}^{\infty}f^{\kappa}$. First let us make sense of this sum: It is an
infinite sum which does not (generally) converge with respect to the discrete
topology. But it converges pointwise, in the sense that for every element
$x\in V$ the sum $\sum\limits_{\kappa=0}^{\infty}f^{\kappa}\left(  x\right)  $
converges with respect to the discrete topology on $V$. In fact, $x\in V$ must
lie in $V_{n}$ for some $n\in\mathbb{N}$, and thus every addend of the
infinite sum $\sum\limits_{\kappa=0}^{\infty}f^{\kappa}\left(  x\right)  $
with $\kappa>n$ is zero (because $f^{\kappa}\left(  V_{n}\right)  =0$ for
every $\kappa\in\mathbb{N}$ and $n\in\mathbb{N}$ satisfying $\kappa>n$), so
that only finitely addends contribute anything nonzero to this sum, and
therefore this sum converges. Hence, the infinite sum $\sum\limits_{\kappa
=0}^{\infty}f^{\kappa}$ converges pointwise and thus is well-defined.

Now we can see that $\left(  \operatorname*{id}-f\right)  \circ\sum
\limits_{\kappa=0}^{\infty}f^{\kappa}=\operatorname*{id}$ and $\sum
\limits_{\kappa=0}^{\infty}f^{\kappa}\circ\left(  \operatorname*{id}-f\right)
=\operatorname*{id}$ (the proof uses the same argument as the standard proof
that $\left(  1-X\right)  \cdot\sum\limits_{\kappa=0}^{\infty}X^{\kappa}=1$
and $\sum\limits_{\kappa=0}^{\infty}X^{\kappa}\cdot\left(  1-X\right)  =1$ in
the power series ring $k\left[  \left[  X\right]  \right]  $, but we also
tacitly use the fact that composition of maps $V\rightarrow V$ is continuous
with respect to the pointwise convergence topology). Thus, the map
$\operatorname*{id}-f$ has an inverse, namely $\sum\limits_{\kappa=0}^{\infty
}f^{\kappa}$. It is clear that both of these maps respect the filtration, so
that Proposition \ref{prop.filtr.iso} is proven.
\end{vershort}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.filtr.iso}.} We notice that every
$n\in\mathbb{N}$ satisfies $V_{n-1}\subseteq V_{n}$\ \ \ \ {\footnotemark}%
\footnotetext{\textit{Proof.} Let $n\in\mathbb{N}$ be arbitrary. Then, two
cases are possible: the case when $n=0$, and the case when $n>0$. In the case
$n=0$, we have $V_{n-1}=V_{0-1}=V_{-1}=0\subseteq V_{n}$. In the case $n>0$,
we have $V_{n-1}\subseteq V_{n}$ (because $\left(  V_{n}\right)  _{n\geq0}$ is
a $k$-module filtration). Hence, in both cases $n=0$ and $n>0$, we have
$V_{n-1}\subseteq V_{n}$. Thus, $V_{n-1}\subseteq V_{n}$ always holds.}.

Now, every $n\in\mathbb{N}$ satisfies $f\left(  V_{n}\right)  \subseteq
V_{n-1}\subseteq V_{n}$. Hence, the map $f$ respects the filtration. Since the
map $\operatorname{id}$ also respects the filtration, it follows from
Proposition \ref{prop.gr.functor} \textbf{(c)} (applied to $\operatorname{id}%
$, $f$, $V$ and $\left(  V_{n}\right)  _{n\geq0}$ instead of $f$, $g$, $W$ and
$\left(  W_{n}\right)  _{n\geq0}$) that the map $\operatorname{id}-f$ also
respects the filtration.

Next, let us show that for every $m\in\mathbb{N}$,%
\begin{equation}
\text{the map }\sum_{\kappa=0}^{m}f^{\kappa}:V\rightarrow V\text{ respects the
filtration.} \label{pf.filtr.iso.0}%
\end{equation}


\textit{Proof of (\ref{pf.filtr.iso.0}).} We are going to verify
(\ref{pf.filtr.iso.0}) by induction over $m$:

\textit{Induction base:} For $m=0$, the map $\sum\limits_{\kappa=0}%
^{m}f^{\kappa}:V\rightarrow V$ equals $\sum\limits_{\kappa=0}^{0}f^{\kappa
}=f^{0}=\operatorname{id}$ and thus respects the filtration (since it is known
that $\operatorname{id}$ respects the filtration). Thus, (\ref{pf.filtr.iso.0}%
) holds for $m=0$. The induction base is thus complete.

\textit{Induction step:} Let $n\in\mathbb{N}_{+}$ be arbitrary. Assume that
(\ref{pf.filtr.iso.0}) holds for $m=n-1$. We then must prove that
(\ref{pf.filtr.iso.0}) holds for $m=n$.

Since (\ref{pf.filtr.iso.0}) holds for $m=n-1$, we know that the map
$\sum\limits_{\kappa=0}^{n-1}f^{\kappa}:V\rightarrow V$ respects the
filtration. Combined with the fact that the map $f$ respects the filtration,
this yields that the map $f\circ\sum\limits_{\kappa=0}^{n-1}f^{\kappa}$
respects the filtration (according to Proposition \ref{prop.gr.functor}
\textbf{(b)}, applied to $f$, $\sum\limits_{\kappa=0}^{n-1}f^{\kappa}$, $V$,
$\left(  V_{n}\right)  _{n\geq0}$, $V$ and $\left(  V_{n}\right)  _{n\geq0}$
instead of $g$, $f$, $U$, $\left(  U_{n}\right)  _{n\geq0}$, $W$ and $\left(
W_{n}\right)  _{n\geq0}$). Combined with the (trivial) fact that the map
$-\operatorname{id}$ respects the filtration, this yields that the map
$f\circ\sum\limits_{\kappa=0}^{n-1}f^{\kappa}-\left(  -\operatorname{id}%
\right)  $ respects the filtration (according to Proposition
\ref{prop.gr.functor} \textbf{(c)}, applied to $f\circ\sum\limits_{\kappa
=0}^{n-1}f^{\kappa}$, $-\operatorname{id}$, $V$ and $\left(  V_{n}\right)
_{n\geq0}$ instead of $f$, $g$, $W$ and $\left(  W_{n}\right)  _{n\geq0}$).
Since%
\begin{align*}
f\circ\sum\limits_{\kappa=0}^{n-1}f^{\kappa}-\left(  -\operatorname{id}%
\right)   &  =\underbrace{f\circ\sum\limits_{\kappa=0}^{n-1}f^{\kappa}%
}_{\substack{=\sum\limits_{\kappa=0}^{n-1}f\circ f^{\kappa}\\\text{(since
}f\text{ is }k\text{-linear)}}}+\underbrace{\operatorname{id}}_{=f^{0}}%
=\sum\limits_{\kappa=0}^{n-1}\underbrace{f\circ f^{\kappa}}_{=f^{\kappa+1}%
}+f^{0}=\sum\limits_{\kappa=0}^{n-1}f^{\kappa+1}+f^{0}\\
&  =\sum\limits_{\kappa=1}^{n}f^{\kappa}+f^{0}\ \ \ \ \ \ \ \ \ \ \left(
\text{here we substituted }\kappa\text{ for }\kappa+1\text{ in the sum}\right)
\\
&  =\sum\limits_{\kappa=0}^{n}f^{\kappa},
\end{align*}
this rewrites as follows: The map $\sum\limits_{\kappa=0}^{n}f^{\kappa}$
respects the filtration. In other words, (\ref{pf.filtr.iso.0}) holds for
$m=n$. This completes the induction step. Thus (\ref{pf.filtr.iso.0}) is
proven for every $m\in\mathbb{N}$.

Next we will prove that every $\ell\in\mathbb{N}$ satisfies%
\begin{equation}
f^{\ell+1}\left(  V_{\ell}\right)  =0. \label{pf.filtr.iso.1}%
\end{equation}


\textit{Proof of (\ref{pf.filtr.iso.1}).} We are going to verify
(\ref{pf.filtr.iso.1}) by induction over $\ell$:

\textit{Induction base:} For $\ell=0$, we have%
\begin{align*}
f^{\ell+1}\left(  V_{\ell}\right)   &  =\underbrace{f^{0+1}}_{=f^{1}=f}\left(
V_{0}\right)  =f\left(  V_{0}\right)  \subseteq V_{0-1}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{due to the condition }\left(  f\left(
V_{n}\right)  \subseteq V_{n-1}\text{ for every }n\in\mathbb{N}\right)
\text{, applied to }n=0\right) \\
&  =V_{-1}=0.
\end{align*}
In other words, for $\ell=0$, we have $f^{\ell+1}\left(  V_{\ell}\right)  =0$.
Thus, (\ref{pf.filtr.iso.1}) holds for $\ell=0$. The induction base is thus complete.

\textit{Induction step:} Let $n\in\mathbb{N}_{+}$ be arbitrary. Assume that
(\ref{pf.filtr.iso.1}) holds for $\ell=n-1$. We then must prove that
(\ref{pf.filtr.iso.1}) holds for $\ell=n$.

Since (\ref{pf.filtr.iso.1}) holds for $\ell=n-1$, we have $f^{\left(
n-1\right)  +1}\left(  V_{n-1}\right)  =0$. Now,%
\[
\underbrace{f^{n+1}}_{=f^{n}\circ f}\left(  V_{n}\right)  =\left(  f^{n}\circ
f\right)  \left(  V_{n}\right)  =\underbrace{f^{n}}_{=f^{\left(  n-1\right)
+1}}\left(  \underbrace{f\left(  V_{n}\right)  }_{\subseteq V_{n-1}}\right)
\subseteq f^{\left(  n-1\right)  +1}\left(  V_{n-1}\right)  =0,
\]
so that $f^{n+1}\left(  V_{n}\right)  =0$. In other words,
(\ref{pf.filtr.iso.1}) holds for $\ell=n$. This completes the induction step.
Thus (\ref{pf.filtr.iso.1}) is proven for every $\ell\in\mathbb{N}$.

Now, let us define a map $R:V\rightarrow V$ as follows:

For every $v\in V$, let us define the element $R\left(  v\right)  $ as follows:

Since $\left(  V_{n}\right)  _{n\geq0}$ is a $k$-module filtration of $V$, we
have $\bigcup\limits_{n\geq0}V_{n}=V$ (by the definition of a filtration).
Now, $v\in V=\bigcup\limits_{n\geq0}V_{n}$ yields that there exists some
$m\in\mathbb{N}$ such that $v\in V_{m}$. Now, the element $\sum\limits_{\kappa
=0}^{m}f^{\kappa}\left(  v\right)  $ does \textit{not} depend on $m$ (but only
depends on $v$).\ \ \ \ {\footnotemark}\footnotetext{\textit{Proof.} Let
$m_{1}$ be an element of $\mathbb{N}$ such that $v\in V_{m_{1}}$. Let $m_{2}$
be an element of $\mathbb{N}$ such that $v\in V_{m_{2}}$. We are now going to
prove that $\sum\limits_{\kappa=0}^{m_{1}}f^{\kappa}\left(  v\right)
=\sum\limits_{\kappa=0}^{m_{2}}f^{\kappa}\left(  v\right)  $.
\par
In fact, we can WLOG assume that $m_{1}\geq m_{2}$ (otherwise, we switch
$m_{1}$ and $m_{2}$). So let us assume this. We have $v\in V_{m_{2}}$ and thus
$f^{m_{2}+1}\left(  v\right)  \in f^{m_{2}+1}\left(  V_{m_{2}}\right)  =0$ (by
(\ref{pf.filtr.iso.1}), applied to $\ell=m_{2}$), so that $f^{m_{2}+1}\left(
v\right)  =0$. Now, every $\kappa\geq m_{2}+1$ satisfies $f^{\kappa
}=f^{\left(  \kappa-\left(  m_{2}+1\right)  \right)  \circ\left(
m_{2}+1\right)  }=f^{\kappa-\left(  m_{2}+1\right)  }\circ f^{m_{2}+1}$.
Hence, every $\kappa\geq m_{2}+1$ satisfies $f^{\kappa}\left(  v\right)
=\left(  f^{\kappa-\left(  m_{2}+1\right)  }\circ f^{m_{2}+1}\right)  \left(
v\right)  =f^{\kappa-\left(  m_{2}+1\right)  }\left(  \underbrace{f^{m_{2}%
+1}\left(  v\right)  }_{=0}\right)  =f^{\kappa-\left(  m_{2}+1\right)
}\left(  0\right)  =0$ (since $f^{\kappa-\left(  m_{2}+1\right)  }$ is
$k$-linear). Now,
\[
\sum\limits_{\kappa=0}^{m_{1}}f^{\kappa}\left(  v\right)  =\sum\limits_{\kappa
=0}^{m_{2}}f^{\kappa}\left(  v\right)  +\sum\limits_{\kappa=m_{2}+1}^{m_{1}%
}\underbrace{f^{\kappa}\left(  v\right)  }_{=0\text{ (since }\kappa\geq
m_{2}+1\text{)}}=\sum\limits_{\kappa=0}^{m_{2}}f^{\kappa}\left(  v\right)
+\underbrace{\sum\limits_{\kappa=m_{2}+1}^{m_{1}}0}_{=0}=\sum\limits_{\kappa
=0}^{m_{2}}f^{\kappa}\left(  v\right)  .
\]
\par
Thus we have proven that, for every element $m_{1}\in\mathbb{N}$ such that
$v\in V_{m_{1}}$, and for every element $m_{2}\in\mathbb{N}$ such that $v\in
V_{m_{2}}$, we have $\sum\limits_{\kappa=0}^{m_{1}}f^{\kappa}\left(  v\right)
=\sum\limits_{\kappa=0}^{m_{2}}f^{\kappa}\left(  v\right)  $. In other words,
we have proven that any two values of $m\in\mathbb{N}$ such that $v\in V_{m}$
lead to the same value for $\sum\limits_{\kappa=0}^{m}f^{\kappa}\left(
v\right)  $. In other words, we have proven that the element $\sum
\limits_{\kappa=0}^{m}f^{\kappa}\left(  v\right)  $ does \textit{not} depend
on $m$ (as long as $m$ satisfies $v\in V_{m}$).} Therefore, we can define an
element $R\left(  v\right)  $ of $V$ (only depending on $v$) by $R\left(
v\right)  =\sum\limits_{\kappa=0}^{m}f^{\kappa}\left(  v\right)  $.

Thus we have defined an element $R\left(  v\right)  \in V$ for every $v\in V$.
In other words, we have thus defined a map $R:V\rightarrow V$.

According to its definition, this map $R$ satisfies%
\begin{equation}
R\left(  v\right)  =\sum\limits_{\kappa=0}^{m}f^{\kappa}\left(  v\right)
\text{ for every }m\in\mathbb{N}\text{ satisfying }v\in V_{m}.
\label{pf.filtr.iso.R}%
\end{equation}


Now we are going to show that $R\circ\left(  \operatorname{id}-f\right)
=\operatorname{id}$.

\textit{Proof of }$R\circ\left(  \operatorname{id}-f\right)
=\operatorname{id}$\textit{.} Let $w\in V$ be arbitrary. Let $v=\left(
\operatorname{id}-f\right)  \left(  w\right)  $.

Since $w\in V=\bigcup\limits_{n\geq0}V_{n}$, there exists some $m\in
\mathbb{N}$ such that $w\in V_{m}$. Consider this $m$. Then,%
\begin{align*}
v  &  =\left(  \operatorname{id}-f\right)  \left(  w\right)
=\underbrace{\operatorname{id}\left(  w\right)  }_{=w\in V_{m}}%
-\underbrace{f\left(  w\right)  }_{\in f\left(  V_{m}\right)  \text{ (since
}w\in V_{m}\text{)}}\\
&  \subseteq V_{m}-\underbrace{f\left(  V_{m}\right)  }_{\substack{\subseteq
V_{m}\text{ (since the map }f\\\text{respects the filtration)}}}\subseteq
V_{m}-V_{m}\subseteq V_{m}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }V_{m}\text{
is a }k\text{-module}\right)  .
\end{align*}


Now,%
\begin{align*}
&  \left(  R\circ\left(  \operatorname{id}-f\right)  \right)  \left(  w\right)
\\
&  =R\left(  \underbrace{\left(  \operatorname{id}-f\right)  \left(  w\right)
}_{=v}\right)  =R\left(  v\right) \\
&  =\sum\limits_{\kappa=0}^{m}f^{\kappa}\left(  \underbrace{v}_{=\left(
\operatorname{id}-f\right)  \left(  w\right)  =w-f\left(  w\right)  }\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{according to (\ref{pf.filtr.iso.R}), since
}v\in V_{m}\right) \\
&  =\sum\limits_{\kappa=0}^{m}\underbrace{f^{\kappa}\left(  w-f\left(
w\right)  \right)  }_{\substack{=f^{\kappa}\left(  w\right)  -f^{\kappa
}\left(  f\left(  w\right)  \right)  \\\text{(since }f^{\kappa}\text{ is
}k\text{-linear)}}}=\sum\limits_{\kappa=0}^{m}\left(  f^{\kappa}\left(
w\right)  -\underbrace{f^{\kappa}\left(  f\left(  w\right)  \right)
}_{=\left(  f^{\kappa}\circ f\right)  \left(  w\right)  }\right)
=\sum\limits_{\kappa=0}^{m}\left(  f^{\kappa}\left(  w\right)
-\underbrace{\left(  f^{\kappa}\circ f\right)  }_{=f^{\kappa+1}}\left(
w\right)  \right) \\
&  =\sum\limits_{\kappa=0}^{m}\left(  f^{\kappa}\left(  w\right)
-f^{\kappa+1}\left(  w\right)  \right)  =\sum\limits_{\kappa=0}^{m}f^{\kappa
}\left(  w\right)  -\sum\limits_{\kappa=0}^{m}f^{\kappa+1}\left(  w\right)
=\underbrace{\sum\limits_{\kappa=0}^{m}f^{\kappa}\left(  w\right)  }%
_{=f^{0}\left(  w\right)  +\sum\limits_{\kappa=1}^{m}f^{\kappa}\left(
w\right)  }-\underbrace{\sum\limits_{\kappa=1}^{m+1}f^{\kappa}\left(
w\right)  }_{=\sum\limits_{\kappa=1}^{m}f^{\kappa}\left(  w\right)
+f^{m+1}\left(  w\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }\kappa\text{ for
}\kappa+1\text{ in the second sum}\right) \\
&  =\left(  f^{0}\left(  w\right)  +\sum\limits_{\kappa=1}^{m}f^{\kappa
}\left(  w\right)  \right)  -\left(  \sum\limits_{\kappa=1}^{m}f^{\kappa
}\left(  w\right)  +f^{m+1}\left(  w\right)  \right)  =f^{0}\left(  w\right)
-f^{m+1}\left(  w\right)  .
\end{align*}
Since $\underbrace{f^{0}}_{=\operatorname{id}}\left(  w\right)
=\operatorname{id}\left(  w\right)  $ and $f^{m+1}\left(  w\right)  =0$
(because $w\in V_{m}$ and thus $f^{m+1}\left(  w\right)  \in f^{m+1}\left(
V_{m}\right)  =0$ (due to (\ref{pf.filtr.iso.1}), applied to $\ell=m$), so
that $f^{m+1}\left(  w\right)  =0$), this rewrites as $\left(  R\circ\left(
\operatorname{id}-f\right)  \right)  \left(  w\right)  =\operatorname{id}%
\left(  w\right)  -0=\operatorname{id}\left(  w\right)  $. Since this holds
for all $w\in V$, we thus conclude that $R\circ\left(  \operatorname{id}%
-f\right)  =\operatorname{id}$.

This completes the proof of $R\circ\left(  \operatorname{id}-f\right)
=\operatorname{id}$.

\textit{Proof of }$\left(  \operatorname{id}-f\right)  \circ
R=\operatorname{id}$\textit{.} Let $v\in V$ be arbitrary.

Since $v\in V=\bigcup\limits_{n\geq0}V_{n}$, there exists some $m\in
\mathbb{N}$ such that $v\in V_{m}$. Consider this $m$. Then,%
\begin{align*}
&  \left(  \left(  \operatorname{id}-f\right)  \circ R\right)  \left(
v\right)  =\left(  \operatorname{id}-f\right)  \left(  R\left(  v\right)
\right)  =\left(  \operatorname{id}-f\right)  \left(  \sum\limits_{\kappa
=0}^{m}f^{\kappa}\left(  v\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since (\ref{pf.filtr.iso.R}) says that
}R\left(  v\right)  =\sum\limits_{\kappa=0}^{m}f^{\kappa}\left(  v\right)
\right) \\
&  =\underbrace{\operatorname{id}\left(  \sum\limits_{\kappa=0}^{m}f^{\kappa
}\left(  v\right)  \right)  }_{=\sum\limits_{\kappa=0}^{m}f^{\kappa}\left(
v\right)  }-\underbrace{f\left(  \sum\limits_{\kappa=0}^{m}f^{\kappa}\left(
v\right)  \right)  }_{\substack{=\sum\limits_{\kappa=0}^{m}f\left(  f^{\kappa
}\left(  v\right)  \right)  \\\text{(since }f\text{ is }k\text{-linear)}%
}}=\sum\limits_{\kappa=0}^{m}f^{\kappa}\left(  v\right)  -\sum\limits_{\kappa
=0}^{m}\underbrace{f\left(  f^{\kappa}\left(  v\right)  \right)  }_{=\left(
f\circ f^{\kappa}\right)  \left(  v\right)  }\\
&  =\sum\limits_{\kappa=0}^{m}f^{\kappa}\left(  v\right)  -\sum\limits_{\kappa
=0}^{m}\underbrace{\left(  f\circ f^{\kappa}\right)  }_{=f^{\kappa+1}}\left(
v\right)  =\sum\limits_{\kappa=0}^{m}f^{\kappa}\left(  v\right)
-\sum\limits_{\kappa=0}^{m}f^{\kappa+1}\left(  v\right) \\
&  =\underbrace{\sum\limits_{\kappa=0}^{m}f^{\kappa}\left(  v\right)
}_{=f^{0}\left(  v\right)  +\sum\limits_{\kappa=1}^{m}f^{\kappa}\left(
v\right)  }-\underbrace{\sum\limits_{\kappa=1}^{m+1}f^{\kappa}\left(
v\right)  }_{=\sum\limits_{\kappa=1}^{m}f^{\kappa}\left(  v\right)
+f^{m+1}\left(  v\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }\kappa\text{ for
}\kappa+1\text{ in the second sum}\right) \\
&  =\left(  f^{0}\left(  v\right)  +\sum\limits_{\kappa=1}^{m}f^{\kappa
}\left(  v\right)  \right)  -\left(  \sum\limits_{\kappa=1}^{m}f^{\kappa
}\left(  v\right)  +f^{m+1}\left(  v\right)  \right)  =f^{0}\left(  v\right)
-f^{m+1}\left(  v\right)  .
\end{align*}
Since $\underbrace{f^{0}}_{=\operatorname{id}}\left(  v\right)
=\operatorname{id}\left(  v\right)  $ and $f^{m+1}\left(  v\right)  =0$
(because $v\in V_{m}$ and thus $f^{m+1}\left(  v\right)  \in f^{m+1}\left(
V_{m}\right)  =0$ (due to (\ref{pf.filtr.iso.1}), applied to $\ell=m$), so
that $f^{m+1}\left(  v\right)  =0$), this rewrites as $\left(  \left(
\operatorname{id}-f\right)  \circ R\right)  \left(  v\right)
=\operatorname{id}\left(  v\right)  -0=\operatorname{id}\left(  v\right)  $.
Since this holds for all $v\in V$, we thus conclude that $\left(
\operatorname{id}-f\right)  \circ R=\operatorname{id}$.

The proof of $\left(  \operatorname{id}-f\right)  \circ R=\operatorname{id}$
is thus complete.

We have thus shown that $R\circ\left(  \operatorname{id}-f\right)
=\operatorname{id}$ and $\left(  \operatorname{id}-f\right)  \circ
R=\operatorname{id}$. In other words, the two maps $\operatorname{id}-f$ and
$R$ are mutually inverse. Hence, the map $\operatorname{id}-f$ is invertible,
and its inverse is $R$.

The $k$-module homomorphism $\operatorname{id}-f$ is a $k$-module isomorphism
(since it is invertible). This proves Proposition \ref{prop.filtr.iso}
\textbf{(a)}.

We already know that the map $\operatorname{id}-f$ respects the filtration.
Let us now prove that so does $R$:

Let $n\in\mathbb{N}$ be arbitrary. Every $v\in V_{n}$ satisfies%
\begin{align*}
R\left(  v\right)   &  =\sum\limits_{\kappa=0}^{n}f^{\kappa}\left(  v\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.filtr.iso.R}), applied to
}m=n\right) \\
&  =\left(  \sum\limits_{\kappa=0}^{n}f^{\kappa}\right)  \left(  v\right)
\in\left(  \sum\limits_{\kappa=0}^{n}f^{\kappa}\right)  \left(  V_{n}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }v\in V_{n}\right) \\
&  \subseteq V_{n}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since (\ref{pf.filtr.iso.0}) (applied to }m=n\text{) says that}\\
\text{the map }\sum\limits_{\kappa=0}^{n}f^{\kappa}\text{ respects the
filtration}%
\end{array}
\right)  .
\end{align*}
In other words, $R\left(  V_{n}\right)  \subseteq V_{n}$. Since this holds for
every $n\in\mathbb{N}$, we thus have proven that the map $R$ respects the
filtration. Since $R=\left(  \operatorname{id}-f\right)  ^{-1}$ (because we
know that $R$ is the inverse of $\operatorname{id}-f$), this yields that the
map $\left(  \operatorname{id}-f\right)  ^{-1}$ respects the filtration.
Combined with the fact that $\operatorname{id}-f$ respects the filtration,
this completes the proof of Proposition \ref{prop.filtr.iso} \textbf{(b)}.
\end{verlong}

\subsection{A consequence about isomorphisms}

We are not going to use Proposition \ref{prop.filtr.iso} directly; instead we
will apply the following corollary:

\begin{corollary}
\label{cor.filtr.iso.cor}Let $k$ be a commutative ring. Let $V$ and $W$ be two
$k$-modules. Let $\left(  V_{n}\right)  _{n\geq0}$ be a $k$-module filtration
of $V$. Let $\left(  W_{n}\right)  _{n\geq0}$ be a $k$-module filtration of
$W$. Let $f:V\rightarrow W$ and $g:W\rightarrow V$ be two $k$-module
homomorphisms such that%
\[
\left(  \left(  \operatorname*{id}-g\circ f\right)  \left(  V_{n}\right)
\subseteq V_{n-1}\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}\right)
,
\]
where $V_{-1}$ denotes the $k$-submodule $0$ of $V$. Assume further that $f$
is a $k$-module isomorphism, and that $f$ and $f^{-1}$ respect the filtration.
Then:\newline\textbf{(a)} The $k$-module homomorphism $g$ is an
isomorphism.\newline\textbf{(b)} Each of the maps $g$ and $g^{-1}$ respects
the filtration.
\end{corollary}

\begin{vershort}
\textit{Proof of Corollary \ref{cor.filtr.iso.cor}.} Since $\left(
\operatorname*{id}-g\circ f\right)  \left(  V_{n}\right)  \subseteq V_{n-1}$
for every $n\in\mathbb{N}$, we can apply Proposition \ref{prop.filtr.iso} to
$\operatorname*{id}-g\circ f$ instead of $f$. As a result, we obtain (from
Proposition \ref{prop.filtr.iso} \textbf{(a)}) that the $k$-module
homomorphism $g\circ f$ is an isomorphism, and (from Proposition
\ref{prop.filtr.iso} \textbf{(b)}) that each of the maps $g\circ f$ and
$\left(  g\circ f\right)  ^{-1}$ respects the filtration.

Since $g\circ f$ and $f^{-1}$ are isomorphisms, so is their composition
$\left(  g\circ f\right)  \circ f^{-1}=g$, and thus Corollary
\ref{cor.filtr.iso.cor} \textbf{(a)} is proven.

Since both maps $\left(  g\circ f\right)  ^{-1}$ and $f$ respect the
filtration, so does their composition $f\circ\left(  g\circ f\right)
^{-1}=g^{-1}$. This handles one half of Corollary \ref{cor.filtr.iso.cor}
\textbf{(b)}.

The other half is proven similarly: Both maps $g\circ f$ and $f^{-1}$ respect
the filtration; hence, so does their composition $\left(  g\circ f\right)
\circ f^{-1}=g$.

The proof of Corollary \ref{cor.filtr.iso.cor} is now complete.
\end{vershort}

\begin{verlong}
\textit{Proof of Corollary \ref{cor.filtr.iso.cor}.} Since $\left(
\operatorname*{id}-g\circ f\right)  \left(  V_{n}\right)  \subseteq V_{n-1}$
for every $n\in\mathbb{N}$, we can apply Proposition \ref{prop.filtr.iso} to
$\operatorname*{id}-g\circ f$ instead of $f$. Hence, Proposition
\ref{prop.filtr.iso} \textbf{(a)} (applied to $\operatorname*{id}-g\circ f$
instead of $f$) yields that the $k$-module homomorphism $\operatorname{id}%
-\left(  \operatorname*{id}-g\circ f\right)  $ is an isomorphism. On the other
hand, Proposition \ref{prop.filtr.iso} \textbf{(b)} (applied to
$\operatorname*{id}-g\circ f$ instead of $f$) yields that each of the maps
$\operatorname*{id}-\left(  \operatorname*{id}-g\circ f\right)  $ and $\left(
\operatorname*{id}-\left(  \operatorname*{id}-g\circ f\right)  \right)  ^{-1}$
respects the filtration.

We know that $\operatorname{id}-\left(  \operatorname*{id}-g\circ f\right)  $
is an isomorphism. Since $\operatorname{id}-\left(  \operatorname*{id}-g\circ
f\right)  =g\circ f$, we thus conclude that $g\circ f$ is an isomorphism. But
since $f^{-1}$ is an isomorphism (this is because $f$ is an isomorphism), the
composition $\left(  g\circ f\right)  \circ f^{-1}$ must also be an
isomorphism (because $g\circ f$ and $f^{-1}$ are isomorphisms, and because the
composition of two isomorphisms is an isomorphism). Since $\left(  g\circ
f\right)  \circ f^{-1}=g$, this yields that $g$ is an isomorphism. This proves
Corollary \ref{cor.filtr.iso.cor} \textbf{(a)}.

Now, we know that each of the maps $\operatorname*{id}-\left(
\operatorname*{id}-g\circ f\right)  $ and $\left(  \operatorname*{id}-\left(
\operatorname*{id}-g\circ f\right)  \right)  ^{-1}$ respects the filtration.
Since $\operatorname*{id}-\left(  \operatorname*{id}-g\circ f\right)  =g\circ
f$ and $\left(  \underbrace{\operatorname*{id}-\left(  \operatorname*{id}%
-g\circ f\right)  }_{=g\circ f}\right)  ^{-1}=\left(  g\circ f\right)
^{-1}=f^{-1}\circ g^{-1}$, this result rewrites as follows: Each of the maps
$g\circ f$ and $f^{-1}\circ g^{-1}$ respects the filtration.

Since each of the maps $f^{-1}\circ g^{-1}$ and $f$ respects the filtration,
Proposition \ref{prop.gr.functor} \textbf{(b)} (applied to $V$, $\left(
V_{n}\right)  _{n\geq0}$, $V$, $\left(  V_{n}\right)  _{n\geq0}$, $W$,
$\left(  W_{n}\right)  _{n\geq0}$, $f^{-1}\circ g^{-1}$ and $f$ instead of
$U$, $\left(  U_{n}\right)  _{n\geq0}$, $V$, $\left(  V_{n}\right)  _{n\geq0}%
$, $W$, $\left(  W_{n}\right)  _{n\geq0}$, $f$ and $g$) yields that the map
$f\circ\left(  f^{-1}\circ g^{-1}\right)  $ respects the filtration. Since
$f\circ\left(  f^{-1}\circ g^{-1}\right)  =g^{-1}$, this rewrites as follows:
The map $g^{-1}$ respects the filtration.

Since each of the maps $g\circ f$ and $f^{-1}$ respects the filtration,
Proposition \ref{prop.gr.functor} \textbf{(b)} (applied to $W$, $\left(
W_{n}\right)  _{n\geq0}$, $V$, $\left(  V_{n}\right)  _{n\geq0}$, $V$,
$\left(  V_{n}\right)  _{n\geq0}$, $f^{-1}$ and $g\circ f$ instead of $U$,
$\left(  U_{n}\right)  _{n\geq0}$, $V$, $\left(  V_{n}\right)  _{n\geq0}$,
$W$, $\left(  W_{n}\right)  _{n\geq0}$, $f$ and $g$) yields that the map
$\left(  g\circ f\right)  \circ f^{-1}$ respects the filtration. Since
$\left(  g\circ f\right)  \circ f^{-1}=g$, this rewrites as follows: The map
$g$ respects the filtration.

This completes the proof of Corollary \ref{cor.filtr.iso.cor}.
\end{verlong}

\subsection{Splitting of filtrations of $\mathfrak{g}$-modules}

Some filtrations are simpler than others. One particular property some
$\mathfrak{g}$-module filtrations have is being $\mathfrak{g}$\textit{-split}:

\begin{definition}
\label{defs.split.filtr-g}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a $k$-Lie algebra. Let $V$ be a $\mathfrak{g}$-module. Let $\left(
V_{n}\right)  _{n\geq0}$ be a $\mathfrak{g}$-module filtration of $V$.\newline
For every integer $n\geq1$, let $\iota_{n}$ denote the canonical inclusion of
$V_{n-1}$ into $V_{n}$, and let $\pi_{n}$ denote the canonical projection of
$V_{n}$ onto $V_{n}\diagup V_{n-1}$.\newline We say that \textit{the
}$\mathfrak{g}$\textit{-module filtration }$\left(  V_{n}\right)  _{n\geq0}$
is $\mathfrak{g}$\textit{-split} if and only if%
\[
\left(
\begin{array}
[c]{c}%
\text{the short exact sequence }\\%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%0 \ar[r] & V_{n-1} \ar[r]^-{\iota_n} & V_n \ar[r]^-{\pi_n} & V_n \diagup
%V_{n-1} \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrix{
0 \ar[r] & V_{n-1} \ar[r]^-{\iota_n} & V_n \ar[r]^-{\pi_n} & V_n \diagup
V_{n-1} \ar[r] & 0
}%
%EndExpansion
\text{ is }\mathfrak{g}\text{-split for every }n\geq1
\end{array}
\right)  .
\]

\end{definition}

The next proposition shows that $\mathfrak{g}$-split filtrations are those
coming from direct sum decompositions of $V$:

\begin{proposition}
\label{prop.split.filtr-g}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a $k$-Lie algebra. Let $V$ be a $\mathfrak{g}$-module. Let $\left(
V_{n}\right)  _{n\geq0}$ be a $\mathfrak{g}$-module filtration of $V$. Then,
the filtration $\left(  V_{n}\right)  _{n\geq0}$ is $\mathfrak{g}$-split if
and only if there exists a family $\left(  W_{n}\right)  _{n\in\mathbb{N}}$ of
$\mathfrak{g}$-submodules of $V$ such that $V=\bigoplus\limits_{n\in
\mathbb{N}}W_{n}$ and such that every $p\in\mathbb{N}$ satisfies
$V_{p}=\bigoplus\limits_{n=0}^{p}W_{n}$.
\end{proposition}

Note that this Proposition \ref{prop.split.filtr-g} depends on the countable
Axiom of Choice.

\begin{remark}
A remark about notation: In Proposition \ref{prop.split.filtr-g}, we could
have just as well written $\left(  W_{n}\right)  _{n\geq0}$ instead of
$\left(  W_{n}\right)  _{n\in\mathbb{N}}$. In fact, $\left(  W_{n}\right)
_{n\geq0}$ and $\left(  W_{n}\right)  _{n\in\mathbb{N}}$ are two equivalent
ways to denote one and the same family. However, we prefer to use the notation
$\left(  W_{n}\right)  _{n\in\mathbb{N}}$ for families of $\mathfrak{g}%
$-submodules of $V$ satisfying $V=\bigoplus\limits_{n\in\mathbb{N}}W_{n}$
(such families are called "$\mathfrak{g}$-module \textit{gradings}" of $V$) as
opposed to the notation $\left(  V_{n}\right)  _{n\geq0}$ for filtrations, in
order to make gradings easier to distinguish from filtrations (there is no
deeper reason behind this).
\end{remark}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.split.filtr-g}.} In order to verify
Proposition \ref{prop.split.filtr-g}, we must show two lemmas:

\begin{lemma}
\label{lem.split.filtr-g.1}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie algebra. Let $V$ be a $\mathfrak{g}$-module. Let $\left(
V_{n}\right)  _{n\geq0}$ be a $\mathfrak{g}$-module filtration of $V$. If the
filtration $\left(  V_{n}\right)  _{n\geq0}$ is $\mathfrak{g}$-split, then
there exists a family $\left(  W_{n}\right)  _{n\in\mathbb{N}}$ of
$\mathfrak{g}$-submodules of $V$ such that $V=\bigoplus\limits_{n\in
\mathbb{N}}W_{n}$ and such that every $p\in\mathbb{N}$ satisfies
$V_{p}=\bigoplus\limits_{n=0}^{p}W_{n}$.
\end{lemma}

\begin{lemma}
\label{lem.split.filtr-g.2}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie algebra. Let $V$ be a $\mathfrak{g}$-module. Let $\left(
V_{n}\right)  _{n\geq0}$ be a $\mathfrak{g}$-module filtration of $V$. If
there exists a family $\left(  W_{n}\right)  _{n\in\mathbb{N}}$ of
$\mathfrak{g}$-submodules of $V$ such that $V=\bigoplus\limits_{n\in
\mathbb{N}}W_{n}$ and such that every $p\in\mathbb{N}$ satisfies
$V_{p}=\bigoplus\limits_{n=0}^{p}W_{n}$, then the filtration $\left(
V_{n}\right)  _{n\geq0}$ is $\mathfrak{g}$-split.
\end{lemma}

\textit{Proof of Lemma \ref{lem.split.filtr-g.1}.} Assume that the filtration
$\left(  V_{n}\right)  _{n\geq0}$ is $\mathfrak{g}$-split.

For every $n\geq1$, let $\iota_{n}$ denote the canonical inclusion of
$V_{n-1}$ into $V_{n}$, and let $\pi_{n}$ denote the canonical projection of
$V_{n}$ onto $V_{n}\diagup V_{n-1}$.

The filtration $\left(  V_{n}\right)  _{n\geq0}$ is $\mathfrak{g}$-split.
According to Definition \ref{defs.split.filtr-g}, this means that the short
exact sequence $%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%0 \ar[r] & V_{n-1} \ar[r]^-{\iota_n} & V_n \ar[r]^-{\pi_n} & V_n \diagup
%V_{n-1} \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrix{
0 \ar[r] & V_{n-1} \ar[r]^-{\iota_n} & V_n \ar[r]^-{\pi_n} & V_n \diagup
V_{n-1} \ar[r] & 0
}%
%EndExpansion
$ is $\mathfrak{g}$-split for every $n\geq1$. Thus, for every $n\geq1$, the
short exact sequence $%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%0 \ar[r] & V_{n-1} \ar[r]^-{\iota_n} & V_n \ar[r]^-{\pi_n} & V_n \diagup
%V_{n-1} \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrix{
0 \ar[r] & V_{n-1} \ar[r]^-{\iota_n} & V_n \ar[r]^-{\pi_n} & V_n \diagup
V_{n-1} \ar[r] & 0
}%
%EndExpansion
$ is $\mathfrak{g}$-split.

Let $n\in\mathbb{N}_{+}$ be arbitrary. Then, $n\geq1$. Applying Corollary
\ref{cor.g.split} to $V_{n-1}$, $V_{n}$, $V_{n}\diagup V_{n-1}$, $\iota_{n}$
and $\pi_{n}$ instead of $A$, $B$, $C$, $f$ and $g$, we see that the exact
sequence $%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%0 \ar[r] & V_{n-1} \ar[r]^-{\iota_n} & V_n \ar[r]^-{\pi_n} & V_n \diagup
%V_{n-1} \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrix{
0 \ar[r] & V_{n-1} \ar[r]^-{\iota_n} & V_n \ar[r]^-{\pi_n} & V_n \diagup
V_{n-1} \ar[r] & 0
}%
%EndExpansion
$ is $\mathfrak{g}$-split if and only if there exists a $\mathfrak{g}%
$-submodule $P$ of $V_{n}$ such that $V_{n}=\iota_{n}\left(  V_{n-1}\right)
\oplus P$. Since we know that the exact sequence $%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%0 \ar[r] & V_{n-1} \ar[r]^-{\iota_n} & V_n \ar[r]^-{\pi_n} & V_n \diagup
%V_{n-1} \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrix{
0 \ar[r] & V_{n-1} \ar[r]^-{\iota_n} & V_n \ar[r]^-{\pi_n} & V_n \diagup
V_{n-1} \ar[r] & 0
}%
%EndExpansion
$ is $\mathfrak{g}$-split, we thus conclude that there exists a $\mathfrak{g}%
$-submodule $P$ of $V_{n}$ such that $V_{n}=\iota_{n}\left(  V_{n-1}\right)
\oplus P$. Denote this $\mathfrak{g}$-submodule $P$ by $W_{n}$. Then,
$V_{n}=\iota_{n}\left(  V_{n-1}\right)  \oplus W_{n}=V_{n-1}\oplus W_{n}$
(since $\iota_{n}$ is the inclusion map, and thus $\iota_{n}\left(
V_{n-1}\right)  =V_{n-1}$).

We have thus defined a $\mathfrak{g}$-submodule $W_{n}$ of $V_{n}$ for every
$n\in\mathbb{N}_{+}$, and this $\mathfrak{g}$-submodule satisfies
$V_{n}=V_{n-1}\oplus W_{n}$.

Let us additionally define a $\mathfrak{g}$-submodule $W_{0}$ of $V_{0}$ by
$W_{0}=V_{0}$. Thus, a $\mathfrak{g}$-submodule $W_{n}$ of $V_{n}$ is defined
for every $n\in\mathbb{N}$ (because we have defined a $\mathfrak{g}$-submodule
$W_{n}$ of $V_{n}$ for every $n\in\mathbb{N}_{+}$, and we have defined a
$\mathfrak{g}$-submodule $W_{0}$ of $V_{0}$, but $\mathbb{N}_{+}\cup\left\{
0\right\}  =\mathbb{N}$). Hence, we have defined a family $\left(
W_{n}\right)  _{n\in\mathbb{N}}$ of $\mathfrak{g}$-submodules of $V$.

Now let us prove that%
\begin{equation}
\text{every }p\in\mathbb{N}\text{ satisfies }V_{p}=\bigoplus\limits_{n=0}%
^{p}W_{n}. \label{pf.lem.split.filtr-g.1.1}%
\end{equation}


\textit{Proof of (\ref{pf.lem.split.filtr-g.1.1}).} We are going to verify
(\ref{pf.lem.split.filtr-g.1.1}) by induction over $p$:

\textit{Induction base:} For $p=0$, we have $V_{p}=V_{0}=W_{0}=\bigoplus
\limits_{n=0}^{0}W_{n}=\bigoplus\limits_{n=0}^{p}W_{n}$ (since $0=p$). In
other words, (\ref{pf.lem.split.filtr-g.1.1}) holds for $p=0$. This completes
the induction base.

\textit{Induction step:} Let $q\in\mathbb{N}_{+}$ be arbitrary. Assume that
(\ref{pf.lem.split.filtr-g.1.1}) is proven for $p=q-1$. Now we must prove
(\ref{pf.lem.split.filtr-g.1.1}) for $p=q$.

Since (\ref{pf.lem.split.filtr-g.1.1}) is proven for $p=q-1$, we have
$V_{q-1}=\bigoplus\limits_{n=0}^{q-1}W_{n}$. Now, we know that every
$n\in\mathbb{N}_{+}$ satisfies $V_{n}=V_{n-1}\oplus W_{n}$. Applying this to
$n=q$, we obtain $V_{q}=V_{q-1}\oplus W_{q}$, and thus
\[
V_{q}=\underbrace{V_{q-1}}_{=\bigoplus\limits_{n=0}^{q-1}W_{n}}\oplus
W_{q}=\left(  \bigoplus\limits_{n=0}^{q-1}W_{n}\right)  \oplus W_{q}%
=\bigoplus\limits_{n=0}^{q}W_{n}.
\]
Thus, we have proven (\ref{pf.lem.split.filtr-g.1.1}) for $p=q$. This
completes the induction step. The proof of (\ref{pf.lem.split.filtr-g.1.1}) is
thus complete.

Next we are going to show that $V=\bigoplus\limits_{n\in\mathbb{N}}W_{n}$.

In fact, we first note that every $p\in\mathbb{N}$ satisfies
\begin{align*}
V_{p}  &  =\bigoplus\limits_{n=0}^{p}W_{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.lem.split.filtr-g.1.1})}\right) \\
&  =\sum\limits_{n=0}^{p}W_{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{since direct
sums are sums}\right) \\
&  \subseteq\sum\limits_{n=0}^{\infty}W_{n},
\end{align*}
and thus%
\begin{align*}
V  &  =\bigcup_{p\in\mathbb{N}}\underbrace{V_{p}}_{\subseteq\sum
\limits_{n=0}^{\infty}W_{n}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left(
V_{n}\right)  _{n\geq0}\text{ is a }\mathfrak{g}\text{-module filtration of
}V\right) \\
&  \subseteq\bigcup_{p\in\mathbb{N}}\sum\limits_{n=0}^{\infty}W_{n}%
=\sum\limits_{n=0}^{\infty}W_{n}.
\end{align*}
Combined with the obvious relation $\sum\limits_{n=0}^{\infty}W_{n}\subseteq
V$, this yields $V=\sum\limits_{n=0}^{\infty}W_{n}$. Now, in order to prove
that $V=\bigoplus\limits_{n\in\mathbb{N}}W_{n}$, we only need to show that the
sum $\sum\limits_{n=0}^{\infty}W_{n}$ is a direct sum.

Let $\left(  a_{n}\right)  _{n\in\mathbb{N}}$ be a sequence of elements of $V$
satisfying $\left(  a_{n}\in W_{n}\text{ for every }n\in\mathbb{N}\right)  $,
$\left(  a_{n}=0\text{ for all but finitely many }n\in\mathbb{N}\right)  $ and
$\sum\limits_{n=0}^{\infty}a_{n}=0$.

Since $\left(  a_{n}=0\text{ for all but finitely many }n\in\mathbb{N}\right)
$, there exists a finite subset $S$ of $\mathbb{N}$ such that $\left(
a_{n}=0\text{ for all }n\in\mathbb{N}\diagdown S\right)  $. Consider this $S$.
Clearly, the set $S$ has an upper bound (because $S$ is finite). Let $M$ be
this upper bound. Then, every $n\in S$ satisfies $n\leq M$ (because $M$ is an
upper bound of the set $S$). Hence, every $n\in\mathbb{N}$ satisfying $n>M$
must satisfy $n\notin S$ (because otherwise, it would satisfy $n\in S$,
therefore $n\leq M$, which would contradict $n>M$). Hence, every
$n\in\mathbb{N}$ satisfying $n>M$ must satisfy $a_{n}=0$ (because $n>M$ yields
$n\notin S$, so that $n\in\mathbb{N}\diagdown S$ and thus $a_{n}=0$). On the
other hand, (\ref{pf.lem.split.filtr-g.1.1}) (applied to $p=M$) yields
$V_{M}=\bigoplus\limits_{n=0}^{M}W_{n}$. Thus, the sum $\sum\limits_{n=0}%
^{M}W_{n}$ is a direct sum.

Now,%
\[
0=\sum\limits_{n=0}^{\infty}a_{n}=\sum\limits_{n=0}^{M}a_{n}+\sum
\limits_{n=M+1}^{\infty}\underbrace{a_{n}}_{=0\text{ (since }n>M\text{)}}%
=\sum\limits_{n=0}^{M}a_{n}+\underbrace{\sum\limits_{n=M+1}^{\infty}0}%
_{=0}=\sum\limits_{n=0}^{M}a_{n}.
\]


Now, $\left(  a_{n}\right)  _{n\in\left\{  0,1,...,M\right\}  }$ is an
$\left(  M+1\right)  $-tuple of elements of $V$ satisfying \newline$\left(
a_{n}\in W_{n}\text{ for every }n\in\left\{  0,1,...,M\right\}  \right)  $ and
$\sum\limits_{n=0}^{M}a_{n}=0$. Therefore, $a_{n}=0$ for every $n\in\left\{
0,1,...,M\right\}  $ (because the sum $\sum\limits_{n=0}^{M}W_{n}$ is a direct sum).

Now, every $n\in\mathbb{N}$ satisfies $a_{n}=0$. \ \ \ \ {\footnotemark}%
\footnotetext{\textit{Proof.} Let $n\in\mathbb{N}$ be arbitrary. Then we must
have one of the following two cases: either $n\leq M$ or $n>M$. But if $n\leq
M$, then $a_{n}=0$ (since $n\leq M$ leads to $n\in\left\{  0,1,...,M\right\}
$ and thus $a_{n}=0$). On the other hand, if $n>M$, then $a_{n}=0$ (as we
already know). Thus, in each of the two cases $n\leq M$ and $n>M$, we have
$a_{n}=0$. Hence, $a_{n}=0$ holds in every case.}

We have thus shown that every sequence $\left(  a_{n}\right)  _{n\in
\mathbb{N}}$ of elements of $V$ satisfying \newline$\left(  a_{n}\in
W_{n}\text{ for every }n\in\mathbb{N}\right)  $, $\left(  a_{n}=0\text{ for
all but finitely many }n\in\mathbb{N}\right)  $ and $\sum\limits_{n=0}%
^{\infty}a_{n}=0$ must also satisfy $\left(  a_{n}=0\text{ for every }%
n\in\mathbb{N}\right)  $. In other words, the sum $\sum\limits_{n=0}^{\infty
}W_{n}$ is a direct sum. Hence, $V=\sum\limits_{n=0}^{\infty}W_{n}$ rewrites
as $V=\bigoplus\limits_{n=0}^{\infty}W_{n}=\bigoplus\limits_{n\in\mathbb{N}%
}W_{n}$.

Altogether, we have now proven that there exists a family $\left(
W_{n}\right)  _{n\in\mathbb{N}}$ of $\mathfrak{g}$-submodules of $V$ such that
$V=\bigoplus\limits_{n\in\mathbb{N}}W_{n}$ and such that every $p\in
\mathbb{N}$ satisfies $V_{p}=\bigoplus\limits_{n=0}^{p}W_{n}$. This proves
Lemma \ref{lem.split.filtr-g.1}.

\textit{Proof of Lemma \ref{lem.split.filtr-g.2}.} Assume that there exists a
family $\left(  W_{n}\right)  _{n\in\mathbb{N}}$ of $\mathfrak{g}$-submodules
of $V$ such that $V=\bigoplus\limits_{n\in\mathbb{N}}W_{n}$ and such that
every $p\in\mathbb{N}$ satisfies $V_{p}=\bigoplus\limits_{n=0}^{p}W_{n}$.

For every $n\geq1$, let $\iota_{n}$ denote the canonical inclusion of
$V_{n-1}$ into $V_{n}$, and let $\pi_{n}$ denote the canonical projection of
$V_{n}$ onto $V_{n}\diagup V_{n-1}$.

We know that every $p\in\mathbb{N}$ satisfies
\begin{equation}
V_{p}=\bigoplus\limits_{n=0}^{p}W_{n}=\bigoplus\limits_{m=0}^{p}W_{m}
\label{pf.lem.split.filtr-g.2.1}%
\end{equation}
(here, we renamed $n$ as $m$ in the sum).

Now let $n\geq1$ be arbitrary. Then, $V_{n}=\bigoplus\limits_{m=0}^{n}W_{m}$
(due to (\ref{pf.lem.split.filtr-g.2.1}), applied to $p=n$) but also
$V_{n-1}=\bigoplus\limits_{m=0}^{n-1}W_{m}$ (due to
(\ref{pf.lem.split.filtr-g.2.1}), applied to $p=n-1$). Thus, $V_{n}%
=\bigoplus\limits_{m=0}^{n}W_{m}=\underbrace{\left(  \bigoplus\limits_{m=0}%
^{n-1}W_{m}\right)  }_{=V_{n-1}}\oplus W_{n}=V_{n-1}\oplus W_{n}$. Since
$V_{n-1}=\iota_{n}\left(  V_{n-1}\right)  $ (because $\iota_{n}$ is the
inclusion map), this rewrites as $V_{n}=\iota_{n}\left(  V_{n-1}\right)
\oplus W_{n}$.

Now, applying Corollary \ref{cor.g.split} to $V_{n-1}$, $V_{n}$, $V_{n}\diagup
V_{n-1}$, $\iota_{n}$ and $\pi_{n}$ instead of $A$, $B$, $C$, $f$ and $g$, we
see that the exact sequence $%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%0 \ar[r] & V_{n-1} \ar[r]^-{\iota_n} & V_n \ar[r]^-{\pi_n} & V_n \diagup
%V_{n-1} \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrix{
0 \ar[r] & V_{n-1} \ar[r]^-{\iota_n} & V_n \ar[r]^-{\pi_n} & V_n \diagup
V_{n-1} \ar[r] & 0
}%
%EndExpansion
$ is $\mathfrak{g}$-split if and only if there exists a $\mathfrak{g}%
$-submodule $P$ of $V_{n}$ such that $V_{n}=\iota_{n}\left(  V_{n-1}\right)
\oplus P$. Since we know that there exists a $\mathfrak{g}$-submodule $P$ of
$V_{n}$ such that $V_{n}=\iota_{n}\left(  V_{n-1}\right)  \oplus P$ (namely,
$P=W_{n}$, because $V_{n}=\iota_{n}\left(  V_{n-1}\right)  \oplus W_{n}$), we
can thus conclude that the exact sequence $%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%0 \ar[r] & V_{n-1} \ar[r]^-{\iota_n} & V_n \ar[r]^-{\pi_n} & V_n \diagup
%V_{n-1} \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrix{
0 \ar[r] & V_{n-1} \ar[r]^-{\iota_n} & V_n \ar[r]^-{\pi_n} & V_n \diagup
V_{n-1} \ar[r] & 0
}%
%EndExpansion
$ is $\mathfrak{g}$-split.

Hence, we have shown that for every $n\geq1$, the exact sequence \newline$%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%0 \ar[r] & V_{n-1} \ar[r]^-{\iota_n} & V_n \ar[r]^-{\pi_n} & V_n \diagup
%V_{n-1} \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrix{
0 \ar[r] & V_{n-1} \ar[r]^-{\iota_n} & V_n \ar[r]^-{\pi_n} & V_n \diagup
V_{n-1} \ar[r] & 0
}%
%EndExpansion
$ is $\mathfrak{g}$-split. According to Definition \ref{defs.split.filtr-g},
this means that the filtration $\left(  V_{n}\right)  _{n\geq0}$ is
$\mathfrak{g}$-split.

We have hence shown that the filtration $\left(  V_{n}\right)  _{n\geq0}$ is
$\mathfrak{g}$-split. Thus, Lemma \ref{lem.split.filtr-g.2} is proven.

Proposition \ref{prop.split.filtr-g} immediately follows from Lemma
\ref{lem.split.filtr-g.1} and Lemma \ref{lem.split.filtr-g.2}.
\end{verlong}

We will apply Proposition \ref{prop.split.filtr-g} in a slightly different
version. First a definition:

\begin{definition}
\label{defs.bifiltered}Let $k$ be a commutative ring. Let $V$ and $W$ be two
$k$-modules. Let $f:V\rightarrow W$ be a $k$-module isomorphism. Let $\left(
V_{n}\right)  _{n\geq0}$ be a $k$-module filtration of $V$, and let $\left(
W_{n}\right)  _{n\geq0}$ be a $k$-module filtration of $W$.\newline The
isomorphism $f$ is said to be \textit{bifiltered} if and only if both maps $f$
and $f^{-1}$ respect the filtration.
\end{definition}

\begin{proposition}
\label{prop.split.filtr-g-2}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie algebra. Let $V$ be a $\mathfrak{g}$-module. Let $\left(
V_{n}\right)  _{n\geq0}$ be a $\mathfrak{g}$-module filtration of $V$.
Consider the $\mathfrak{g}$-module $\bigoplus\limits_{p\in\mathbb{N}%
}\operatorname*{gr}\nolimits_{p}V$, and the filtration $\left(  \bigoplus
\limits_{p=0}^{n}\operatorname*{gr}\nolimits_{p}V\right)  _{n\geq0}$ of this
$\mathfrak{g}$-module $\bigoplus\limits_{p\in\mathbb{N}}\operatorname*{gr}%
\nolimits_{p}V$.\newline Then, the filtration $\left(  V_{n}\right)  _{n\geq
0}$ is $\mathfrak{g}$-split if and only if there exists a bifiltered
$\mathfrak{g}$-module isomorphism $V\rightarrow\bigoplus\limits_{p\in
\mathbb{N}}\operatorname*{gr}\nolimits_{p}V$.
\end{proposition}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.split.filtr-g-2}.} In order to verify
Proposition \ref{prop.split.filtr-g-2}, we must show two lemmas:

\begin{lemma}
\label{lem.split.filtr-g.3}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie algebra. Let $V$ be a $\mathfrak{g}$-module. Let $\left(
V_{n}\right)  _{n\geq0}$ be a $\mathfrak{g}$-module filtration of $V$.
Consider the $\mathfrak{g}$-module $\bigoplus\limits_{p\in\mathbb{N}%
}\operatorname*{gr}\nolimits_{p}V$, and the filtration $\left(  \bigoplus
\limits_{p=0}^{n}\operatorname*{gr}\nolimits_{p}V\right)  _{n\geq0}$ of this
$\mathfrak{g}$-module $\bigoplus\limits_{p\in\mathbb{N}}\operatorname*{gr}%
\nolimits_{p}V$.\newline If the filtration $\left(  V_{n}\right)  _{n\geq0}$
is $\mathfrak{g}$-split, then there exists a bifiltered $\mathfrak{g}$-module
isomorphism $V\rightarrow\bigoplus\limits_{p\in\mathbb{N}}\operatorname*{gr}%
\nolimits_{p}V$.
\end{lemma}

\begin{lemma}
\label{lem.split.filtr-g.4}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie algebra. Let $V$ be a $\mathfrak{g}$-module. Let $\left(
V_{n}\right)  _{n\geq0}$ be a $\mathfrak{g}$-module filtration of $V$.
Consider the $\mathfrak{g}$-module $\bigoplus\limits_{p\in\mathbb{N}%
}\operatorname*{gr}\nolimits_{p}V$, and the filtration $\left(  \bigoplus
\limits_{p=0}^{n}\operatorname*{gr}\nolimits_{p}V\right)  _{n\geq0}$ of this
$\mathfrak{g}$-module $\bigoplus\limits_{p\in\mathbb{N}}\operatorname*{gr}%
\nolimits_{p}V$.\newline If there exists a bifiltered $\mathfrak{g}$-module
isomorphism $V\rightarrow\bigoplus\limits_{p\in\mathbb{N}}\operatorname*{gr}%
\nolimits_{p}V$, then the filtration $\left(  V_{n}\right)  _{n\geq0}$ is
$\mathfrak{g}$-split.
\end{lemma}

\textit{Proof of Lemma \ref{lem.split.filtr-g.3}.} Assume that the filtration
$\left(  V_{n}\right)  _{n\geq0}$ is $\mathfrak{g}$-split. Then, Lemma
\ref{lem.split.filtr-g.1} yields that there exists a family $\left(
W_{n}\right)  _{n\in\mathbb{N}}$ of $\mathfrak{g}$-submodules of $V$ such that
$V=\bigoplus\limits_{n\in\mathbb{N}}W_{n}$ and such that every $p\in
\mathbb{N}$ satisfies $V_{p}=\bigoplus\limits_{n=0}^{p}W_{n}$. Consider this
family $\left(  W_{n}\right)  _{n\in\mathbb{N}}$.

Every $q\in\mathbb{N}$ satisfies $V_{q-1}=\bigoplus\limits_{n=0}^{q-1}W_{n}%
$.\ \ \ \ {\footnotemark}\footnotetext{\textit{Proof.} Let $q\in\mathbb{N}%
$.\ We distinguish between two cases:
\par
\textit{Case 1:} We have $q=0$.
\par
\textit{Case 2:} We have $q\geq1$.
\par
In Case 1, we have $V_{q-1}=V_{0-1}=V_{-1}=0$ and $\bigoplus\limits_{n=0}%
^{q-1}W_{n}=\bigoplus\limits_{n=0}^{0-1}W_{n}=\left(  \text{empty sum}\right)
=0$, so that $V_{q-1}=\bigoplus\limits_{n=0}^{q-1}W_{n}$ in Case 1.
\par
In Case 2, we have $V_{q-1}=\bigoplus\limits_{n=0}^{q-1}W_{n}$ (this follows
from $V_{p}=\bigoplus\limits_{n=0}^{p}W_{n}$, applied to $p=q-1$).
\par
Therefore, we have proven that $V_{q-1}=\bigoplus\limits_{n=0}^{q-1}W_{n}$
holds in both possible cases 1 and 2. Thus, $V_{q-1}=\bigoplus\limits_{n=0}%
^{q-1}W_{n}$ is proven.} Renaming $q$ as $p$ in this fact, we thus obtain:
Every $p\in\mathbb{N}$ satisfies $V_{p-1}=\bigoplus\limits_{n=0}^{p-1}W_{n}$.
Thus, every $p\in\mathbb{N}$ satisfies%
\[
V_{p}=\bigoplus\limits_{n=0}^{p}W_{n}=\underbrace{\left(  \bigoplus
\limits_{n=0}^{p-1}W_{n}\right)  }_{=V_{p-1}}\oplus W_{p}=V_{p-1}\oplus
W_{p}.
\]


Let $p\in\mathbb{N}$.

Let $\rho_{p}$ be the canonical projection from $V_{p}$ to $V_{p}\diagup
V_{p-1}$. Clearly, $\rho_{p}$ is a surjective $\mathfrak{g}$-module
homomorphism satisfying $\rho_{p}\left(  V_{p-1}\right)  =0$.

Let $\pi_{p}$ be the projection from the direct sum $V_{p-1}\oplus W_{p}$ onto
its addend $W_{p}$. Then, $\pi_{p}$ is a surjective $\mathfrak{g}$-module
homomorphism $V_{p-1}\oplus W_{p}\rightarrow W_{p}$. Since $V_{p-1}\oplus
W_{p}=V_{p}$, we thus conclude that $\pi_{p}$ is a surjective $\mathfrak{g}%
$-module homomorphism $V_{p}\rightarrow W_{p}$.

Let $\iota_{p}$ be the injection of the $\mathfrak{g}$-module $W_{p}$ in the
direct sum $V_{p-1}\oplus W_{p}$. Then, $\iota_{p}$ is an injective
$\mathfrak{g}$-module homomorphism $W_{p}\rightarrow V_{p-1}\oplus W_{p}$.
Since $V_{p-1}\oplus W_{p}=V_{p}$, we thus conclude that $\iota_{p}$ is an
injective $\mathfrak{g}$-module homomorphism $W_{p}\rightarrow V_{p}$.

Since $\pi_{p}$ is the projection from the direct sum $V_{p-1}\oplus W_{p}$
onto its addend $W_{p}$, it must send the other addend $V_{p-1}$ to $0$. In
other words, $\pi_{p}\left(  V_{p-1}\right)  =0$. Therefore, by the
homomorphism theorem, the homomorphism $\pi_{p}$ factors through $V_{p}\diagup
V_{p-1}$. In other words, there exists a $\mathfrak{g}$-module homomorphism
$\overline{\pi_{p}}:V_{p}\diagup V_{p-1}\rightarrow W_{p}$ such that
$\overline{\pi_{p}}\circ\rho_{p}=\pi_{p}$ (since $\rho_{p}$ is the canonical
projection from $V_{p}$ to $V_{p}\diagup V_{p-1}$).

We have thus defined four $\mathfrak{g}$-module homomorphisms $\rho_{p}$,
$\pi_{p}$, $\iota_{p}$ and $\overline{\pi_{p}}$ for every $p\in\mathbb{N}$.

Let $p\in\mathbb{N}$ again.

Every $v\in W_{p}$ satisfies $\iota_{p}\left(  v\right)  =v$ (since $\iota
_{p}$ is the injection of the $\mathfrak{g}$-module $W_{p}$ in the direct sum
$V_{p-1}\oplus W_{p}$) and $\pi_{p}\left(  v\right)  =v$ (since $\pi_{p}$ is a
projection onto $W_{p}$, but $v\in W_{p}$). Thus, every $v\in W_{p}$ satisfies
$\left(  \pi_{p}\circ\iota_{p}\right)  \left(  v\right)  =\pi_{p}\left(
\underbrace{\iota_{p}\left(  v\right)  }_{=v}\right)  =\pi_{p}\left(
v\right)  =v=\operatorname*{id}\left(  v\right)  $. In other words, $\pi
_{p}\circ\iota_{p}=\operatorname*{id}$.

Now,%
\[
\overline{\pi_{p}}\circ\left(  \rho_{p}\circ\iota_{p}\right)
=\underbrace{\overline{\pi_{p}}\circ\rho_{p}}_{=\pi_{p}}\circ\iota_{p}=\pi
_{p}\circ\iota_{p}=\operatorname*{id}.
\]


On the other hand, let us prove that $\left(  \rho_{p}\circ\iota_{p}\right)
\circ\overline{\pi_{p}}=\operatorname*{id}$. In fact, let $w\in V_{p}\diagup
V_{p-1}$ be arbitrary. Then, there exists some $v\in V_{p}$ such that
$w=\rho_{p}\left(  v\right)  $ (since $\rho_{p}$ is surjective). Consider this
$v$. Since $w=\rho_{p}\left(  v\right)  $, we have%
\begin{align*}
\left(  \left(  \rho_{p}\circ\iota_{p}\right)  \circ\overline{\pi_{p}}\right)
\left(  w\right)   &  =\left(  \left(  \rho_{p}\circ\iota_{p}\right)
\circ\overline{\pi_{p}}\right)  \left(  \rho_{p}\left(  v\right)  \right)
=\left(  \left(  \rho_{p}\circ\iota_{p}\right)  \circ\underbrace{\overline
{\pi_{p}}\circ\rho_{p}}_{=\pi_{p}}\right)  \left(  v\right) \\
&  =\left(  \left(  \rho_{p}\circ\iota_{p}\right)  \circ\pi_{p}\right)
\left(  v\right)  =\rho_{p}\left(  \iota_{p}\left(  \pi_{p}\left(  v\right)
\right)  \right)  .
\end{align*}
Now, we can write the element $v\in V_{p}$ in the form $v=v_{1}+v_{2}$ for
some $v_{1}\in V_{p-1}$ and $v_{2}\in W_{p}$ (because $v\in V_{p}%
=V_{p-1}\oplus W_{p}$). Consider these $v_{1}$ and $v_{2}$. Since $\pi_{p}$ is
the projection from the direct sum $V_{p-1}\oplus W_{p}$ onto its addend
$W_{p}$, it satisfies $\pi_{p}\mid_{V_{p-1}}=0$ and $\pi_{p}\mid_{W_{p}%
}=\operatorname*{id}$. Now, $v=v_{1}+v_{2}$ yields
\begin{align*}
\pi_{p}\left(  v\right)   &  =\pi_{p}\left(  v_{1}+v_{2}\right)
=\underbrace{\pi_{p}\left(  v_{1}\right)  }_{\substack{=\left(  \pi_{p}%
\mid_{V_{p-1}}\right)  \left(  v_{1}\right)  \\\text{(since }v_{1}\in
V_{p-1}\text{)}}}+\underbrace{\pi_{p}\left(  v_{2}\right)  }%
_{\substack{=\left(  \pi_{p}\mid_{W_{p}}\right)  \left(  v_{2}\right)
\\\text{(since }v_{2}\in W_{p}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\pi_{p}\text{ is linear}\right) \\
&  =\underbrace{\left(  \pi_{p}\mid_{V_{p-1}}\right)  }_{=0}\left(
v_{1}\right)  +\underbrace{\left(  \pi_{p}\mid_{W_{p}}\right)  }%
_{=\operatorname*{id}}\left(  v_{2}\right)  =\underbrace{0\left(
v_{1}\right)  }_{=0}+\underbrace{\operatorname*{id}\left(  v_{2}\right)
}_{=v_{2}}=v_{2}.
\end{align*}
Besides, $\iota_{p}\left(  v_{2}\right)  =v_{2}$ (since $\iota_{p}$ is the
injection of the $\mathfrak{g}$-module $W_{p}$ in the direct sum
$V_{p-1}\oplus W_{p}$). Thus,%
\[
\left(  \left(  \rho_{p}\circ\iota_{p}\right)  \circ\overline{\pi_{p}}\right)
\left(  w\right)  =\rho_{p}\left(  \iota_{p}\left(  \underbrace{\pi_{p}\left(
v\right)  }_{=v_{2}}\right)  \right)  =\rho_{p}\left(  \underbrace{\iota
_{p}\left(  v_{2}\right)  }_{=v_{2}}\right)  =\rho_{p}\left(  v_{2}\right)  .
\]
Comparing this with%
\begin{align*}
\operatorname*{id}\left(  w\right)   &  =w=\rho_{p}\left(  \underbrace{v}%
_{=v_{1}+v_{2}}\right)  =\rho_{p}\left(  v_{1}+v_{2}\right)  =\underbrace{\rho
_{p}\left(  v_{1}\right)  }_{\substack{=0\text{ (since }v_{1}\in
V_{p-1}\\\text{and }\rho_{p}\left(  V_{p-1}\right)  =0\text{)}}}+\rho
_{p}\left(  v_{2}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\rho
_{p}\text{ is }k\text{-linear}\right) \\
&  =\rho_{p}\left(  v_{2}\right)  ,
\end{align*}
we obtain%
\[
\left(  \left(  \rho_{p}\circ\iota_{p}\right)  \circ\overline{\pi_{p}}\right)
\left(  w\right)  =\operatorname*{id}\left(  w\right)  .
\]
Since this holds for every $w\in V_{p}\diagup V_{p-1}$, we have thus shown
that $\left(  \rho_{p}\circ\iota_{p}\right)  \circ\overline{\pi_{p}%
}=\operatorname*{id}$.

We have now proven that $\overline{\pi_{p}}\circ\left(  \rho_{p}\circ\iota
_{p}\right)  =\operatorname*{id}$ and $\left(  \rho_{p}\circ\iota_{p}\right)
\circ\overline{\pi_{p}}=\operatorname*{id}$ for every $p\in\mathbb{N}$.

For every $p\in\mathbb{N}$, the $\mathfrak{g}$-module homomorphism
$\overline{\pi_{p}}:V_{p}\diagup V_{p-1}\rightarrow W_{p}$ is a $\mathfrak{g}%
$-module homomorphism $\operatorname*{gr}\nolimits_{p}V\rightarrow W_{p}$
(since $\operatorname*{gr}\nolimits_{p}V=V_{p}\diagup V_{p-1}$ by the
definition of $\operatorname*{gr}\nolimits_{p}V$). Therefore, the direct sum
$\bigoplus\limits_{p\in\mathbb{N}}\overline{\pi_{p}}$ is a $\mathfrak{g}%
$-module homomorphism $\bigoplus\limits_{p\in\mathbb{N}}\operatorname*{gr}%
\nolimits_{p}V\rightarrow\bigoplus\limits_{p\in\mathbb{N}}W_{p}$. Since
$V=\bigoplus\limits_{n\in\mathbb{N}}W_{n}=\bigoplus\limits_{p\in\mathbb{N}%
}W_{p}$ (here, we renamed $n$ as $p$), we can thus conclude that
$\bigoplus\limits_{p\in\mathbb{N}}\overline{\pi_{p}}$ is a $\mathfrak{g}%
$-module homomorphism $\bigoplus\limits_{p\in\mathbb{N}}\operatorname*{gr}%
\nolimits_{p}V\rightarrow V$.

For every $p\in\mathbb{N}$, the $\mathfrak{g}$-module homomorphism $\rho
_{p}\circ\iota_{p}:W_{p}\rightarrow V_{p}\diagup V_{p-1}$ is a $\mathfrak{g}%
$-module homomorphism $W_{p}\rightarrow\operatorname*{gr}\nolimits_{p}V$
(since $\operatorname*{gr}\nolimits_{p}V=V_{p}\diagup V_{p-1}$ by the
definition of $\operatorname*{gr}\nolimits_{p}V$). Therefore, the direct sum
$\bigoplus\limits_{p\in\mathbb{N}}\rho_{p}\circ\iota_{p}$ is a $\mathfrak{g}%
$-module homomorphism $\bigoplus\limits_{p\in\mathbb{N}}W_{p}\rightarrow
\bigoplus\limits_{p\in\mathbb{N}}\operatorname*{gr}\nolimits_{p}V$. Since
$V=\bigoplus\limits_{n\in\mathbb{N}}W_{n}=\bigoplus\limits_{p\in\mathbb{N}%
}W_{p}$ (here, we renamed $n$ as $p$), we can thus conclude that
$\bigoplus\limits_{p\in\mathbb{N}}\rho_{p}\circ\iota_{p}$ is a $\mathfrak{g}%
$-module homomorphism $V\rightarrow\bigoplus\limits_{p\in\mathbb{N}%
}\operatorname*{gr}\nolimits_{p}V$.

Since%
\[
\left(  \bigoplus\limits_{p\in\mathbb{N}}\overline{\pi_{p}}\right)
\circ\left(  \bigoplus\limits_{p\in\mathbb{N}}\rho_{p}\circ\iota_{p}\right)
=\bigoplus\limits_{p\in\mathbb{N}}\underbrace{\overline{\pi_{p}}\circ\left(
\rho_{p}\circ\iota_{p}\right)  }_{=\operatorname*{id}}=\bigoplus
\limits_{p\in\mathbb{N}}\operatorname*{id}=\operatorname*{id}%
\]
and%
\[
\left(  \bigoplus\limits_{p\in\mathbb{N}}\rho_{p}\circ\iota_{p}\right)
\circ\left(  \bigoplus\limits_{p\in\mathbb{N}}\overline{\pi_{p}}\right)
=\bigoplus\limits_{p\in\mathbb{N}}\underbrace{\left(  \rho_{p}\circ\iota
_{p}\right)  \circ\overline{\pi_{p}}}_{=\operatorname*{id}}=\bigoplus
\limits_{p\in\mathbb{N}}\operatorname*{id}=\operatorname*{id},
\]
the two $\mathfrak{g}$-module homomorphisms $\bigoplus\limits_{p\in\mathbb{N}%
}\overline{\pi_{p}}$ and $\bigoplus\limits_{p\in\mathbb{N}}\rho_{p}\circ
\iota_{p}$ are mutually inverse. Thus, these two $\mathfrak{g}$-module
homomorphisms $\bigoplus\limits_{p\in\mathbb{N}}\overline{\pi_{p}}$ and
$\bigoplus\limits_{p\in\mathbb{N}}\rho_{p}\circ\iota_{p}$ are $\mathfrak{g}%
$-module isomorphisms and satisfy $\left(  \bigoplus\limits_{p\in\mathbb{N}%
}\rho_{p}\circ\iota_{p}\right)  ^{-1}=\bigoplus\limits_{p\in\mathbb{N}%
}\overline{\pi_{p}}$.

Every $m\in\mathbb{N}$ satisfies $\left(  \bigoplus\limits_{p\in\mathbb{N}%
}\rho_{p}\circ\iota_{p}\right)  \left(  V_{m}\right)  \subseteq\bigoplus
\limits_{p=0}^{m}\operatorname*{gr}\nolimits_{p}V$.\ \ \ \ {\footnotemark}%
\footnotetext{\textit{Proof.} Let $m\in\mathbb{N}$ be arbitrary. Let $v\in
V_{m}$ be arbitrary. Then, $v\in V_{m}=\bigoplus\limits_{n=0}^{m}W_{n}$ (due
to $V_{p}=\bigoplus\limits_{n=0}^{p}W_{n}$, applied to $p=m$). Thus, there
exist $m+1$ elements $w_{0}$, $w_{1}$, $...$, $w_{m}$ of $V$ such that
$\left(  w_{i}\in W_{i}\text{ for every }i\in\left\{  0,1,...,m\right\}
\right)  $ and $v=\sum\limits_{i=0}^{m}w_{i}$. Consider these elements $w_{0}%
$, $w_{1}$, $...$, $w_{m}$. Then,%
\begin{align*}
\left(  \bigoplus\limits_{p\in\mathbb{N}}\rho_{p}\circ\iota_{p}\right)
\left(  v\right)   &  =\left(  \bigoplus\limits_{p\in\mathbb{N}}\rho_{p}%
\circ\iota_{p}\right)  \left(  \sum\limits_{i=0}^{m}w_{i}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }v=\sum\limits_{i=0}^{m}w_{i}\right)
\\
&  =\sum\limits_{i=0}^{m}\underbrace{\left(  \bigoplus\limits_{p\in\mathbb{N}%
}\rho_{p}\circ\iota_{p}\right)  \left(  w_{i}\right)  }_{\substack{=\left(
\rho_{i}\circ\iota_{i}\right)  \left(  w_{i}\right)  \\\text{(since }w_{i}\in
W_{i}\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\bigoplus
\limits_{p\in\mathbb{N}}\rho_{p}\circ\iota_{p}\text{ is }k\text{-linear}%
\right) \\
&  =\sum\limits_{i=0}^{m}\underbrace{\left(  \rho_{i}\circ\iota_{i}\right)
\left(  w_{i}\right)  }_{\substack{\in\operatorname*{gr}\nolimits_{i}%
V\\\text{(since }\rho_{i}\circ\iota_{i}:W_{i}\rightarrow\operatorname*{gr}%
\nolimits_{i}V\text{)}}}\in\sum\limits_{i=0}^{m}\operatorname*{gr}%
\nolimits_{i}V=\sum\limits_{p=0}^{m}\operatorname*{gr}\nolimits_{p}%
V\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed }i\text{ as }p\right) \\
&  =\bigoplus\limits_{p=0}^{m}\operatorname*{gr}\nolimits_{p}%
V\ \ \ \ \ \ \ \ \ \ \left(  \text{since the sum }\sum\limits_{p=0}%
^{m}\operatorname*{gr}\nolimits_{p}V\text{ is a direct sum}\right)  .
\end{align*}
Since this holds for every $v\in V_{m}$, we obtain $\left(  \bigoplus
\limits_{p\in\mathbb{N}}\rho_{p}\circ\iota_{p}\right)  \left(  V_{m}\right)
\subseteq\bigoplus\limits_{p=0}^{m}\operatorname*{gr}\nolimits_{p}V$, qed.} If
we rename $m$ as $n$ in this fact, we obtain the following result: Every
$n\in\mathbb{N}$ satisfies $\left(  \bigoplus\limits_{p\in\mathbb{N}}\rho
_{p}\circ\iota_{p}\right)  \left(  V_{n}\right)  \subseteq\bigoplus
\limits_{p=0}^{n}\operatorname*{gr}\nolimits_{p}V$. This yields that the map
$\bigoplus\limits_{p\in\mathbb{N}}\rho_{p}\circ\iota_{p}$ respects the
filtration (since the filtration on $V$ is $\left(  V_{n}\right)  _{n\geq0}$,
while the filtration on $\bigoplus\limits_{p\in\mathbb{N}}\operatorname*{gr}%
\nolimits_{p}V$ is $\left(  \bigoplus\limits_{p=0}^{n}\operatorname*{gr}%
\nolimits_{p}V\right)  _{n\geq0}$).

On the other hand, every $m\in\mathbb{N}$ satisfies $\left(  \bigoplus
\limits_{p\in\mathbb{N}}\overline{\pi_{p}}\right)  \left(  \bigoplus
\limits_{p=0}^{m}\operatorname*{gr}\nolimits_{p}V\right)  \subseteq V_{m}%
$.\ \ \ \ {\footnotemark}\footnotetext{\textit{Proof.} Let $m\in\mathbb{N}$ be
arbitrary. Let $v\in\bigoplus\limits_{p=0}^{m}\operatorname*{gr}%
\nolimits_{p}V$ be arbitrary. Then, there exist $m+1$ elements $v_{0}$,
$v_{1}$, $...$, $v_{m}$ of $\bigoplus\limits_{p\in\mathbb{N}}%
\operatorname*{gr}\nolimits_{p}V$ such that $\left(  v_{i}\in
\operatorname*{gr}\nolimits_{i}V\text{ for every }i\in\left\{
0,1,...,m\right\}  \right)  $ and $v=\sum\limits_{i=0}^{m}v_{i}$. Consider
these elements $v_{0}$, $v_{1}$, $...$, $v_{m}$. Then,%
\begin{align*}
\left(  \bigoplus\limits_{p\in\mathbb{N}}\overline{\pi_{p}}\right)  \left(
v\right)   &  =\left(  \bigoplus\limits_{p\in\mathbb{N}}\overline{\pi_{p}%
}\right)  \left(  \sum\limits_{i=0}^{m}v_{i}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }v=\sum\limits_{i=0}^{m}v_{i}\right)
\\
&  =\sum\limits_{i=0}^{m}\underbrace{\left(  \bigoplus\limits_{p\in\mathbb{N}%
}\overline{\pi_{p}}\right)  \left(  v_{i}\right)  }_{\substack{=\overline
{\pi_{i}}\left(  v_{i}\right)  \\\text{(since }v_{i}\in\operatorname*{gr}%
\nolimits_{i}V\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\bigoplus\limits_{p\in\mathbb{N}}\overline{\pi_{p}}\text{ is }k\text{-linear}%
\right) \\
&  =\sum\limits_{i=0}^{m}\underbrace{\overline{\pi_{i}}\left(  v_{i}\right)
}_{\substack{\in W_{i}\\\text{(since }\overline{\pi_{i}}:\operatorname*{gr}%
\nolimits_{i}V\rightarrow W_{i}\text{)}}}\in\sum\limits_{i=0}^{m}W_{i}%
=\sum\limits_{n=0}^{m}W_{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed
}i\text{ as }n\right)
\end{align*}
and%
\begin{align*}
V_{m}  &  =\bigoplus\limits_{n=0}^{m}W_{n}\ \ \ \ \ \ \ \ \ \ \left(
\text{due to }V_{p}=\bigoplus\limits_{n=0}^{p}W_{n}\text{, applied to
}p=m\right) \\
&  =\sum\limits_{n=0}^{m}W_{n}\ \ \ \ \ \ \ \ \ \ \left(  \text{since direct
sums are sums}\right)  ,
\end{align*}
so that%
\[
\left(  \bigoplus\limits_{p\in\mathbb{N}}\overline{\pi_{p}}\right)  \left(
v\right)  \in\sum\limits_{n=0}^{m}W_{n}=V_{m}.
\]
Since this holds for every $v\in\bigoplus\limits_{p=0}^{m}\operatorname*{gr}%
\nolimits_{p}V$, we obtain $\left(  \bigoplus\limits_{p\in\mathbb{N}}%
\overline{\pi_{p}}\right)  \left(  \bigoplus\limits_{p=0}^{m}%
\operatorname*{gr}\nolimits_{p}V\right)  \subseteq V_{m}$, qed.} If we rename
$m$ as $n$ in this fact, we obtain the following result: Every $n\in
\mathbb{N}$ satisfies $\left(  \bigoplus\limits_{p\in\mathbb{N}}\overline
{\pi_{p}}\right)  \left(  \bigoplus\limits_{p=0}^{n}\operatorname*{gr}%
\nolimits_{p}V\right)  \subseteq V_{n}$. This yields that the map
$\bigoplus\limits_{p\in\mathbb{N}}\overline{\pi_{p}}$ respects the filtration
(since the filtration on $V$ is $\left(  V_{n}\right)  _{n\geq0}$, while the
filtration on $\bigoplus\limits_{p\in\mathbb{N}}\operatorname*{gr}%
\nolimits_{p}V$ is $\left(  \bigoplus\limits_{p=0}^{n}\operatorname*{gr}%
\nolimits_{p}V\right)  _{n\geq0}$). Since $\left(  \bigoplus\limits_{p\in
\mathbb{N}}\rho_{p}\circ\iota_{p}\right)  ^{-1}=\bigoplus\limits_{p\in
\mathbb{N}}\overline{\pi_{p}}$, this rewrites as follows: The map $\left(
\bigoplus\limits_{p\in\mathbb{N}}\rho_{p}\circ\iota_{p}\right)  ^{-1}$
respects the filtration.

According to Definition \ref{defs.bifiltered} (applied to $\bigoplus
\limits_{p\in\mathbb{N}}\operatorname*{gr}\nolimits_{p}V$, $\left(
\bigoplus\limits_{p=0}^{n}\operatorname*{gr}\nolimits_{p}V\right)  _{n\geq0}$
and $\bigoplus\limits_{p\in\mathbb{N}}\rho_{p}\circ\iota_{p}$ instead of $W$,
$\left(  W_{n}\right)  _{n\geq0}$ and $f$), the isomorphism $\bigoplus
\limits_{p\in\mathbb{N}}\rho_{p}\circ\iota_{p}$ is bifiltered if and only if
both maps $\bigoplus\limits_{p\in\mathbb{N}}\rho_{p}\circ\iota_{p}$ and
$\left(  \bigoplus\limits_{p\in\mathbb{N}}\rho_{p}\circ\iota_{p}\right)
^{-1}$ respect the filtration. Since we know that both maps $\bigoplus
\limits_{p\in\mathbb{N}}\rho_{p}\circ\iota_{p}$ and $\left(  \bigoplus
\limits_{p\in\mathbb{N}}\rho_{p}\circ\iota_{p}\right)  ^{-1}$ respect the
filtration, we thus conclude that the isomorphism $\bigoplus\limits_{p\in
\mathbb{N}}\rho_{p}\circ\iota_{p}$ is bifiltered.

Thus we know that there exists a bifiltered $\mathfrak{g}$-module isomorphism
$V\rightarrow\bigoplus\limits_{p\in\mathbb{N}}\operatorname*{gr}%
\nolimits_{p}V$ (namely, the isomorphism $\bigoplus\limits_{p\in\mathbb{N}%
}\rho_{p}\circ\iota_{p}$). This proves Lemma \ref{lem.split.filtr-g.3}.

\textit{Proof of Lemma \ref{lem.split.filtr-g.4}.} Assume that there exists a
bifiltered $\mathfrak{g}$-module isomorphism $V\rightarrow\bigoplus
\limits_{p\in\mathbb{N}}\operatorname*{gr}\nolimits_{p}V$. Let $f$ be such an isomorphism.

For each $n\in\mathbb{N}$, let $W_{n}$ be the $\mathfrak{g}$-submodule
$f^{-1}\left(  \operatorname*{gr}\nolimits_{n}V\right)  $ of $V$ (here, of
course, $\operatorname*{gr}\nolimits_{n}V$ is considered as a $\mathfrak{g}%
$-submodule of $\bigoplus\limits_{p\in\mathbb{N}}\operatorname*{gr}%
\nolimits_{p}V$).

Then,%
\begin{align*}
V  &  =f^{-1}\left(  \bigoplus\limits_{p\in\mathbb{N}}\operatorname*{gr}%
\nolimits_{p}V\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is a
}k\text{-module isomorphism}\right) \\
&  =f^{-1}\left(  \bigoplus\limits_{n\in\mathbb{N}}\operatorname*{gr}%
\nolimits_{n}V\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed
}p\text{ as }n\right) \\
&  =\bigoplus\limits_{n\in\mathbb{N}}\underbrace{f^{-1}\left(
\operatorname*{gr}\nolimits_{n}V\right)  }_{=W_{n}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }f\text{ is a }k\text{-module isomorphism}\right) \\
&  =\bigoplus\limits_{n\in\mathbb{N}}W_{n}.
\end{align*}


Also, Definition \ref{defs.bifiltered} (applied to $\bigoplus\limits_{p\in
\mathbb{N}}\operatorname*{gr}\nolimits_{p}V$ and $\left(  \bigoplus
\limits_{p=0}^{n}\operatorname*{gr}\nolimits_{p}V\right)  _{n\geq0}$ instead
of $W$ and $\left(  W_{n}\right)  _{n\geq0}$) yields that the map $f$ is
bifiltered if and only if both maps $f$ and $f^{-1}$ respect the filtration.
Since we know that $f$ is bifiltered, we thus conclude that both maps $f$ and
$f^{-1}$ respect the filtration.

Every $n\in\mathbb{N}$ satisfies $f\left(  V_{n}\right)  \subseteq
\bigoplus\limits_{p=0}^{n}\operatorname*{gr}\nolimits_{p}V$ (since $f$
respects the filtration) and $f^{-1}\left(  \bigoplus\limits_{p=0}%
^{n}\operatorname*{gr}\nolimits_{p}V\right)  \subseteq V_{n}$ (since $f^{-1}$
respects the filtration). Thus, every $n\in\mathbb{N}$ satisfies
$V_{n}\subseteq f^{-1}\left(  \bigoplus\limits_{p=0}^{n}\operatorname*{gr}%
\nolimits_{p}V\right)  $ (since $f\left(  V_{n}\right)  \subseteq
\bigoplus\limits_{p=0}^{n}\operatorname*{gr}\nolimits_{p}V$). Combined with
$f^{-1}\left(  \bigoplus\limits_{p=0}^{n}\operatorname*{gr}\nolimits_{p}%
V\right)  \subseteq V_{n}$, this yields that $V_{n}=f^{-1}\left(
\bigoplus\limits_{p=0}^{n}\operatorname*{gr}\nolimits_{p}V\right)  $ for every
$n\in\mathbb{N}$. Renaming $n$ and $p$ as $p$ and $n$ in this fact, we obtain
the following result: $V_{p}=f^{-1}\left(  \bigoplus\limits_{n=0}%
^{p}\operatorname*{gr}\nolimits_{n}V\right)  $ for every $p\in\mathbb{N}$.

Thus, every $p\in\mathbb{N}$ satisfies%
\begin{align*}
V_{p}  &  =f^{-1}\left(  \bigoplus\limits_{n=0}^{p}\operatorname*{gr}%
\nolimits_{n}V\right)  =\bigoplus\limits_{n=0}^{p}\underbrace{f^{-1}\left(
\operatorname*{gr}\nolimits_{n}V\right)  }_{=W_{n}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is a }k\text{-module
isomorphism}\right) \\
&  =\bigoplus\limits_{n=0}^{p}W_{n}.
\end{align*}


Thus we have found a family $\left(  W_{n}\right)  _{n\in\mathbb{N}}$ of
$\mathfrak{g}$-submodules of $V$ such that $V=\bigoplus\limits_{n\in
\mathbb{N}}W_{n}$ and such that every $p\in\mathbb{N}$ satisfies
$V_{p}=\bigoplus\limits_{n=0}^{p}W_{n}$. Therefore, we can apply Lemma
\ref{lem.split.filtr-g.2}, and as a result we obtain that the filtration
$\left(  V_{n}\right)  _{n\geq0}$ is $\mathfrak{g}$-split. This proves Lemma
\ref{lem.split.filtr-g.4}.

Proposition \ref{prop.split.filtr-g-2} immediately follows from Lemma
\ref{lem.split.filtr-g.3} and Lemma \ref{lem.split.filtr-g.4}.
\end{verlong}

\subsection{A trivial lemma}

Here is one lemma, hardly worth the name, that we are going to use later in
the proof:

\begin{lemma}
\label{lem.surj}Let $k$ be a commutative ring. Let $\mathfrak{h}$ be a $k$-Lie
algebra. Let $A$, $B$ and $C$ be three $\mathfrak{h}$-modules. Let
$f:A\rightarrow B$ and $g:B\rightarrow C$ be two $k$-module homomorphisms such
that $f$ is a surjective $\mathfrak{h}$-module homomorphism. Assume that
$g\circ f$ is an $\mathfrak{h}$-module homomorphism. Then, $g$ is an
$\mathfrak{h}$-module homomorphism.
\end{lemma}

\textit{Proof of Lemma \ref{lem.surj}.} Let $a\in\mathfrak{h}$ and $v\in B$ be arbitrary.

Since $f$ is surjective, there is a $u\in A$ such that $v=f\left(  u\right)
$. Thus,%
\[
a\rightharpoonup v=a\rightharpoonup\left(  f\left(  u\right)  \right)
=f\left(  a\rightharpoonup u\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}f\text{ is an }\mathfrak{h}\text{-module homomorphism}\right)  ,
\]
and thus%
\begin{align*}
g\left(  a\rightharpoonup v\right)   &  =g\left(  f\left(  a\rightharpoonup
u\right)  \right)  =\left(  g\circ f\right)  \left(  a\rightharpoonup u\right)
\\
&  =a\rightharpoonup\underbrace{\left(  \left(  g\circ f\right)  \left(
u\right)  \right)  }_{=g\left(  f\left(  u\right)  \right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }g\circ f\text{ is an }\mathfrak{h}%
\text{-module homomorphism}\right) \\
&  =a\rightharpoonup\left(  g\left(  \underbrace{f\left(  u\right)  }%
_{=v}\right)  \right)  =a\rightharpoonup\left(  g\left(  v\right)  \right)  .
\end{align*}
Thus, we have shown that every $a\in\mathfrak{h}$ and $v\in B$ satisfy
$g\left(  a\rightharpoonup v\right)  =a\rightharpoonup\left(  g\left(
v\right)  \right)  $. In other words, $g$ is an $\mathfrak{h}$-module
homomorphism. This proves Lemma \ref{lem.surj}.

\subsection{A variation on the nine lemma}

The following fact is completely unrelated to Lie algebras. It is one of
several algebraic statements related to the nine lemma, but having both weaker
assertions and weaker conditions. We record it here to use it in Section
\ref{sect.pbw}:

\begin{proposition}
\label{prop.9l}Let $k$ be a commutative ring. Let $A$, $B$, $C$ and $D$ be
$k$-modules, and let $x:A\rightarrow B$, $y:A\rightarrow C$, $z:B\rightarrow
D$ and $w:C\rightarrow D$ be $k$-linear maps such that the diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%A \ar[r]^x \ar[d]_y & B \ar[d]^z \\
%C \ar[r]_w & D
%}} }%
%BeginExpansion
\xymatrix{
A \ar[r]^x \ar[d]_y & B \ar[d]^z \\
C \ar[r]_w & D
}
%EndExpansion
\label{prop.9l.diag}%
\end{equation}
commutes. Assume that $\operatorname*{Ker}z\subseteq x\left(
\operatorname*{Ker}y\right)  $. Further assume that $y$ is surjective. Then,
$\operatorname*{Ker}w=y\left(  \operatorname*{Ker}x\right)  $.
\end{proposition}

\textit{Proof of Proposition \ref{prop.9l}.} We know that the diagram%
\[%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrix{
%A \ar[r]^x \ar[d]_y & B \ar[d]^z \\
%C \ar[r]_w & D
%}}}%
%BeginExpansion
\xymatrix{
A \ar[r]^x \ar[d]_y & B \ar[d]^z \\
C \ar[r]_w & D
}%
%EndExpansion
\]
commutes. In other words, $w\circ y=z\circ x$.

We have%
\begin{align*}
w\left(  y\left(  \operatorname*{Ker}x\right)  \right)   &
=\underbrace{\left(  w\circ y\right)  }_{=z\circ x}\left(  \operatorname*{Ker}%
x\right)  =\left(  z\circ x\right)  \left(  \operatorname*{Ker}x\right)
=z\left(  \underbrace{x\left(  \operatorname*{Ker}x\right)  }_{=0}\right)
=z\left(  0\right)  =0\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }z\text{ is }k\text{-linear}%
\right)  ,
\end{align*}
and thus $y\left(  \operatorname*{Ker}x\right)  \subseteq\operatorname*{Ker}%
w$. We will now prove that $\operatorname*{Ker}w\subseteq y\left(
\operatorname*{Ker}x\right)  $:

Let $c\in\operatorname*{Ker}w$ be arbitrary. Then, $w\left(  c\right)  =0$.
Now, since $y$ is surjective, there exists some $a\in A$ such that $c=y\left(
a\right)  $. Consider this $a$. Then,
\[
0=w\left(  \underbrace{c}_{=y\left(  a\right)  }\right)  =w\left(  y\left(
a\right)  \right)  =\underbrace{\left(  w\circ y\right)  }_{=z\circ x}\left(
a\right)  =\left(  z\circ x\right)  \left(  a\right)  =z\left(  x\left(
a\right)  \right)  ,
\]
so that $x\left(  a\right)  \in\operatorname*{Ker}z\subseteq x\left(
\operatorname*{Ker}y\right)  $. Thus, there exists some $a^{\prime}%
\in\operatorname*{Ker}y$ such that $x\left(  a\right)  =x\left(  a^{\prime
}\right)  $. Consider this $a^{\prime}$. Since $x$ is $k$-linear, we have
$x\left(  a-a^{\prime}\right)  =\underbrace{x\left(  a\right)  }_{=x\left(
a^{\prime}\right)  }-x\left(  a^{\prime}\right)  =x\left(  a^{\prime}\right)
-x\left(  a^{\prime}\right)  =0$, so that $a-a^{\prime}\in\operatorname*{Ker}%
x$. Thus, $y\left(  a-a^{\prime}\right)  \in y\left(  \operatorname*{Ker}%
x\right)  $. But since%
\begin{align*}
y\left(  a-a^{\prime}\right)   &  =\underbrace{y\left(  a\right)  }%
_{=c}-\underbrace{y\left(  a^{\prime}\right)  }_{=0\text{ (since }a^{\prime
}\in\operatorname*{Ker}y\text{)}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}y\text{ is }k\text{-linear}\right) \\
&  =c-0=c,
\end{align*}
this rewrites as $c\in y\left(  \operatorname*{Ker}x\right)  $.

We have thus shown that every $c\in\operatorname*{Ker}w$ satisfies $c\in
y\left(  \operatorname*{Ker}x\right)  $. Thus, $\operatorname*{Ker}w\subseteq
y\left(  \operatorname*{Ker}x\right)  $. Combined with $y\left(
\operatorname*{Ker}x\right)  \subseteq\operatorname*{Ker}w$, this yields
$\operatorname*{Ker}w=y\left(  \operatorname*{Ker}x\right)  $. This proves
Proposition \ref{prop.9l}.

Note that we would not lose any generality if we would replace $k$ by
$\mathbb{Z}$ in the statement of Proposition \ref{prop.9l}, because every
$k$-module is an abelian group, i. e., a $\mathbb{Z}$-module (with additional
structure). We could actually generalize Proposition \ref{prop.9l} by
replacing "$k$-modules" by "groups" (not necessarily abelian), but we will not
have any use for Proposition \ref{prop.9l} in this generality here.

\section{\label{sect.L25}The isomorphism $\operatorname{gr}_{n}\left(  \left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  \right)  \protect\cong\mathfrak{n}^{\otimes
n}$}

\subsection{Statement of the theorem}

In this chapter, we are going to show the following fact (which generalizes
Lemma 3.4 in \cite{calaque1}):

\begin{theorem}
\label{thm.l25}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie
algebra, and let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Assume
that the inclusion $\mathfrak{h}\hookrightarrow\mathfrak{g}$ splits \textit{as
a }$k$\textit{-module inclusion} (but not necessarily as an $\mathfrak{h}%
$-module inclusion).\newline Let $J$ be the two-sided ideal%
\[
\left(  \otimes\mathfrak{g}\right)  \cdot\left\langle v\otimes w-w\otimes
v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}%
\times\mathfrak{h}\right\rangle \cdot\left(  \otimes\mathfrak{g}\right)
\]
of the $k$-algebra $\otimes\mathfrak{g}$.\newline As we know, $\mathfrak{g}$
is a $\mathfrak{g}$-module, and thus also an $\mathfrak{h}$-module (by
Definition \ref{defs.g-as-h-module}). Let $\mathfrak{n}=\mathfrak{g}%
\diagup\mathfrak{h}$. This $\mathfrak{n}$ is an $\mathfrak{h}$-module (because
both $\mathfrak{g}$ and $\mathfrak{h}$ are $\mathfrak{h}$-modules).\newline
Let $\pi:\mathfrak{g}\rightarrow\mathfrak{n}$ be the canonical projection with
kernel $\mathfrak{h}$. Obviously, $\pi$ is an $\mathfrak{h}$-module
homomorphism. Thus, $\otimes\pi:\otimes\mathfrak{g}\rightarrow\otimes
\mathfrak{n}$ is also an $\mathfrak{h}$-module homomorphism (according to
Proposition \ref{prop.(X)-hom-is-hom}).\newline We consider $\mathfrak{h}$ as
an $\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$ by means of the embedding
$\mathfrak{h}\hookrightarrow\mathfrak{g}\hookrightarrow\otimes\mathfrak{g}%
$.\newline\textbf{(a)} Both $J$ and $\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}$ are $\mathfrak{h}$-submodules of $\otimes\mathfrak{g}$.
Thus, $\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ is an $\mathfrak{h}%
$-module. Let $\zeta:\otimes\mathfrak{g}\rightarrow\left(  \otimes
\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  $ be the canonical projection. Then, $\zeta$ is an
$\mathfrak{h}$-module homomorphism.\newline\textbf{(b)} For every
$n\in\mathbb{N}$, let $F_{n}$ be the $\mathfrak{h}$-submodule $\zeta\left(
\mathfrak{g}^{\otimes\leq n}\right)  $ of $\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
$.\ \ \ \ {\footnotemark} Also define an $\mathfrak{h}$-submodule $F_{-1}$ of
$\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ by $F_{-1}=0$. Then, $\left(
F_{n}\right)  _{n\geq0}$ is an $\mathfrak{h}$-module filtration of $\left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  $ and satisfies $F_{n}\diagup F_{n-1}%
\cong\mathfrak{n}^{\otimes n}$ as $\mathfrak{h}$-modules for every
$n\in\mathbb{N}$.\newline\textbf{(c)} Let $n\in\mathbb{N}$. There exists one
and only one $k$-module homomorphism $\Omega_{n}:F_{n}\diagup F_{n-1}%
\rightarrow\operatorname*{gr}\nolimits_{n}\left(  \otimes\mathfrak{n}\right)
$ for which the diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{9pc}
%\xymatrix{
%\operatorname*{gr}_n\left(\otimes\mathfrak{g}\right) \ar[d]_{\operatorname
%*{gr}_n\zeta} \ar[dr]^{\operatorname*{gr}_n\left(\otimes\pi\right)} & \\
%\operatorname*{gr}_n\left(\left(\otimes\mathfrak{g}\right)\diagup
%\left(J + \left(\otimes\mathfrak{g}\right)\cdot\mathfrak{h}\right
%) \right) = F_n \diagup F_{n-1} \ar[r]_-{\Omega_n} & \operatorname*{gr}%
%_n\left(\otimes\mathfrak{n}\right)
%}} }%
%BeginExpansion
\xymatrixcolsep{9pc}
\xymatrix{
\operatorname*{gr}_n\left(\otimes\mathfrak{g}\right) \ar[d]_{\operatorname
*{gr}_n\zeta} \ar[dr]^{\operatorname*{gr}_n\left(\otimes\pi\right)} & \\
\operatorname*{gr}_n\left(\left(\otimes\mathfrak{g}\right)\diagup
\left(J + \left(\otimes\mathfrak{g}\right)\cdot\mathfrak{h}\right
) \right) = F_n \diagup F_{n-1} \ar[r]_-{\Omega_n} & \operatorname*{gr}%
_n\left(\otimes\mathfrak{n}\right)
}
%EndExpansion
\label{thm.l25.c.diag}%
\end{equation}
commutes. Denote this homomorphism $\Omega_{n}$ by $\omega_{n}$. Then,
$\omega_{n}$ is an $\mathfrak{h}$-module isomorphism, and the diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{9pc}
%\xymatrix{
%\operatorname*{gr}_n\left(\otimes\mathfrak{g}\right) \ar[d]_{\operatorname
%*{gr}_n\zeta} \ar[dr]^{\operatorname*{gr}_n\left(\otimes\pi\right)} & \\
%\operatorname*{gr}_n\left(\left(\otimes\mathfrak{g}\right)\diagup
%\left(J + \left(\otimes\mathfrak{g}\right)\cdot\mathfrak{h}\right
%) \right) = F_n \diagup F_{n-1} \ar[r]_-{\omega_n} & \operatorname*{gr}%
%_n\left(\otimes\mathfrak{n}\right)
%}} }%
%BeginExpansion
\xymatrixcolsep{9pc}
\xymatrix{
\operatorname*{gr}_n\left(\otimes\mathfrak{g}\right) \ar[d]_{\operatorname
*{gr}_n\zeta} \ar[dr]^{\operatorname*{gr}_n\left(\otimes\pi\right)} & \\
\operatorname*{gr}_n\left(\left(\otimes\mathfrak{g}\right)\diagup
\left(J + \left(\otimes\mathfrak{g}\right)\cdot\mathfrak{h}\right
) \right) = F_n \diagup F_{n-1} \ar[r]_-{\omega_n} & \operatorname*{gr}%
_n\left(\otimes\mathfrak{n}\right)
}
%EndExpansion
\label{thm.l25.c.diag'}%
\end{equation}
commutes.\newline Applying Definition \ref{defs.filtr.grad} to $\mathfrak{n}$
and $n$ instead of $V$ and $p$, we obtain a map $\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}:\mathfrak{n}^{\otimes n}\rightarrow
\operatorname*{gr}\nolimits_{n}\left(  \otimes\mathfrak{n}\right)  $.
According to Proposition \ref{prop.filtr.grad.g} (applied to $\mathfrak{h}$,
$n$ and $\mathfrak{n}$ instead of $\mathfrak{g}$, $p$ and $V$), this map
$\operatorname*{grad}\nolimits_{\mathfrak{n},n}$ is a canonical $\mathfrak{h}%
$-module isomorphism. Thus, its inverse $\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}^{-1}$ is an $\mathfrak{h}$-module isomorphism as
well. The composition $\operatorname*{grad}\nolimits_{\mathfrak{n},n}%
^{-1}\circ\omega_{n}:F_{n}\diagup F_{n-1}\rightarrow\mathfrak{n}^{\otimes n}$
is an $\mathfrak{h}$-module isomorphism (because $\omega_{n}$ and
$\operatorname*{grad}\nolimits_{\mathfrak{n},n}^{-1}$ are $\mathfrak{h}%
$-module isomorphisms).
\end{theorem}

\footnotetext{In fact, $\zeta\left(  \mathfrak{g}^{\otimes\leq n}\right)  $ is
indeed an $\mathfrak{h}$-submodule (because $\zeta$ is an $\mathfrak{h}%
$-module homomorphism and because $\mathfrak{g}^{\otimes\leq n}$ is an
$\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$).}

We are going to prove this theorem at the end of Section \ref{sect.L25} (by
means of explicitly constructing an isomorphism $F_{n}\diagup F_{n-1}%
\rightarrow\mathfrak{n}^{\otimes n}$), after paving our way with some
auxiliary results. But let us first remark what this theorem actually says:

Theorem \ref{thm.l25} \textbf{(a)} is a result of computational nature and
does not require the condition that the inclusion $\mathfrak{h}\hookrightarrow
\mathfrak{g}$ splits as a $k$-module inclusion. We are going to prove it in
Proposition \ref{prop.l25a} \textbf{(a)}. This is \textit{not} the difficult
part of Theorem \ref{thm.l25}.

Theorem \ref{thm.l25} \textbf{(b)} (or, more precisely, its "$F_{n}\diagup
F_{n-1}\cong\mathfrak{n}^{\otimes n}$ as $\mathfrak{h}$-modules" part) is what
Lemma 3.4 in \cite{calaque1} states (except that \cite{calaque1} only
considers the case when $k$ is a field, and of course the condition that the
inclusion $\mathfrak{h}\hookrightarrow\mathfrak{g}$ splits as a $k$-module
becomes void in this case).

\begin{vershort}
Theorem \ref{thm.l25} \textbf{(c)}, however, is what Lemma 3.4 in
\cite{calaque1} actually \textbf{means to state}, and what is actually used
later on. For example, the proof of Theorem 4.4 in \cite{calaque1} requires
Theorem \ref{thm.l25} \textbf{(c)}; just having Theorem \ref{thm.l25}
\textbf{(b)} would not be enough for it.
\end{vershort}

\begin{verlong}
Theorem \ref{thm.l25} \textbf{(c)}, however, is what Lemma 3.4 in
\cite{calaque1} actually \textbf{means to state}. In \cite{calaque1}, as in
many papers on algebra, there is a slight discrepancy between the statements
of some results and the actual meaning that the authors give to these results.
This discrepancy manifests itself in the fact that Lemma 3.4 in
\cite{calaque1} formally only states that $F_{n}\diagup F_{n-1}\cong%
\mathfrak{n}^{\otimes n}$ as $\mathfrak{h}$-modules, i. e., that there exists
some $\mathfrak{h}$-module isomorphism $F_{n}\diagup F_{n-1}\rightarrow
\mathfrak{n}^{\otimes n}$, but the authors actually \textbf{want to say} that
\textbf{a very particular map }$F_{n}\diagup F_{n-1}\rightarrow\mathfrak{n}%
^{\otimes n}$ is an $\mathfrak{h}$-module isomorphism (namely, the map
$\operatorname*{grad}\nolimits_{\mathfrak{n},n}^{-1}\circ\omega_{n}$ of our
Theorem \ref{thm.l25} \textbf{(c)}). This assertion (that
$\operatorname*{grad}\nolimits_{\mathfrak{n},n}^{-1}\circ\omega_{n}$ is an
$\mathfrak{h}$-module isomorphism) is a stronger assertion than just the
existence of an $\mathfrak{h}$-module isomorphism $F_{n}\diagup F_{n-1}%
\rightarrow\mathfrak{n}^{\otimes n}$, and \textbf{this stronger assertion is
the one actually used later on} (for example, in the proof of Theorem 4.4 in
\cite{calaque1}). Therefore, if we would only prove Theorem \ref{thm.l25}
\textbf{(b)} (but not Theorem \ref{thm.l25} \textbf{(c)}), we would miss some
of the tools we would need later on.
\end{verlong}

Also note that Theorem \ref{thm.l25} \textbf{(c)} shows that not only does
there exist an $\mathfrak{h}$-module isomorphism $F_{n}\diagup F_{n-1}%
\rightarrow\mathfrak{n}^{\otimes n}$, but also that there exists an
$\mathfrak{h}$-module isomorphism $F_{n}\diagup F_{n-1}\rightarrow
\mathfrak{n}^{\otimes n}$ independent of choice of the splitting of the
$k$-module inclusion $\mathfrak{h}\hookrightarrow\mathfrak{g}$. See
Proposition \ref{prop.l25.ind} for the details of this.

\begin{remark}
The condition that the inclusion $\mathfrak{h}\hookrightarrow\mathfrak{g}$
split \textit{as a }$k$\textit{-module inclusion }\textbf{cannot be removed
from Theorem \ref{thm.l25}}, as the counterexample $k=\mathbb{Z}$,
$\mathfrak{g}=\mathfrak{sl}_{2}\mathbb{Q}$, $\mathfrak{h}=\left(
\mathfrak{sl}_{2}\mathbb{Q}\right)  \cap\mathbb{Z}^{2\times2}$ shows. (In this
counterexample, $\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}$
contains $\mathfrak{g}^{\otimes i}$ for all $i\geq2$, and thus we can easily
see that $J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}$ contains
$\mathfrak{g}^{\otimes i}$ for all $i\geq1$, so that $\left(  \otimes
\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  \cong\mathbb{Z}$ and $F_{1}=F_{0}$, whereas
$\mathfrak{n}^{\otimes1}=\left(  \mathfrak{sl}_{2}\mathbb{Q}\right)
\diagup\left(  \left(  \mathfrak{sl}_{2}\mathbb{Q}\right)  \cap\mathbb{Z}%
^{2\times2}\right)  \neq0$.)\newline I do not know whether a flatness
hypothesis of some kind (like requiring $\mathfrak{g}\diagup\mathfrak{h}$ to
be a flat $k$-module) could be used instead of this splitting condition.
\end{remark}

\subsection{$J$ and $\left(  \otimes\mathfrak{g}\right)  \mathfrak{h}$ are
$\mathfrak{h}$-submodules of $\otimes\mathfrak{g}$}

Let us start by proving Theorem \ref{thm.l25} \textbf{(a)} without the
condition that the inclusion $\mathfrak{h}\hookrightarrow\mathfrak{g}$ splits
as a $k$-module inclusion. More precisely let us prove the following:

\begin{proposition}
\label{prop.l25a}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra, and let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$.\newline Let $J$ be the two-sided ideal%
\[
\left(  \otimes\mathfrak{g}\right)  \cdot\left\langle v\otimes w-w\otimes
v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}%
\times\mathfrak{h}\right\rangle \cdot\left(  \otimes\mathfrak{g}\right)
\]
of the $k$-algebra $\otimes\mathfrak{g}$.\newline As we know, $\mathfrak{g}$
is a $\mathfrak{g}$-module, and thus also an $\mathfrak{h}$-module (by
Definition \ref{defs.g-as-h-module}). We consider $\mathfrak{h}$ as an
$\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$ by means of the embedding
$\mathfrak{h}\hookrightarrow\mathfrak{g}\hookrightarrow\otimes\mathfrak{g}%
$.\newline\textbf{(a)} Both $J$ and $\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}$ are $\mathfrak{h}$-submodules of $\otimes\mathfrak{g}$.
Thus, $\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ is an $\mathfrak{h}%
$-module. Let $\zeta:\otimes\mathfrak{g}\rightarrow\left(  \otimes
\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  $ be the canonical projection. Then, $\zeta$ is an
$\mathfrak{h}$-module homomorphism.\newline\textbf{(b)} Let $J_{0}$ denote the
$k$-submodule $\left\langle v\otimes w-w\otimes v-\left[  v,w\right]
\ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}\right\rangle $
of $\otimes\mathfrak{g}$. Then, $J=\left(  \otimes\mathfrak{g}\right)  \cdot
J_{0}\cdot\left(  \otimes\mathfrak{g}\right)  $. Both $J_{0}$ and $J_{0}%
\cdot\left(  \otimes\mathfrak{g}\right)  $ are $\mathfrak{h}$-submodules of
$\otimes\mathfrak{g}$.
\end{proposition}

\begin{vershort}
\textit{Proof of Proposition \ref{prop.l25a}.} \textbf{(b)} From the
definitions of $J$ and $J_{0}$, it is immediately clear that $J=\left(
\otimes\mathfrak{g}\right)  \cdot J_{0}\cdot\left(  \otimes\mathfrak{g}%
\right)  $. Also, it is clear that once $J_{0}$ is proven to be an
$\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$, then $J_{0}\cdot\left(
\otimes\mathfrak{g}\right)  $ also must be an $\mathfrak{h}$-submodule of
$\otimes\mathfrak{g}$ (by Proposition \ref{prop.g-alg.PQ}, applied to $J_{0}$,
$\otimes\mathfrak{g}$, $\otimes\mathfrak{g}$ and $\mathfrak{h}$ instead of
$P$, $Q$, $A$ and $\mathfrak{g}$). Therefore, in order to prove Proposition
\ref{prop.l25a} \textbf{(b)}, it only remains to show that $J_{0}$ is an
$\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$. This means proving that
$a\rightharpoonup x\in J_{0}$ for every $a\in\mathfrak{g}$ and $x\in J_{0}$.

For this, it is enough to prove that $a\rightharpoonup\left(  p\otimes
q-q\otimes p-\left[  p,q\right]  \right)  \in J_{0}$ for every $a\in
\mathfrak{g}$ and $\left(  p,q\right)  \in\mathfrak{g}\times\mathfrak{h}$
(because $J_{0}=\left\langle v\otimes w-w\otimes v-\left[  v,w\right]
\ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}\right\rangle $
yields that every $x\in J_{0}$ is a linear combination of terms of the form
$p\otimes q-q\otimes p-\left[  p,q\right]  $ with $\left(  p,q\right)
\in\mathfrak{g}\times\mathfrak{h}$). But this can be done by direct
computation:%
\begin{align*}
&  a\rightharpoonup\left(  p\otimes q-q\otimes p-\left[  p,q\right]  \right)
\\
&  =\underbrace{a\rightharpoonup\left(  p\otimes q\right)  }%
_{\substack{=\left(  a\rightharpoonup p\right)  \otimes q+p\otimes\left(
a\rightharpoonup q\right)  }}-\underbrace{a\rightharpoonup\left(  q\otimes
p\right)  }_{\substack{=\left(  a\rightharpoonup q\right)  \otimes
p+q\otimes\left(  a\rightharpoonup p\right)  }}-a\rightharpoonup\left[
p,q\right] \\
&  =\underbrace{\left(  a\rightharpoonup p\right)  }_{\substack{=\left[
a,p\right]  }}\otimes q+p\otimes\underbrace{\left(  a\rightharpoonup q\right)
}_{\substack{=\left[  a,q\right]  }}-\underbrace{\left(  a\rightharpoonup
q\right)  }_{\substack{=\left[  a,q\right]  }}\otimes p-q\otimes
\underbrace{\left(  a\rightharpoonup p\right)  }_{\substack{=\left[
a,p\right]  }}-\underbrace{a\rightharpoonup\left[  p,q\right]  }%
_{\substack{=\left[  a,\left[  p,q\right]  \right]  =\left[  p,\left[
a,q\right]  \right]  +\left[  \left[  a,p\right]  ,q\right]  \\\text{(by
antisymmetry and Jacobi)}}}\\
&  =\left[  a,p\right]  \otimes q+p\otimes\left[  a,q\right]  -\left[
a,q\right]  \otimes p-q\otimes\left[  a,p\right]  -\left[  p,\left[
a,q\right]  \right]  -\left[  \left[  a,p\right]  ,q\right] \\
&  =\underbrace{\left(  \left[  a,p\right]  \otimes q-q\otimes\left[
a,p\right]  -\left[  \left[  a,p\right]  ,q\right]  \right)  }_{\in
J_{0}\text{ (since }\left[  a,p\right]  \in\mathfrak{h}\text{)}}%
+\underbrace{\left(  p\otimes\left[  a,q\right]  -\left[  a,q\right]  \otimes
p-\left[  p,\left[  a,q\right]  \right]  \right)  }_{\in J_{0}\text{ (since
}p\in\mathfrak{h}\text{)}}\\
&  \in J_{0}+J_{0}\subseteq J_{0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}J_{0}\text{ is a }k\text{-module}\right)  .
\end{align*}
This proves Proposition \ref{prop.l25a} \textbf{(b).}

\textbf{(a)} Since both $\otimes\mathfrak{g}$ and $\mathfrak{h}$ are
$\mathfrak{h}$-submodules of $\otimes\mathfrak{g}$, it follows from
Proposition \ref{prop.g-alg.PQ} (applied to $\otimes\mathfrak{g}$,
$\mathfrak{h}$, $\otimes\mathfrak{g}$ and $\mathfrak{h}$ instead of $P$, $Q$,
$A$ and $\mathfrak{g}$) that $\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}$ is an $\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$.

Since $\otimes\mathfrak{g}$ and $J_{0}\cdot\left(  \otimes\mathfrak{g}\right)
$ are $\mathfrak{h}$-submodules of $\otimes\mathfrak{g}$, Proposition
\ref{prop.g-alg.PQ} (applied to $\otimes\mathfrak{g}$, $J_{0}\cdot\left(
\otimes\mathfrak{g}\right)  $, $\otimes\mathfrak{g}$ and $\mathfrak{h}$
instead of $P$, $Q$, $A$ and $\mathfrak{g}$) yields that $\left(
\otimes\mathfrak{g}\right)  \cdot J_{0}\cdot\left(  \otimes\mathfrak{g}%
\right)  $ is an $\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$. Since
$\left(  \otimes\mathfrak{g}\right)  \cdot J_{0}\cdot\left(  \otimes
\mathfrak{g}\right)  =J$ by Proposition \ref{prop.l25a} \textbf{(b)}, we
conclude that $J$ is an $\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$.

Since we now know that both $J$ and $\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}$ are $\mathfrak{h}$-submodules of $\otimes\mathfrak{g}$, the
sum $J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}$ must also be an
$\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$. Thus, $\left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  $ is an $\mathfrak{h}$-module. Therefore,
the canonical projection $\zeta:\otimes\mathfrak{g}\rightarrow\left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  $ is an $\mathfrak{h}$-module homomorphism.
This completes the proof of Proposition \ref{prop.l25a} \textbf{(a)}.
\end{vershort}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.l25a}.} \textbf{(b)} Clearly,%
\[
J=\left(  \otimes\mathfrak{g}\right)  \cdot\underbrace{\left\langle v\otimes
w-w\otimes v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)  \in
\mathfrak{g}\times\mathfrak{h}\right\rangle }_{=J_{0}}\cdot\left(
\otimes\mathfrak{g}\right)  =\left(  \otimes\mathfrak{g}\right)  \cdot
J_{0}\cdot\left(  \otimes\mathfrak{g}\right)  .
\]


Next, we are going to prove that $J_{0}$ is an $\mathfrak{h}$-submodule of
$\otimes\mathfrak{g}$.

Clearly, $J_{0}=\left\langle v\otimes w-w\otimes v-\left[  v,w\right]
\ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}\right\rangle $
yields that%
\begin{equation}
v\otimes w-w\otimes v-\left[  v,w\right]  \in J_{0}%
\ \ \ \ \ \ \ \ \ \ \text{for every }v\in\mathfrak{g}\text{ and }%
w\in\mathfrak{h}. \label{pf.l25a.J0}%
\end{equation}


Let $a\in\mathfrak{h}$ be arbitrary. Let us denote by $L_{a}$ the map
$\otimes\mathfrak{g}\rightarrow\otimes\mathfrak{g},$ $x\mapsto
a\rightharpoonup x$. This map $L_{a}$ is $k$-linear (since the Lie action of
$\otimes\mathfrak{g}$ is $k$-bilinear).

We are now going to prove that%
\begin{equation}
a\rightharpoonup\left(  p\otimes q-q\otimes p-\left[  p,q\right]  \right)  \in
J_{0}\ \ \ \ \ \ \ \ \ \ \text{for every }p\in\mathfrak{g}\text{ and }%
q\in\mathfrak{h}. \label{pf.l25a.1}%
\end{equation}


\textit{Proof of (\ref{pf.l25a.1}).} Let $p\in\mathfrak{g}$ and $q\in
\mathfrak{h}$ be arbitrary. Then, $\left[  q,a\right]  =-\left[  a,q\right]  $
(by (\ref{[]-3}), applied to $q$ and $a$ instead of $v$ and $w$) and $\left[
q,\left[  a,p\right]  \right]  =-\left[  \left[  a,p\right]  ,q\right]  $ (by
(\ref{[]-3}), applied to $q$ and $\left[  a,p\right]  $ instead of $v$ and
$w$). Also, note that $\left[  a,p\right]  \in\mathfrak{h}$ (because
$a\in\mathfrak{h}$ and $p\in\mathfrak{h}$, and because $\mathfrak{h}$ is a Lie
subalgebra of $\mathfrak{g}$). Furthermore, (\ref{prop.g-is-g-module.form})
(applied to $a$ and $p$ instead of $v$ and $w$) yields $a\rightharpoonup
p=\left[  a,p\right]  $. On the other hand, (\ref{prop.g-is-g-module.form})
(applied to $a$ and $q$ instead of $v$ and $w$) yields $a\rightharpoonup
q=\left[  a,q\right]  $. Besides, (\ref{prop.g-is-g-module.form}) (applied to
$a$ and $\left[  p,q\right]  $ instead of $v$ and $w$) yields%
\begin{align*}
a\rightharpoonup\left[  p,q\right]   &  =\left[  a,\left[  p,q\right]
\right]  =-\left[  p,\underbrace{\left[  q,a\right]  }_{\substack{=-\left[
a,q\right]  }}\right]  -\underbrace{\left[  q,\left[  a,p\right]  \right]
}_{=-\left[  \left[  a,p\right]  ,q\right]  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since (\ref{[]-2}) (applied to }a\text{, }p\text{, }q\text{ instead of
}u\text{, }v\text{, }w\text{) yields}\\
\left[  a,\left[  p,q\right]  \right]  +\left[  p,\left[  q,a\right]  \right]
+\left[  q,\left[  a,p\right]  \right]  =0
\end{array}
\right) \\
&  =\underbrace{-\left[  p,-\left[  a,q\right]  \right]  }_{=\left[  p,\left[
a,q\right]  \right]  }-\left(  -\left[  \left[  a,p\right]  ,q\right]
\right)  =\left[  p,\left[  a,q\right]  \right]  +\left[  \left[  a,p\right]
,q\right]  .
\end{align*}


Now,%
\begin{align*}
&  a\rightharpoonup\left(  p\otimes q-q\otimes p-\left[  p,q\right]  \right)
\\
&  =\underbrace{a\rightharpoonup\left(  p\otimes q\right)  }%
_{\substack{=\left(  a\rightharpoonup p\right)  \otimes q+p\otimes\left(
a\rightharpoonup q\right)  \\\text{(by (\ref{prop.lie(X).form}), applied
to}\\p\text{ and }q\text{ instead of }v\text{ and }w\text{)}}%
}-\underbrace{a\rightharpoonup\left(  q\otimes p\right)  }_{\substack{=\left(
a\rightharpoonup q\right)  \otimes p+q\otimes\left(  a\rightharpoonup
p\right)  \\\text{(by (\ref{prop.lie(X).form}), applied to}\\q\text{ and
}p\text{ instead of }v\text{ and }w\text{)}}}-a\rightharpoonup\left[
p,q\right] \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Lie action of }\otimes
\mathfrak{g}\text{ is }k\text{-bilinear}\right) \\
&  =\underbrace{\left(  a\rightharpoonup p\right)  }_{\substack{=\left[
a,p\right]  }}\otimes q+p\otimes\underbrace{\left(  a\rightharpoonup q\right)
}_{\substack{=\left[  a,q\right]  }}-\underbrace{\left(  a\rightharpoonup
q\right)  }_{\substack{=\left[  a,q\right]  }}\otimes p-q\otimes
\underbrace{\left(  a\rightharpoonup p\right)  }_{\substack{=\left[
a,p\right]  }}-\underbrace{a\rightharpoonup\left[  p,q\right]  }_{=\left[
p,\left[  a,q\right]  \right]  +\left[  \left[  a,p\right]  ,q\right]  }\\
&  =\left[  a,p\right]  \otimes q+p\otimes\left[  a,q\right]  -\left[
a,q\right]  \otimes p-q\otimes\left[  a,p\right]  -\left[  p,\left[
a,q\right]  \right]  -\left[  \left[  a,p\right]  ,q\right] \\
&  =\underbrace{\left(  \left[  a,p\right]  \otimes q-q\otimes\left[
a,p\right]  -\left[  \left[  a,p\right]  ,q\right]  \right)  }_{\in
J_{0}\text{ (by (\ref{pf.l25a.J0}) (applied to }v=\left[  a,p\right]  \text{
and }w=q\text{), since }\left[  a,p\right]  \in\mathfrak{h}\text{)}%
}+\underbrace{\left(  p\otimes\left[  a,q\right]  -\left[  a,q\right]  \otimes
p-\left[  p,\left[  a,q\right]  \right]  \right)  }_{\in J_{0}\text{ (by
(\ref{pf.l25a.J0}) (applied to }v=p\text{ and }w=\left[  a,q\right]  \text{),
since }p\in\mathfrak{h}\text{)}}\\
&  \in J_{0}+J_{0}\subseteq J_{0}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}J_{0}\text{ is a }k\text{-module}\right)  .
\end{align*}
This proves (\ref{pf.l25a.1}).

Now we have:%
\begin{equation}
\left\{  a\rightharpoonup\left(  v\otimes w-w\otimes v-\left[  v,w\right]
\right)  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}%
\right\}  \subseteq J_{0}. \label{pf.l25a.2}%
\end{equation}
\textit{Proof of (\ref{pf.l25a.2}).} Let $\gamma\in\left\{  a\rightharpoonup
\left(  v\otimes w-w\otimes v-\left[  v,w\right]  \right)  \ \mid\ \left(
v,w\right)  \in\mathfrak{g}\times\mathfrak{h}\right\}  $ be arbitrary. Then,
$\gamma$ has the form $\gamma=a\rightharpoonup\left(  p\otimes q-q\otimes
p-\left[  p,q\right]  \right)  $ for some $\left(  p,q\right)  \in
\mathfrak{g}\times\mathfrak{h}$. Hence, $\gamma=a\rightharpoonup\left(
p\otimes q-q\otimes p-\left[  p,q\right]  \right)  \in J_{0}$ (by
(\ref{pf.l25a.1})). Thus, we have shown that every $\gamma\in\left\{
a\rightharpoonup\left(  v\otimes w-w\otimes v-\left[  v,w\right]  \right)
\ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}\right\}  $
satisfies $\gamma\in J_{0}$. In other words, $\left\{  a\rightharpoonup\left(
v\otimes w-w\otimes v-\left[  v,w\right]  \right)  \ \mid\ \left(  v,w\right)
\in\mathfrak{g}\times\mathfrak{h}\right\}  \subseteq J_{0}$. This proves
(\ref{pf.l25a.2}).

Now,
\begin{align*}
J_{0}  &  =\left\langle v\otimes w-w\otimes v-\left[  v,w\right]
\ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}\right\rangle \\
&  =\left\langle \left\{  v\otimes w-w\otimes v-\left[  v,w\right]
\ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}\right\}
\right\rangle \ \ \ \ \ \ \ \ \ \ \left(  \text{by Convention \ref{conv.<>}
\textbf{(b)}}\right)
\end{align*}
and thus%
\begin{align*}
L_{a}\left(  J_{0}\right)   &  =L_{a}\left(  \left\langle \left\{  v\otimes
w-w\otimes v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)  \in
\mathfrak{g}\times\mathfrak{h}\right\}  \right\rangle \right) \\
&  =\left\langle \underbrace{L_{a}\left(  \left\{  v\otimes w-w\otimes
v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}%
\times\mathfrak{h}\right\}  \right)  }_{=\left\{  L_{a}\left(  v\otimes
w-w\otimes v-\left[  v,w\right]  \right)  \ \mid\ \left(  v,w\right)
\in\mathfrak{g}\times\mathfrak{h}\right\}  }\right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{according to Proposition \ref{prop.<>} \textbf{(b)} (applied to
}M=\otimes\mathfrak{g}\text{, }R=\otimes\mathfrak{g}\text{,}\\
f=L_{a}\text{ and }S=\left\{  v\otimes w-w\otimes v-\left[  v,w\right]
\ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}\right\}
\text{)}%
\end{array}
\right) \\
&  =\left\langle \left\{  L_{a}\left(  v\otimes w-w\otimes v-\left[
v,w\right]  \right)  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}%
\times\mathfrak{h}\right\}  \right\rangle \\
&  =\left\langle \left\{  a\rightharpoonup\left(  v\otimes w-w\otimes
v-\left[  v,w\right]  \right)  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}%
\times\mathfrak{h}\right\}  \right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since every }\left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}\text{
satisfies}\\
L_{a}\left(  v\otimes w-w\otimes v-\left[  v,w\right]  \right)
=a\rightharpoonup\left(  v\otimes w-w\otimes v-\left[  v,w\right]  \right) \\
\text{by the definition of }L_{a}%
\end{array}
\right) \\
&  \subseteq J_{0}%
\end{align*}
(according to Proposition \ref{prop.<>} \textbf{(a)} (applied to
$M=\otimes\mathfrak{g}$, $Q=J_{0}$ and \newline$S=\left\{  a\rightharpoonup
\left(  v\otimes w-w\otimes v-\left[  v,w\right]  \right)  \ \mid\ \left(
v,w\right)  \in\mathfrak{g}\times\mathfrak{h}\right\}  $), because of
(\ref{pf.l25a.2})).

Thus, every $a\in\mathfrak{h}$ and every $x\in J_{0}$ satisfy
$a\rightharpoonup x\in J_{0}$ (since $L_{a}\left(  x\right)  =a\rightharpoonup
x$ by the definition of $L_{a}$, and thus $a\rightharpoonup x=L_{a}\left(
\underbrace{x}_{\in J_{0}}\right)  \in L_{a}\left(  J_{0}\right)  \subseteq
J_{0}$). In other words, $J_{0}$ is an $\mathfrak{h}$-submodule of
$\otimes\mathfrak{g}$. Now, Corollary \ref{cor.biideal} \textbf{(a)} (applied
to $V=\mathfrak{g}$, $P=J_{0}$ and $Q=\otimes\mathfrak{g}$) yields that
$J_{0}\cdot\left(  \otimes\mathfrak{g}\right)  $ is an $\mathfrak{h}%
$-submodule of $\otimes\mathfrak{g}$. This completes the proof of Proposition
\ref{prop.l25a} \textbf{(b)}.

\textbf{(a)} Since both $\otimes\mathfrak{g}$ and $\mathfrak{h}$ are
$\mathfrak{h}$-submodules of $\otimes\mathfrak{g}$, it follows from Corollary
\ref{cor.biideal} \textbf{(a)} (applied to $V=\mathfrak{g}$, $P=\otimes
\mathfrak{g}$ and $Q=\mathfrak{h}$) that $\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}$ is an $\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$.

Since $J_{0}$ is an $\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$,
Corollary \ref{cor.biideal} \textbf{(b)} (applied to $V=\mathfrak{g}$ and
$R=J_{0}$) yields that $\left(  \otimes\mathfrak{g}\right)  \cdot J_{0}%
\cdot\left(  \otimes\mathfrak{g}\right)  $ is an $\mathfrak{h}$-submodule of
$\otimes\mathfrak{g}$. Since $\left(  \otimes\mathfrak{g}\right)  \cdot
J_{0}\cdot\left(  \otimes\mathfrak{g}\right)  =J$ by Proposition
\ref{prop.l25a} \textbf{(b)}, we conclude that $J$ is an $\mathfrak{h}%
$-submodule of $\otimes\mathfrak{g}$.

Since we now know that both $J$ and $\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}$ are $\mathfrak{h}$-submodules of $\otimes\mathfrak{g}$, the
sum $J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}$ must also be an
$\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$. Thus, $\left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  $ is an $\mathfrak{h}$-module. Therefore,
the canonical projection $\zeta:\otimes\mathfrak{g}\rightarrow\left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  $ is an $\mathfrak{h}$-module homomorphism.
This completes the proof of Proposition \ref{prop.l25a} \textbf{(a)}.
\end{verlong}

Thus we have proved Theorem \ref{thm.l25} \textbf{(a)} (because it trivially
follows from Proposition \ref{prop.l25a} \textbf{(a)}).

\subsection{Planning the proof of Theorem \ref{thm.l25} \textbf{(b)} and
\textbf{(c)}}

Now that we have proven part \textbf{(a)} of Theorem \ref{thm.l25} without
even using the whole assumptions, it is time to sketch our further procedure
to prove the hard one - part \textbf{(b)} and \textbf{(c)}.

In the situation of Theorem \ref{thm.l25}, there exists a $k$-submodule $N$ of
$\mathfrak{g}$ such that $\mathfrak{g}=\mathfrak{h}\oplus N$ (because the
inclusion $\mathfrak{h}\hookrightarrow\mathfrak{g}$ splits as a $k$-module
inclusion). The projection $\pi:\mathfrak{g}\rightarrow\mathfrak{n}$ has the
property that $\pi\mid_{N}:N\rightarrow\mathfrak{n}$ is an isomorphism (this
is easily seen). We are going to define a projection $\varphi:\otimes
\mathfrak{g}\rightarrow\otimes N$ which, composed with the isomorphism
$\otimes\left(  \pi\mid_{N}\right)  :\otimes N\rightarrow\otimes\mathfrak{n}$,
will entail a homomorphism $\otimes\mathfrak{g}\rightarrow\otimes\mathfrak{n}$
(of $k$-modules). This homomorphism respects the filtration and (as we will
prove) satisfies $\varphi\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  =0$, so we will conclude that $\varphi$ induces a
homomorphism $\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \rightarrow
\otimes\mathfrak{n}$. This homomorphism is a $k$-module isomorphism (as we
will, again, prove), but is not an $\mathfrak{h}$-module isomorphism in
general. But as it respects the filtration, it yields a homomorphism from
$F_{n}\diagup F_{n-1}$ to $\mathfrak{n}^{\otimes n}$, which will prove to be
an $\mathfrak{h}$-module isomorphism. Additionally it turns out that this
isomorphism does not depend on the (non-canonical) choice of $N$ and can be
described by a simple formula.

We now proceed to the execution of this plan.

\subsection{Definitions and basic properties of $N$ and $\varphi$}

\begin{definition}
\label{defs.l25.core}Consider the situation of Theorem \ref{thm.l25}%
.\newline\textbf{(a)} Since the inclusion $\mathfrak{h}\hookrightarrow
\mathfrak{g}$ splits \textit{as a }$k$\textit{-module inclusion, there exists
a }$k$-submodule $N$ of $\mathfrak{g}$ such that $\mathfrak{g}=\mathfrak{h}%
\oplus N$ (as $k$-modules).\newline Let $s:\mathfrak{g}\rightarrow
\mathfrak{h}$ be the projection from $\mathfrak{g}$ on $\mathfrak{h}$ with
kernel $N$ (this projection exists since $\mathfrak{g}=\mathfrak{h}\oplus
N$).\newline Let $t:\mathfrak{g}\rightarrow N$ be the projection from
$\mathfrak{g}$ on $N$ with kernel $\mathfrak{h}$ (this projection exists since
$\mathfrak{g}=\mathfrak{h}\oplus N$).\newline Note that both $s$ and $t$ are
$k$-module maps (but not necessarily $\mathfrak{h}$-module maps; actually this
would not even make sense for $t$).\newline\textbf{(b)} Let us now define a
$k$-linear map $\varphi_{p}:\mathfrak{g}^{\otimes p}\rightarrow\otimes N$ for
every $p\in\mathbb{N}$. We are going to define this map $\varphi_{p}$ by
induction over $p$:\newline\textit{Induction base:} For $p=0$, define the map
$\varphi_{p}:\mathfrak{g}^{\otimes p}\rightarrow\otimes N$ by%
\begin{equation}
\left(  \varphi_{0}\left(  \lambda\right)  =\lambda
\ \ \ \ \ \ \ \ \ \ \text{for every }\lambda\in\mathfrak{g}^{\otimes0}\right)
\label{defs.alpha.0}%
\end{equation}
(this definition makes sense since $\mathfrak{g}^{\otimes0}=k\subseteq\otimes
N$).\newline\textit{Induction step:} For any $p>0$, we assume that the map
$\varphi_{p-1}:\mathfrak{g}^{\otimes\left(  p-1\right)  }\rightarrow\otimes N$
is already defined, and now we define a map $\varphi_{p}:\mathfrak{g}^{\otimes
p}\rightarrow\otimes N$ as follows: The map%
\[
\mathfrak{g}\times\mathfrak{g}^{\otimes\left(  p-1\right)  }\rightarrow\otimes
N,\ \ \ \ \ \ \ \ \ \ \left(  u,U\right)  \mapsto t\left(  u\right)
\cdot\varphi_{p-1}\left(  U\right)  +\varphi_{p-1}\left(  s\left(  u\right)
\rightharpoonup U\right)
\]
is $k$-bilinear (because the maps $\varphi_{p-1}$, $t$ and $s$ are $k$-linear
and the Lie action of $\otimes\mathfrak{g}$ is $k$-bilinear). Thus, by the
universal property of the tensor product, this map gives rise to a $k$-linear
map $\mathfrak{g}\otimes\mathfrak{g}^{\otimes\left(  p-1\right)  }%
\rightarrow\otimes N$ which sends $u\otimes U$ to $t\left(  u\right)
\cdot\varphi_{p-1}\left(  U\right)  +\varphi_{p-1}\left(  s\left(  u\right)
\rightharpoonup U\right)  $ for every $\left(  u,U\right)  \in\mathfrak{g}%
\times\mathfrak{g}^{\otimes\left(  p-1\right)  }$. This $k$-linear map is
going to be denoted by $\varphi_{p}$. It is a map from $\mathfrak{g}^{\otimes
p}$ to $\otimes N$ because $\mathfrak{g}\otimes\mathfrak{g}^{\otimes\left(
p-1\right)  }=\mathfrak{g}^{\otimes p}$.\newline This completes the inductive
definition of $\varphi_{p}$ for every $p\in\mathbb{N}$.\newline\textbf{(c)}
Now, we define a $k$-linear map $\varphi:\otimes\mathfrak{g}\rightarrow\otimes
N$ as follows: The sum $\sum\limits_{i\in\mathbb{N}}\varphi_{i}$ of the maps
$\varphi_{i}:\mathfrak{g}^{\otimes i}\rightarrow\otimes N$ is a map from
$\bigoplus\limits_{i\in\mathbb{N}}\mathfrak{g}^{\otimes i}$ to $\otimes N$.
Since $\bigoplus\limits_{i\in\mathbb{N}}\mathfrak{g}^{\otimes i}%
=\otimes\mathfrak{g}$, the sum $\sum\limits_{i\in\mathbb{N}}\varphi_{i}$ of
the maps $\varphi_{i}:\mathfrak{g}^{\otimes i}\rightarrow\otimes N$ is thus a
map from $\otimes\mathfrak{g}$ to $\otimes N$. Denote this map by $\varphi$.
\end{definition}

\begin{condition}
\label{conv.l25.core}Throughout the rest of Section \ref{sect.L25}, we are
going to work in the situation of Definition \ref{defs.l25.core}. So, for
example, when we refer to $\mathfrak{h}$, we mean the Lie subalgebra
$\mathfrak{h}$ of Theorem \ref{thm.l25}, and when we refer to $t$, we mean the
map $t$ of Definition \ref{defs.l25.core}.
\end{condition}

\begin{remark}
\label{rmk.l25.alpha}\textbf{(a)} As a consequence of the inductive step in
the definition of $\varphi_{p}$ (in Definition \ref{defs.l25.core}
\textbf{(b)}), we know that for every $p>0$, the map $\varphi_{p}$ is the
$k$-linear map $\mathfrak{g}\otimes\mathfrak{g}^{\otimes\left(  p-1\right)
}\rightarrow\otimes N$ which sends $u\otimes U$ to $t\left(  u\right)
\cdot\varphi_{p-1}\left(  U\right)  +\varphi_{p-1}\left(  s\left(  u\right)
\rightharpoonup U\right)  $ for every $\left(  u,U\right)  \in\mathfrak{g}%
\times\mathfrak{g}^{\otimes\left(  p-1\right)  }$. In other words,%
\begin{equation}
\varphi_{p}\left(  u\otimes U\right)  =t\left(  u\right)  \cdot\varphi
_{p-1}\left(  U\right)  +\varphi_{p-1}\left(  s\left(  u\right)
\rightharpoonup U\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(
u,U\right)  \in\mathfrak{g}\times\mathfrak{g}^{\otimes\left(  p-1\right)  }.
\label{defs.alpha.p}%
\end{equation}
\newline\textbf{(b)} According to Definition \ref{defs.l25.core} \textbf{(c)},
the map $\varphi:\otimes\mathfrak{g}\rightarrow\otimes N$ is the sum
$\sum\limits_{i\in\mathbb{N}}\varphi_{i}$ of the maps $\varphi_{i}%
:\mathfrak{g}^{\otimes i}\rightarrow\otimes N$. Hence,
\begin{equation}
\varphi\left(  T\right)  =\left(  \sum\limits_{i\in\mathbb{N}}\varphi
_{i}\right)  \left(  T\right)  =\varphi_{p}\left(  T\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }p\in\mathbb{N}\text{ and every }%
T\in\mathfrak{g}^{\otimes p}. \label{rmk.alpha-p}%
\end{equation}
\textbf{(c)} Every $\lambda\in k$ satisfies%
\begin{align}
\varphi\left(  \lambda\right)   &  =\varphi_{0}\left(  \lambda\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{rmk.alpha-p}) (applied to
}p=0\text{ and }T=\lambda\text{), because }\lambda\in k=\mathfrak{g}%
^{\otimes0}\right) \nonumber\\
&  =\lambda\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{defs.alpha.0})}\right)
. \label{rmk.alpha-0}%
\end{align}
This yields, in particular, that $\varphi\left(  k\right)  =k$.\newline%
\textbf{(d)} Every $u\in\mathfrak{g}$ satisfies%
\begin{align}
\varphi\left(  u\right)   &  =\varphi_{1}\left(  u\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{rmk.alpha-p}) (applied to
}p=1\text{ and }T=u\text{), because }u\in\mathfrak{g}=\mathfrak{g}^{\otimes
1}\right) \nonumber\\
&  =\varphi_{1}\left(  u\otimes1\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }u=u\otimes1\text{ under the identification }\mathfrak{g}%
\cong\mathfrak{g}\otimes k\right) \nonumber\\
&  =t\left(  u\right)  \cdot\underbrace{\varphi_{1-1}}_{=\varphi_{0}}\left(
1\right)  +\underbrace{\varphi_{1-1}}_{=\varphi_{0}}\left(
\underbrace{s\left(  u\right)  \rightharpoonup1}_{\substack{=0\text{ (since
the}\\\text{Lie action of }k\text{ is }0\text{)}}}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{defs.alpha.p}), applied to
}p=1\text{ and }U=1\right) \nonumber\\
&  =t\left(  u\right)  \cdot\underbrace{\varphi_{0}\left(  1\right)
}_{\substack{=1\text{ (by (\ref{defs.alpha.0}),}\\\text{applied to }%
\lambda=1\text{)}}}+\underbrace{\varphi_{0}\left(  0\right)  }%
_{\substack{=0\text{ (since}\\\varphi_{0}\text{ is linear)}}}=t\left(
u\right)  \cdot1+0=t\left(  u\right)  . \label{rmk.alpha-1}%
\end{align}
This yields $\varphi\left(  \mathfrak{g}\right)  =t\left(  \mathfrak{g}%
\right)  =N$ (since $t$ is a projection of $\mathfrak{g}$ on $N$%
).\newline\textbf{(e)} We now know that $\varphi\left(  k\right)  =k$ and
$\varphi\left(  \mathfrak{g}\right)  \subseteq N$. But in general, we cannot
generalize this to $\varphi\left(  \mathfrak{g}^{\otimes p}\right)  \subseteq
N^{\otimes p}$ for all $p\in\mathbb{N}$. However, Proposition
\ref{prop.l25.alpha.filt} will give us a weaker result that is actually true.
\end{remark}

First let us show an extension of (\ref{defs.alpha.p}) to all of
$\otimes\mathfrak{g}$:

\begin{proposition}
\label{prop.l25.alpha.rec}Every $u\in\mathfrak{g}$ and $U\in\otimes
\mathfrak{g}$ satisfy%
\begin{equation}
\varphi\left(  u\cdot U\right)  =t\left(  u\right)  \cdot\varphi\left(
U\right)  +\varphi\left(  s\left(  u\right)  \rightharpoonup U\right)  .
\label{prop.alpha.rec}%
\end{equation}

\end{proposition}

\begin{vershort}
\textit{Proof of Proposition \ref{prop.l25.alpha.rec}.} Obviously, the
equation that we want to prove - that is, the equation (\ref{prop.alpha.rec})
- is linear in $U$. Thus, we can WLOG assume that $U$ is a homogeneous tensor,
i. e., that $U\in\mathfrak{g}^{\otimes m}$ for some $m\in\mathbb{N}$.

Assuming this, we have $u\cdot U=u\otimes U$ (due to (\ref{*=(X)})). On the
other hand, $u\otimes U\in\mathfrak{g}^{\otimes\left(  m+1\right)  }$, so that
$\varphi\left(  u\otimes U\right)  =\varphi_{m+1}\left(  u\otimes U\right)  $
(since $\varphi=\sum\limits_{i\in\mathbb{N}}\varphi_{i}$). On the other hand,
$U\in\mathfrak{g}^{\otimes m}$ yields $s\left(  u\right)  \rightharpoonup
U\in\mathfrak{g}^{\otimes m}$, so that $\varphi\left(  s\left(  u\right)
\rightharpoonup U\right)  =\varphi_{m}\left(  s\left(  u\right)
\rightharpoonup U\right)  $ (again because $\varphi=\sum\limits_{i\in
\mathbb{N}}\varphi_{i}$). Also, $\varphi\left(  U\right)  =\varphi_{m}\left(
U\right)  $ (for the same reason). Hence,%
\begin{align*}
\varphi\left(  \underbrace{u\cdot U}_{=u\otimes U}\right)   &  =\varphi\left(
u\otimes U\right)  =\varphi_{m+1}\left(  u\otimes U\right)  =t\left(
u\right)  \cdot\underbrace{\varphi_{m}\left(  U\right)  }_{=\varphi\left(
U\right)  }+\underbrace{\varphi_{m}\left(  s\left(  u\right)  \rightharpoonup
U\right)  }_{=\varphi\left(  s\left(  u\right)  \rightharpoonup U\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{due to (\ref{defs.alpha.p}), applied to
}p=m+1\right) \\
&  =t\left(  u\right)  \cdot\varphi\left(  U\right)  +\varphi\left(  s\left(
u\right)  \rightharpoonup U\right)  .
\end{align*}
This proves (\ref{prop.alpha.rec}) and, with it, Proposition
\ref{prop.l25.alpha.rec}.
\end{vershort}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.l25.alpha.rec}.} Let $u\in\mathfrak{g}$
and $U\in\otimes\mathfrak{g}$ be arbitrary. Since $U\in\otimes\mathfrak{g}%
=\bigoplus\limits_{i\in\mathbb{N}}\mathfrak{g}^{\otimes i}$, we can write $U$
as a family $U=\left(  U_{i}\right)  _{i\in\mathbb{N}}$, where $\left(
U_{i}\in\mathfrak{g}^{\otimes i}\text{ for every }i\in\mathbb{N}\right)  $.
Taking into account that we consider $\mathfrak{g}^{\otimes p}$ as a
$k$-submodule of $\bigoplus\limits_{i\in\mathbb{N}}\mathfrak{g}^{\otimes i}$
for every $p\in\mathbb{N}$, we have $\left(  U_{i}\right)  _{i\in\mathbb{N}%
}=\sum\limits_{i\in\mathbb{N}}U_{i}$, and thus $U=\left(  U_{i}\right)
_{i\in\mathbb{N}}=\sum\limits_{i\in\mathbb{N}}U_{i}$. Thus,
\begin{align*}
s\left(  u\right)   &  \rightharpoonup U=s\left(  u\right)  \rightharpoonup
\left(  \sum\limits_{i\in\mathbb{N}}U_{i}\right)  =\sum\limits_{i\in
\mathbb{N}}\left(  s\left(  u\right)  \rightharpoonup U_{i}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Lie action of }\otimes
\mathfrak{g}\text{ is }k\text{-bilinear}\right)  .
\end{align*}
On the other hand, every $i\in\mathbb{N}$ satisfies $s\left(  u\right)
\rightharpoonup U_{i}\in\mathfrak{g}^{\otimes i}$ (because $U_{i}%
\in\mathfrak{g}^{\otimes i}$ and because $\mathfrak{g}^{\otimes i}$ is a
$\mathfrak{g}$-module) and $\underbrace{u}_{\in\mathfrak{g}}\otimes
\underbrace{U_{i}}_{\in\mathfrak{g}^{\otimes i}}\in\mathfrak{g}\otimes
\mathfrak{g}^{\otimes i}=\mathfrak{g}^{\otimes\left(  i+1\right)  }$. Also,
since $\varphi=\sum\limits_{i\in\mathbb{N}}\varphi_{i}=\sum\limits_{p\in
\mathbb{N}}\varphi_{p}$ is the sum of the maps $\varphi_{p}:\mathfrak{g}%
^{\otimes p}\rightarrow\otimes N$ for all $p\in\mathbb{N}$, we have
\begin{equation}
\varphi\left(  v\right)  =\varphi_{i}\left(  v\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }v\in\mathfrak{g}^{\otimes i}.
\label{pf.l25.alpha.rec.1}%
\end{equation}
Similarly,%
\begin{equation}
\varphi\left(  v\right)  =\varphi_{i+1}\left(  v\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }v\in\mathfrak{g}^{\otimes\left(
i+1\right)  }. \label{pf.l25.alpha.rec.2}%
\end{equation}


Every $i\in\mathbb{N}$ satisfies $\varphi\left(  u\otimes U_{i}\right)
=\varphi_{i+1}\left(  u\otimes U_{i}\right)  $ according to
(\ref{pf.l25.alpha.rec.2}) (applied to $u\otimes U_{i}$ instead of $v$),
because $u\otimes U_{i}\in\mathfrak{g}^{\otimes\left(  i+1\right)  }$.

Every $i\in\mathbb{N}$ satisfies $\varphi\left(  U_{i}\right)  =\varphi
_{i}\left(  U_{i}\right)  $ according to (\ref{pf.l25.alpha.rec.1}) (applied
to $U_{i}$ instead of $v$), since $U_{i}\in\mathfrak{g}^{\otimes i}$.

Every $i\in\mathbb{N}$ satisfies $\varphi\left(  s\left(  u\right)
\rightharpoonup U_{i}\right)  =\varphi_{i}\left(  s\left(  u\right)
\rightharpoonup U_{i}\right)  $ according to (\ref{pf.l25.alpha.rec.1})
(applied to $s\left(  u\right)  \rightharpoonup U_{i}$ instead of $v$), since
$s\left(  u\right)  \rightharpoonup U_{i}\in\mathfrak{g}^{\otimes i}$.

Now, $U=\sum\limits_{i\in\mathbb{N}}U_{i}$ leads to
\begin{align*}
&  u\cdot U\\
&  =u\cdot\sum\limits_{i\in\mathbb{N}}U_{i}=\sum\limits_{i\in\mathbb{N}}u\cdot
U_{i}=\sum\limits_{i\in\mathbb{N}}u\otimes U_{i}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since every }i\in\mathbb{N}\text{ satisfies }u\in\mathfrak{g}%
=\mathfrak{g}^{\otimes1}\text{ and }U_{i}\in\mathfrak{g}^{\otimes i}\text{ and
therefore}\\
u\cdot U_{i}=u\otimes U_{i}\text{ (due to (\ref{*=(X)}), applied to }u\text{,
}U_{i}\text{, }1\text{ and }i\text{ instead of }a\text{, }b\text{, }n\text{
and }m\text{)}%
\end{array}
\right)  ,
\end{align*}
and therefore%
\begin{align*}
\varphi\left(  u\cdot U\right)   &  =\varphi\left(  \sum\limits_{i\in
\mathbb{N}}u\otimes U_{i}\right)  =\sum\limits_{i\in\mathbb{N}}%
\underbrace{\varphi\left(  u\otimes U_{i}\right)  }_{=\varphi_{i+1}\left(
u\otimes U_{i}\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varphi\text{
is }k\text{-linear}\right) \\
&  =\sum_{i\in\mathbb{N}}\varphi_{i+1}\left(  u\otimes U_{i}\right)
=\sum_{i\in\mathbb{N}}\left(  t\left(  u\right)  \cdot\underbrace{\varphi
_{\left(  i+1\right)  -1}}_{=\varphi_{i}}\left(  U_{i}\right)
+\underbrace{\varphi_{\left(  i+1\right)  -1}}_{=\varphi_{i}}\left(  s\left(
u\right)  \rightharpoonup U_{i}\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since every }i\in\mathbb{N}\text{ satisfies}\\
\varphi_{i+1}\left(  u\otimes U_{i}\right)  =t\left(  u\right)  \cdot
\varphi_{\left(  i+1\right)  -1}\left(  U_{i}\right)  +\varphi_{\left(
i+1\right)  -1}\left(  s\left(  u\right)  \rightharpoonup U_{i}\right) \\
\text{according to (\ref{defs.alpha.p}) (applied to }i+1\text{ and }%
U_{i}\text{ instead of }p\text{ and }U\text{)}%
\end{array}
\right) \\
&  =\sum_{i\in\mathbb{N}}\left(  t\left(  u\right)  \cdot\varphi_{i}\left(
U_{i}\right)  +\varphi_{i}\left(  s\left(  u\right)  \rightharpoonup
U_{i}\right)  \right)  =\sum_{i\in\mathbb{N}}t\left(  u\right)  \cdot
\underbrace{\varphi_{i}\left(  U_{i}\right)  }_{=\varphi\left(  U_{i}\right)
}+\sum_{i\in\mathbb{N}}\underbrace{\varphi_{i}\left(  s\left(  u\right)
\rightharpoonup U_{i}\right)  }_{=\varphi\left(  s\left(  u\right)
\rightharpoonup U_{i}\right)  }\\
&  =\sum_{i\in\mathbb{N}}t\left(  u\right)  \cdot\varphi\left(  U_{i}\right)
+\sum_{i\in\mathbb{N}}\varphi\left(  s\left(  u\right)  \rightharpoonup
U_{i}\right)  .
\end{align*}
Since%
\[
\sum_{i\in\mathbb{N}}t\left(  u\right)  \cdot\varphi\left(  U_{i}\right)
=t\left(  u\right)  \cdot\underbrace{\sum_{i\in\mathbb{N}}\varphi\left(
U_{i}\right)  }_{\substack{=\varphi\left(  \sum\limits_{i\in\mathbb{N}}%
U_{i}\right)  \\\text{(since }\varphi\text{ is }k\text{-linear)}}}=t\left(
u\right)  \cdot\varphi\left(  \underbrace{\sum\limits_{i\in\mathbb{N}}U_{i}%
}_{=U}\right)  =t\left(  u\right)  \cdot\varphi\left(  U\right)
\]
and%
\begin{align*}
\sum_{i\in\mathbb{N}}\varphi\left(  s\left(  u\right)  \rightharpoonup
U_{i}\right)   &  =\varphi\left(  \underbrace{\sum_{i\in\mathbb{N}}\left(
s\left(  u\right)  \rightharpoonup U_{i}\right)  }_{=s\left(  u\right)
\rightharpoonup U}\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\varphi\text{ is }k\text{-linear}\right) \\
&  =\varphi\left(  s\left(  u\right)  \rightharpoonup U\right)  ,
\end{align*}
this becomes%
\[
\varphi\left(  u\cdot U\right)  =\underbrace{\sum_{i\in\mathbb{N}}t\left(
u\right)  \cdot\varphi\left(  U_{i}\right)  }_{=t\left(  u\right)
\cdot\varphi\left(  U\right)  }+\underbrace{\sum_{i\in\mathbb{N}}%
\varphi\left(  s\left(  u\right)  \rightharpoonup U_{i}\right)  }%
_{=\varphi\left(  s\left(  u\right)  \rightharpoonup U\right)  }=t\left(
u\right)  \cdot\varphi\left(  U\right)  +\varphi\left(  s\left(  u\right)
\rightharpoonup U\right)  .
\]
This proves Proposition \ref{prop.l25.alpha.rec}.
\end{verlong}

Two obvious corollaries of Proposition \ref{prop.l25.alpha.rec}:

\begin{corollary}
\label{cor.l25.alpha.rec.h}Every $u\in\mathfrak{h}$ and $U\in\otimes
\mathfrak{g}$ satisfy%
\begin{equation}
\varphi\left(  u\cdot U\right)  =\varphi\left(  u\rightharpoonup U\right)  .
\label{prop.alpha.rec.h}%
\end{equation}

\end{corollary}

\begin{corollary}
\label{cor.l25.alpha.rec.N}Every $u\in N$ and $U\in\otimes\mathfrak{g}$
satisfy%
\begin{equation}
\varphi\left(  u\cdot U\right)  =u\cdot\varphi\left(  U\right)  .
\label{prop.alpha.rec.N}%
\end{equation}

\end{corollary}

\textit{Proof of Corollary \ref{cor.l25.alpha.rec.h}.} Since $u\in
\mathfrak{h}$, we have $t\left(  u\right)  =0$ (since $t$ is a projection with
kernel $\mathfrak{h}$) and $s\left(  u\right)  =u$ (since $s$ is a projection
on $\mathfrak{h}$). Thus, (\ref{prop.alpha.rec}) yields%
\[
\varphi\left(  u\cdot U\right)  =\underbrace{t\left(  u\right)  }_{=0}%
\cdot\varphi\left(  U\right)  +\varphi\left(  \underbrace{s\left(  u\right)
}_{=u}\rightharpoonup U\right)  =\underbrace{0\cdot\varphi\left(  U\right)
}_{=0}+\varphi\left(  u\rightharpoonup U\right)  =\varphi\left(
u\rightharpoonup U\right)  .
\]
This proves Corollary \ref{cor.l25.alpha.rec.h}.

\begin{vershort}
The \textit{proof of Corollary \ref{cor.l25.alpha.rec.N}} is analogous to the
proof of Corollary \ref{cor.l25.alpha.rec.h} just given ($\mathfrak{h}$ and
$N$ change their roles).
\end{vershort}

\begin{verlong}
\textit{Proof of Corollary \ref{cor.l25.alpha.rec.N}.} Since $u\in N$, we have
$s\left(  u\right)  =0$ (since $s$ is a projection with kernel $N$) and
$t\left(  u\right)  =u$ (since $t$ is a projection on $N$). Thus,
(\ref{prop.alpha.rec}) yields%
\[
\varphi\left(  u\cdot U\right)  =\underbrace{t\left(  u\right)  }_{=u}%
\cdot\varphi\left(  U\right)  +\varphi\left(  \underbrace{s\left(  u\right)
}_{=0}\rightharpoonup U\right)  =u\cdot\varphi\left(  U\right)  +\varphi
\left(  \underbrace{0\rightharpoonup U}_{=0}\right)  =u\cdot\varphi\left(
U\right)  +\underbrace{\varphi\left(  0\right)  }_{=0}=u\cdot\varphi\left(
U\right)  .
\]
This proves Corollary \ref{cor.l25.alpha.rec.N}.
\end{verlong}

Before we go on, let us make another convention:

\begin{condition}
\label{conv.left-ind2}Let $V$ be any $k$-module. Let $p\in\mathbb{N}_{+}%
$.\newline According to Convention \ref{conv.left-ind} (applied to $V_{1}=V$,
$V_{2}=V$, $...$, $V_{n}=V$ and $n=p$), a \textit{left-induced tensor} in
$\underbrace{V\otimes V\otimes...\otimes V}_{p\text{ times}}$ means an element
of the form $v\otimes T$ for some $v\in V$ and some $T\in\underbrace{V\otimes
V\otimes...\otimes V}_{p-1\text{ times}}$.\newline Since $\underbrace{V\otimes
V\otimes...\otimes V}_{p\text{ times}}=V^{\otimes p}$ and
$\underbrace{V\otimes V\otimes...\otimes V}_{p-1\text{ times}}\in
V^{\otimes\left(  p-1\right)  }$, we can rewrite this fact as follows: A
\textit{left-induced tensor} in $V^{\otimes p}$ means an element of the form
$v\otimes T$ for some $v\in V$ and some $T\in V^{\otimes\left(  p-1\right)  }%
$.\newline We denote by $V_{\operatorname*{lind}}^{\otimes p}$ the subset of
$V^{\otimes p}$ consisting of all left-induced tensors in $V^{\otimes p}$.
\end{condition}

\begin{proposition}
\label{prop.left-ind2} Let $V$ be any $k$-module. Let $p\in\mathbb{N}_{+}$.
Then, $V^{\otimes p}=\left\langle V_{\operatorname*{lind}}^{\otimes
p}\right\rangle $.
\end{proposition}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.left-ind2}.} According to what we said
in Convention \ref{conv.left-ind}, we know the following fact: For every $n$
arbitrary $k$-modules $V_{1}$, $V_{2}$, $...$, $V_{n}$, where $n\geq1$, the
$k$-module $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ is generated by its
left-induced tensors. Applying this to $V_{1}=V$, $V_{2}=V$, $...$, $V_{n}=V$
and $n=p$, we conclude that the $k$-module $\underbrace{V\otimes
V\otimes...\otimes V}_{p\text{ times}}$ is generated by its left-induced
tensors. Since $\underbrace{V\otimes V\otimes...\otimes V}_{p\text{ times}%
}=V^{\otimes p}$, this rewrites as follows: The $k$-module $V^{\otimes p}$ is
generated by its left-induced tensors. In other words: The $k$-module
$V^{\otimes p}$ is generated by $V_{\operatorname*{lind}}^{\otimes p}$
(because the set of all left-induced tensors in $V^{\otimes p}$ is
$V_{\operatorname*{lind}}^{\otimes p}$). In other words, $V^{\otimes
p}=\left\langle V_{\operatorname*{lind}}^{\otimes p}\right\rangle $. This
proves Proposition \ref{prop.left-ind2}.
\end{verlong}

The next (rather trivial) property will again be proven by induction:

\begin{proposition}
\label{prop.l25.alpha.filt}The map $\varphi:\otimes\mathfrak{g}\rightarrow
\otimes N$ respects the filtration. Here, the filtration on $\otimes
\mathfrak{g}$ is the degree filtration $\left(  \mathfrak{g}^{\otimes\leq
n}\right)  _{n\geq0}$, and the filtration on $\otimes N$ is the degree
filtration $\left(  N^{\otimes\leq n}\right)  _{n\geq0}$.
\end{proposition}

\textit{Proof of Proposition \ref{prop.l25.alpha.filt}.} We are going to show
that%
\begin{equation}
\varphi\left(  \mathfrak{g}^{\otimes p}\right)  \subseteq N^{\otimes\leq
p}\text{ for every }p\in\mathbb{N}. \label{pf.l25.alpha.filt.1}%
\end{equation}


\textit{Proof of (\ref{pf.l25.alpha.filt.1}).} We will prove
(\ref{pf.l25.alpha.filt.1}) by induction over $p$:

\textit{Induction base:} We know that $\mathfrak{g}^{\otimes0}=k$ and
$N^{\otimes\leq0}=\bigoplus\limits_{i=0}^{0}N^{\otimes i}=N^{\otimes0}=k$, and
we know (from Remark \ref{rmk.l25.alpha} \textbf{(c)}) that $\varphi\left(
k\right)  =k$. Thus, $\mathfrak{g}^{\otimes0}=k$ yields $\varphi\left(
\mathfrak{g}^{\otimes0}\right)  =\varphi\left(  k\right)  =k=N^{\otimes\leq0}%
$. In other words, (\ref{pf.l25.alpha.filt.1}) holds for $p=0$. This completes
the induction base.

\textit{Induction step:} Let $q\in\mathbb{N}_{+}$. Assume that
(\ref{pf.l25.alpha.filt.1}) holds for $p=q-1$. Now we must prove that
(\ref{pf.l25.alpha.filt.1}) holds for $p=q$.

We have assumed that (\ref{pf.l25.alpha.filt.1}) holds for $p=q-1$. In other
words, $\varphi\left(  \mathfrak{g}^{\otimes\left(  q-1\right)  }\right)
\subseteq N^{\otimes\leq\left(  q-1\right)  }$.

Now we are going to prove that every $U\in\varphi\left(  \mathfrak{g}%
_{\operatorname*{lind}}^{\otimes q}\right)  $ satisfies $U\in N^{\otimes\leq
q}$.

In fact, let $U\in\varphi\left(  \mathfrak{g}_{\operatorname*{lind}}^{\otimes
q}\right)  $ be arbitrary. Then, there exists a $U^{\prime}\in\mathfrak{g}%
_{\operatorname*{lind}}^{\otimes q}$ such that $U=\varphi\left(  U^{\prime
}\right)  $ (since $U\in\varphi\left(  \mathfrak{g}_{\operatorname*{lind}%
}^{\otimes q}\right)  $). Consider this $U^{\prime}$. This $U^{\prime}$ is a
left-induced tensor in $\mathfrak{g}^{\otimes q}$ (since $U^{\prime}%
\in\mathfrak{g}_{\operatorname*{lind}}^{\otimes q}$); this means that there
exists some $v\in\mathfrak{g}$ and $T\in\mathfrak{g}^{\otimes\left(
q-1\right)  }$ such that $U^{\prime}=v\otimes T$. Consider these $v$ and $T$.
Since $v\in\mathfrak{g}=\mathfrak{g}^{\otimes1}$ and $T\in\mathfrak{g}%
^{\otimes\left(  q-1\right)  }$, we have $v\cdot T=v\otimes T$ (due to
(\ref{*=(X)}), applied to $v$, $T$, $1$ and $q-1$ instead of $a$, $b$, $n$ and
$m$).

\begin{vershort}
Now,%
\begin{align*}
U  &  =\varphi\left(  \underbrace{U^{\prime}}_{=v\otimes T=v\cdot T}\right)
=\varphi\left(  v\cdot T\right)  =\underbrace{t\left(  v\right)  }_{\in
N=N^{\otimes1}\subseteq N^{\otimes\leq1}}\cdot\varphi\left(  \underbrace{T}%
_{\in\mathfrak{g}^{\otimes\left(  q-1\right)  }}\right)  +\varphi\left(
\underbrace{s\left(  v\right)  \rightharpoonup T}_{\substack{\in
\mathfrak{g}^{\otimes\left(  q-1\right)  }\text{ (since }T\in\mathfrak{g}%
^{\otimes\left(  q-1\right)  }\\\text{and since }\mathfrak{g}^{\otimes\left(
q-1\right)  }\text{ is a }\mathfrak{g}\text{-module)}}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{prop.alpha.rec}), applied to
}v\text{ and }T\text{ instead of }u\text{ and }U\right) \\
&  \in N^{\otimes\leq1}\cdot\underbrace{\varphi\left(  \mathfrak{g}%
^{\otimes\left(  q-1\right)  }\right)  }_{\subseteq N^{\otimes\leq\left(
q-1\right)  }}+\underbrace{\varphi\left(  \mathfrak{g}^{\otimes\left(
q-1\right)  }\right)  }_{\substack{\subseteq N^{\otimes\leq\left(  q-1\right)
}\subseteq N^{\otimes\leq q}\\\text{(since }\left(  N^{\otimes\leq n}\right)
_{n\geq0}\text{ is a filtration)}}}\subseteq\underbrace{N^{\otimes\leq1}\cdot
N^{\otimes\leq\left(  q-1\right)  }}_{\substack{\subseteq N^{\otimes\leq
q}\\\text{(by Proposition \ref{prop.tensor.filt-is-alg-filt} \textbf{(b)})}%
}}+N^{\otimes\leq q}\\
&  \subseteq N^{\otimes\leq q}+N^{\otimes\leq q}=N^{\otimes\leq q}.
\end{align*}

\end{vershort}

\begin{verlong}
Now,%
\begin{align*}
U  &  =\varphi\left(  \underbrace{U^{\prime}}_{=v\otimes T=v\cdot T}\right)
=\varphi\left(  v\cdot T\right)  =\underbrace{t\left(  v\right)  }_{\in
N=N^{\otimes1}\subseteq N^{\otimes\leq1}}\cdot\varphi\left(  \underbrace{T}%
_{\in\mathfrak{g}^{\otimes\left(  q-1\right)  }}\right)  +\varphi\left(
\underbrace{s\left(  v\right)  \rightharpoonup T}_{\substack{\in
\mathfrak{g}^{\otimes\left(  q-1\right)  }\text{ (since }T\in\mathfrak{g}%
^{\otimes\left(  q-1\right)  }\\\text{and since }\mathfrak{g}^{\otimes\left(
q-1\right)  }\text{ is a }\mathfrak{g}\text{-module)}}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{prop.alpha.rec}), applied to
}v\text{ and }T\text{ instead of }u\text{ and }U\right) \\
&  \in N^{\otimes\leq1}\cdot\underbrace{\varphi\left(  \mathfrak{g}%
^{\otimes\left(  q-1\right)  }\right)  }_{\subseteq N^{\otimes\leq\left(
q-1\right)  }}+\underbrace{\varphi\left(  \mathfrak{g}^{\otimes\left(
q-1\right)  }\right)  }_{\substack{\subseteq N^{\otimes\leq\left(  q-1\right)
}\subseteq N^{\otimes\leq q}\\\text{(since }\left(  N^{\otimes\leq n}\right)
_{n\geq0}\text{ is a filtration)}}}\subseteq N^{\otimes\leq1}\cdot
N^{\otimes\leq\left(  q-1\right)  }+N^{\otimes\leq q}.
\end{align*}
Since Proposition \ref{prop.tensor.filt-is-alg-filt} \textbf{(b)} (applied to
$N$, $1$ and $q-1$ instead of $V$, $n$ and $m$) yields $N^{\otimes\leq1}\cdot
N^{\otimes\leq\left(  q-1\right)  }\subseteq N^{\otimes\leq\left(  1+\left(
q-1\right)  \right)  }=N^{\otimes\leq q}$, this becomes%
\[
U\in\underbrace{N^{\otimes\leq1}\cdot N^{\otimes\leq\left(  q-1\right)  }%
}_{\subseteq N^{\otimes\leq q}}+N^{\otimes\leq q}\subseteq N^{\otimes\leq
q}+N^{\otimes\leq q}\subseteq N^{\otimes\leq q}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }N^{\otimes\leq q}\text{ is a }k\text{-module}\right)  .
\]

\end{verlong}

We have thus proven that every $U\in\varphi\left(  \mathfrak{g}%
_{\operatorname*{lind}}^{\otimes q}\right)  $ satisfies $U\in N^{\otimes\leq
q}$. In other words, $\varphi\left(  \mathfrak{g}_{\operatorname*{lind}%
}^{\otimes q}\right)  \subseteq N^{\otimes\leq q}$.

\begin{vershort}
Now, $\mathfrak{g}^{\otimes q}=\left\langle \mathfrak{g}_{\operatorname*{lind}%
}^{\otimes q}\right\rangle $ (by Proposition \ref{prop.left-ind2}) and thus%
\begin{align*}
\varphi\left(  \mathfrak{g}^{\otimes q}\right)   &  =\varphi\left(
\left\langle \mathfrak{g}_{\operatorname*{lind}}^{\otimes q}\right\rangle
\right)  =\left\langle \varphi\left(  \mathfrak{g}_{\operatorname*{lind}%
}^{\otimes q}\right)  \right\rangle \subseteq N^{\otimes\leq q}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varphi\left(  \mathfrak{g}%
_{\operatorname*{lind}}^{\otimes q}\right)  \subseteq N^{\otimes\leq q}\text{
and since }N^{\otimes\leq q}\text{ is a }k\text{-module}\right)  .
\end{align*}
In other words, (\ref{pf.l25.alpha.filt.1}) holds for $p=q$. This completes
the induction step, and thus the induction proof of (\ref{pf.l25.alpha.filt.1}%
) is done.
\end{vershort}

\begin{verlong}
Now, Proposition \ref{prop.left-ind2} (applied to $q$ and $\mathfrak{g}$
instead of $p$ and $V$) says $\mathfrak{g}^{\otimes q}=\left\langle
\mathfrak{g}_{\operatorname*{lind}}^{\otimes q}\right\rangle $, and thus%
\begin{align*}
\varphi\left(  \mathfrak{g}^{\otimes q}\right)   &  =\varphi\left(
\left\langle \mathfrak{g}_{\operatorname*{lind}}^{\otimes q}\right\rangle
\right)  =\left\langle \varphi\left(  \mathfrak{g}_{\operatorname*{lind}%
}^{\otimes q}\right)  \right\rangle \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Proposition \ref{prop.<>} \textbf{(b)} (applied to}\\
\mathfrak{g}^{\otimes q}\text{, }\mathfrak{g}_{\operatorname*{lind}}^{\otimes
q}\text{, }\otimes N\text{ and }\varphi\text{ instead of }M\text{, }S\text{,
}R\text{ and }f\text{)}%
\end{array}
\right) \\
&  \subseteq N^{\otimes\leq q}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Proposition \ref{prop.<>} \textbf{(a)} (applied to }\otimes N\text{,
}\varphi\left(  \mathfrak{g}_{\operatorname*{lind}}^{\otimes q}\right)  \text{
and }N^{\otimes\leq q}\\
\text{instead of }M\text{, }S\text{ and }Q\text{), since }\varphi\left(
\mathfrak{g}_{\operatorname*{lind}}^{\otimes q}\right)  \subseteq
N^{\otimes\leq q}%
\end{array}
\right)  .
\end{align*}
In other words, (\ref{pf.l25.alpha.filt.1}) holds for $p=q$. This completes
the induction step, and thus the induction proof of (\ref{pf.l25.alpha.filt.1}%
) is done.
\end{verlong}

\begin{vershort}
Now, let $m\in\mathbb{N}$. The definition of $\mathfrak{g}^{\otimes\leq m}$
says $\mathfrak{g}^{\otimes\leq m}=\bigoplus\limits_{i=0}^{m}\mathfrak{g}%
^{\otimes i}$, so that%
\begin{align*}
\varphi\left(  \mathfrak{g}^{\otimes\leq m}\right)   &  =\varphi\left(
\bigoplus\limits_{i=0}^{m}\mathfrak{g}^{\otimes i}\right)  =\sum
\limits_{i=0}^{m}\underbrace{\varphi\left(  \mathfrak{g}^{\otimes i}\right)
}_{\substack{\subseteq N^{\otimes\leq i}\\\text{(by (\ref{pf.l25.alpha.filt.1}%
), applied}\\\text{to }i\text{ instead of }p\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varphi\text{ is }k\text{-linear}%
\right) \\
&  \subseteq\sum\limits_{i=0}^{m}N^{\otimes\leq i}=N^{\otimes\leq m}.
\end{align*}
Since this holds for every $m\in\mathbb{N}$, this yields that $\varphi$
respects the filtration. This proves Proposition \ref{prop.l25.alpha.filt}.
\end{vershort}

\begin{verlong}
Now, let $m\in\mathbb{N}$. The definition of $\mathfrak{g}^{\otimes\leq m}$
says $\mathfrak{g}^{\otimes\leq m}=\bigoplus\limits_{i=0}^{m}\mathfrak{g}%
^{\otimes i}$, so that $\mathfrak{g}^{\otimes\leq m}=\bigoplus\limits_{i=0}%
^{m}\mathfrak{g}^{\otimes i}=\sum\limits_{i=0}^{m}\mathfrak{g}^{\otimes i}$
(since direct sums are sums) and thus%
\begin{align*}
\varphi\left(  \mathfrak{g}^{\otimes\leq m}\right)   &  =\varphi\left(
\sum\limits_{i=0}^{m}\mathfrak{g}^{\otimes i}\right)  =\sum\limits_{i=0}%
^{m}\underbrace{\varphi\left(  \mathfrak{g}^{\otimes i}\right)  }%
_{\substack{\subseteq N^{\otimes\leq i}\\\text{(by (\ref{pf.l25.alpha.filt.1}%
), applied}\\\text{to }i\text{ instead of }p\text{)}}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varphi\text{ is }k\text{-linear}%
\right) \\
&  \subseteq\sum\limits_{i=0}^{m}\underbrace{N^{\otimes\leq i}}%
_{\substack{\subseteq N^{\otimes\leq m}\text{ (since }i\leq m\\\text{and since
}\left(  N^{\otimes\leq n}\right)  _{n\geq0}\text{ is}\\\text{a filtration)}%
}}\subseteq\sum\limits_{i=0}^{m}N^{\otimes\leq m}\subseteq N^{\otimes\leq
m}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }N^{\otimes\leq m}\text{ is a
}k\text{-module}\right)  .
\end{align*}
Since this holds for every $m\in\mathbb{N}$, this yields that $\varphi$
respects the filtration. This proves Proposition \ref{prop.l25.alpha.filt}.
\end{verlong}

The next, still very trivial, result shows how $\varphi$ acts on $\otimes N$
(when $\otimes N$ is considered as a $k$-subalgebra of $\otimes\mathfrak{g}$):

\begin{proposition}
\label{prop.l25.alpha.(X)N}Let us consider $\otimes N$ as a $k$-subalgebra of
$\otimes\mathfrak{g}$ (since $N$ is a $k$-submodule of $\mathfrak{g}$). Then,
$\varphi\mid_{\otimes N}=\operatorname*{id}\nolimits_{\otimes N}$.
\end{proposition}

\textit{Proof of Proposition \ref{prop.l25.alpha.(X)N}.} We are going to prove
that%
\begin{equation}
N^{\otimes p}\subseteq\operatorname*{Ker}\left(  \varphi\mid_{\otimes
N}-\operatorname*{id}\nolimits_{\otimes N}\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }p\in\mathbb{N}\text{.}
\label{pf.l25.alpha.(X)N.1}%
\end{equation}
\textit{Proof of (\ref{pf.l25.alpha.(X)N.1}).} We will prove
(\ref{pf.l25.alpha.(X)N.1}) by induction over $p$:

\begin{vershort}
\textit{Induction base:} Every $\lambda\in N^{\otimes0}$ satisfies
$\varphi\left(  \lambda\right)  =\lambda$ (by (\ref{rmk.alpha-0}), because
$\lambda\in N^{\otimes0}=k$). Thus, (\ref{pf.l25.alpha.(X)N.1}) holds for
$p=0$. This completes the induction base.
\end{vershort}

\begin{verlong}
\textit{Induction base:} Every $\lambda\in N^{\otimes0}$ satisfies%
\[
\left(  \varphi\mid_{\otimes N}-\operatorname*{id}\nolimits_{\otimes
N}\right)  \left(  \lambda\right)  =\underbrace{\left(  \varphi\mid_{\otimes
N}\right)  \left(  \lambda\right)  }_{\substack{=\varphi\left(  \lambda
\right)  =\lambda\text{ (by (\ref{rmk.alpha-0}),}\\\text{since }\lambda\in
N^{\otimes0}=k\text{)}}}-\underbrace{\operatorname*{id}\nolimits_{\otimes
N}\left(  \lambda\right)  }_{=\lambda}=\lambda-\lambda=0.
\]
Thus, every $\lambda\in N^{\otimes0}$ satisfies $\lambda\in\operatorname*{Ker}%
\left(  \varphi\mid_{\otimes N}-\operatorname*{id}\nolimits_{\otimes
N}\right)  $. In other words, $N^{\otimes0}\subseteq\operatorname*{Ker}\left(
\varphi\mid_{\otimes N}-\operatorname*{id}\nolimits_{\otimes N}\right)  $.
Hence, (\ref{pf.l25.alpha.(X)N.1}) holds for $p=0$. This completes the
induction base.
\end{verlong}

\textit{Induction step:} Let $q\in\mathbb{N}_{+}$. Assume that
(\ref{pf.l25.alpha.(X)N.1}) holds for $p=q-1$. Now we must prove that
(\ref{pf.l25.alpha.(X)N.1}) holds for $p=q$.

We have assumed that (\ref{pf.l25.alpha.(X)N.1}) holds for $p=q-1$. In other
words, $N^{\otimes\left(  q-1\right)  }\subseteq\operatorname*{Ker}\left(
\varphi\mid_{\otimes N}-\operatorname*{id}\nolimits_{\otimes N}\right)  $.

Now we are going to prove that every $T\in N_{\operatorname*{lind}}^{\otimes
q}$ satisfies $T\in\operatorname*{Ker}\left(  \varphi\mid_{\otimes
N}-\operatorname*{id}\nolimits_{\otimes N}\right)  $.

In fact, let $T\in N_{\operatorname*{lind}}^{\otimes q}$ be arbitrary. Then,
$T$ is a left-induced tensor in $N^{\otimes q}$ (since $T\in
N_{\operatorname*{lind}}^{\otimes q}$); this means that there exist some $u\in
N$ and $U\in N^{\otimes\left(  q-1\right)  }$ such that $T=u\otimes U$.
Consider these $u$ and $U$. Since $u\in N\subseteq\mathfrak{g}=\mathfrak{g}%
^{\otimes1}$ and $U\in N^{\otimes\left(  q-1\right)  }\subseteq\mathfrak{g}%
^{\otimes\left(  q-1\right)  }$, we have $u\cdot U=u\otimes U$ (due to
(\ref{*=(X)}), applied to $u$, $U$, $1$ and $q-1$ instead of $a$, $b$, $n$ and
$m$).

\begin{vershort}
On the other hand, $U\in N^{\otimes\left(  q-1\right)  }\subseteq
\operatorname*{Ker}\left(  \varphi\mid_{\otimes N}-\operatorname*{id}%
\nolimits_{\otimes N}\right)  $ yields $\left(  \varphi\mid_{\otimes
N}-\operatorname*{id}\nolimits_{\otimes N}\right)  \left(  U\right)  =0$,
which means that $\varphi\left(  U\right)  =U$.
\end{vershort}

\begin{verlong}
On the other hand, $U\in N^{\otimes\left(  q-1\right)  }\subseteq
\operatorname*{Ker}\left(  \varphi\mid_{\otimes N}-\operatorname*{id}%
\nolimits_{\otimes N}\right)  $ yields $\left(  \varphi\mid_{\otimes
N}-\operatorname*{id}\nolimits_{\otimes N}\right)  \left(  U\right)  =0$, so
that%
\[
0=\left(  \varphi\mid_{\otimes N}-\operatorname*{id}\nolimits_{\otimes
N}\right)  \left(  U\right)  =\underbrace{\left(  \varphi\mid_{\otimes
N}\right)  \left(  U\right)  }_{=\varphi\left(  U\right)  }%
-\underbrace{\operatorname*{id}\nolimits_{\otimes N}\left(  U\right)  }%
_{=U}=\varphi\left(  U\right)  -U
\]
and thus $\varphi\left(  U\right)  =U$.
\end{verlong}

\begin{vershort}
Now, $T=u\otimes U=u\cdot U$ yields%
\begin{align*}
\varphi\left(  T\right)   &  =\varphi\left(  u\cdot U\right)  =u\cdot
\underbrace{\varphi\left(  U\right)  }_{=U}\ \ \ \ \ \ \ \ \ \ \left(
\text{due to (\ref{prop.alpha.rec.N})}\right) \\
&  =u\cdot U=T.
\end{align*}
In other words, $\left(  \varphi\mid_{\otimes N}-\operatorname*{id}%
\nolimits_{\otimes N}\right)  \left(  T\right)  =0$, so that $T\in
\operatorname*{Ker}\left(  \varphi\mid_{\otimes N}-\operatorname*{id}%
\nolimits_{\otimes N}\right)  $.
\end{vershort}

\begin{verlong}
Now, $T=u\otimes U=u\cdot U$ yields%
\begin{align*}
\varphi\left(  T\right)   &  =\varphi\left(  u\cdot U\right)  =u\cdot
\underbrace{\varphi\left(  U\right)  }_{=U}\ \ \ \ \ \ \ \ \ \ \left(
\text{due to (\ref{prop.alpha.rec.N})}\right) \\
&  =u\cdot U=T.
\end{align*}
Hence,%
\[
\left(  \varphi\mid_{\otimes N}-\operatorname*{id}\nolimits_{\otimes
N}\right)  \left(  T\right)  =\underbrace{\left(  \varphi\mid_{\otimes
N}\right)  \left(  T\right)  }_{=\varphi\left(  T\right)  =T}%
-\underbrace{\operatorname*{id}\nolimits_{\otimes N}\left(  T\right)  }%
_{=T}=T-T=0,
\]
and thus $T\in\operatorname*{Ker}\left(  \varphi\mid_{\otimes N}%
-\operatorname*{id}\nolimits_{\otimes N}\right)  $.
\end{verlong}

We have thus proven that every $T\in N_{\operatorname*{lind}}^{\otimes q}$
satisfies $T\in\operatorname*{Ker}\left(  \varphi\mid_{\otimes N}%
-\operatorname*{id}\nolimits_{\otimes N}\right)  $. In other words,
$N_{\operatorname*{lind}}^{\otimes q}\subseteq\operatorname*{Ker}\left(
\varphi\mid_{\otimes N}-\operatorname*{id}\nolimits_{\otimes N}\right)  $.

\begin{vershort}
This yields $\left\langle N_{\operatorname*{lind}}^{\otimes q}\right\rangle
\subseteq\operatorname*{Ker}\left(  \varphi\mid_{\otimes N}-\operatorname*{id}%
\nolimits_{\otimes N}\right)  $. Since $N^{\otimes q}=\left\langle
N_{\operatorname*{lind}}^{\otimes q}\right\rangle $ (by Proposition
\ref{prop.left-ind2}), we thus have $N^{\otimes q}\subseteq\operatorname*{Ker}%
\left(  \varphi\mid_{\otimes N}-\operatorname*{id}\nolimits_{\otimes
N}\right)  $. In other words, (\ref{pf.l25.alpha.(X)N.1}) holds for $p=q$.
This completes the induction step, and thus the induction proof of
(\ref{pf.l25.alpha.(X)N.1}) is done.
\end{vershort}

\begin{verlong}
Thus, Proposition \ref{prop.<>} \textbf{(a)} (applied to $\otimes N$,
$N_{\operatorname*{lind}}^{\otimes q}$ and $\operatorname*{Ker}\left(
\varphi\mid_{\otimes N}-\operatorname*{id}\nolimits_{\otimes N}\right)  $
instead of $M$, $S$ and $Q$) yields $\left\langle N_{\operatorname*{lind}%
}^{\otimes q}\right\rangle \subseteq\operatorname*{Ker}\left(  \varphi
\mid_{\otimes N}-\operatorname*{id}\nolimits_{\otimes N}\right)  $. Since
Proposition \ref{prop.left-ind2} (applied to $q$ and $N$ instead of $p$ and
$V$) says $N^{\otimes q}=\left\langle N_{\operatorname*{lind}}^{\otimes
q}\right\rangle $, we thus conclude that $N^{\otimes q}=\left\langle
N_{\operatorname*{lind}}^{\otimes q}\right\rangle \subseteq\operatorname*{Ker}%
\left(  \varphi\mid_{\otimes N}-\operatorname*{id}\nolimits_{\otimes
N}\right)  $. In other words, (\ref{pf.l25.alpha.(X)N.1}) holds for $p=q$.
This completes the induction step, and thus the induction proof of
(\ref{pf.l25.alpha.(X)N.1}) is done.
\end{verlong}

\begin{vershort}
Since $\otimes N=\bigoplus\limits_{p\in\mathbb{N}}N^{\otimes p}$, we can
conclude from (\ref{pf.l25.alpha.(X)N.1}) that $\otimes N\subseteq
\operatorname*{Ker}\left(  \varphi\mid_{\otimes N}-\operatorname*{id}%
\nolimits_{\otimes N}\right)  $. Ergo, $\varphi\mid_{\otimes N}%
=\operatorname*{id}\nolimits_{\otimes N}$. This proves Proposition
\ref{prop.l25.alpha.(X)N}.
\end{vershort}

\begin{verlong}
Now,%
\begin{align*}
\otimes N  &  =\bigoplus_{p\in\mathbb{N}}N^{\otimes p}=\sum_{p\in\mathbb{N}%
}\underbrace{N^{\otimes p}}_{\substack{\subseteq\operatorname*{Ker}\left(
\varphi\mid_{\otimes N}-\operatorname*{id}\nolimits_{\otimes N}\right)
\\\text{(by (\ref{pf.l25.alpha.(X)N.1}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since direct sums are sums}\right) \\
&  \subseteq\sum_{p\in\mathbb{N}}\operatorname*{Ker}\left(  \varphi
\mid_{\otimes N}-\operatorname*{id}\nolimits_{\otimes N}\right)
\subseteq\operatorname*{Ker}\left(  \varphi\mid_{\otimes N}-\operatorname*{id}%
\nolimits_{\otimes N}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname*{Ker}\left(
\varphi\mid_{\otimes N}-\operatorname*{id}\nolimits_{\otimes N}\right)  \text{
is a }k\text{-module}\right)  ,
\end{align*}
so that $\varphi\mid_{\otimes N}-\operatorname*{id}\nolimits_{\otimes N}=0$
and thus $\varphi\mid_{\otimes N}=\operatorname*{id}\nolimits_{\otimes N}$.
This proves Proposition \ref{prop.l25.alpha.(X)N}.
\end{verlong}

As a consequence of Proposition \ref{prop.l25.alpha.(X)N}, we can easily see:

\begin{corollary}
The map $\varphi$ satisfies $\varphi^{2}=\varphi$.
\end{corollary}

But we are not going to use this.

\subsection{A lemma on $\varphi$ and $\mathfrak{h}$-submodules of
$\otimes\mathfrak{g}$}

Our next plan is showing that $\varphi\left(  J\right)  =0$ and $\varphi
\left(  \left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  =0$. We
will simplify this task by the following lemma:

\begin{lemma}
\label{lem.l25.C}Let $C$ be an $\mathfrak{h}$-submodule of $\otimes
\mathfrak{g}$ satisfying $\varphi\left(  C\right)  =0$. Then, $\varphi\left(
\left(  \otimes\mathfrak{g}\right)  \cdot C\right)  =0$.
\end{lemma}

\textit{Proof of Lemma \ref{lem.l25.C}.} We are going to prove that%
\begin{equation}
\varphi\left(  \mathfrak{g}^{\otimes p}\cdot C\right)
=0\ \ \ \ \ \ \ \ \ \ \text{for every }p\in\mathbb{N}\text{.}
\label{pf.l25.C.1}%
\end{equation}
\textit{Proof of (\ref{pf.l25.C.1}).} We will prove (\ref{pf.l25.C.1}) by
induction over $p$:

\textit{Induction base:} We have $\mathfrak{g}^{\otimes0}=k$ and thus
$\mathfrak{g}^{\otimes0}\cdot C=k\cdot C=C$ (since $C$ is a $k$-submodule), so
that $\varphi\left(  \mathfrak{g}^{\otimes0}\cdot C\right)  =\varphi\left(
C\right)  =0$. Thus, (\ref{pf.l25.C.1}) holds for $p=0$. This completes the
induction base.

\begin{vershort}
\textit{Induction step:} Let $q\in\mathbb{N}_{+}$. Assume that
(\ref{pf.l25.C.1}) holds for $p=q-1$. Now we must prove that (\ref{pf.l25.C.1}%
) holds for $p=q$.

We have assumed that (\ref{pf.l25.C.1}) holds for $p=q-1$. In other words,
$\varphi\left(  \mathfrak{g}^{\otimes\left(  q-1\right)  }\cdot C\right)  =0$.

Let $D=\mathfrak{g}^{\otimes\left(  q-1\right)  }\cdot C$. Proposition
\ref{prop.g-alg.PQ} (applied to $\mathfrak{g}^{\otimes\left(  q-1\right)  }$,
$C$, $\otimes\mathfrak{g}$ and $\mathfrak{h}$ instead of $P$, $Q$, $A$ and
$\mathfrak{g}$) yields that $\mathfrak{g}^{\otimes\left(  q-1\right)  }\cdot
C$ is an $\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$ (since
$\mathfrak{g}^{\otimes\left(  q-1\right)  }$ and $C$ are $\mathfrak{h}%
$-submodules of $\otimes\mathfrak{g}$). Thus, $D$ is an $\mathfrak{h}%
$-submodule of $\otimes\mathfrak{g}$ (since $D=\mathfrak{g}^{\otimes\left(
q-1\right)  }\cdot C$). Also, $D=\mathfrak{g}^{\otimes\left(  q-1\right)
}\cdot C$ yields $\varphi\left(  D\right)  =\varphi\left(  \mathfrak{g}%
^{\otimes\left(  q-1\right)  }\cdot C\right)  =0$.

On the other hand, Proposition \ref{prop.tensor.filt-is-alg-filt} \textbf{(a)}
yields $\mathfrak{g}^{\otimes1}\cdot\mathfrak{g}^{\otimes\left(  q-1\right)
}=\mathfrak{g}^{\otimes q}$, so that $\underbrace{\mathfrak{g}^{\otimes q}%
}_{=\mathfrak{g}^{\otimes1}\cdot\mathfrak{g}^{\otimes\left(  q-1\right)  }%
}\cdot C=\underbrace{\mathfrak{g}^{\otimes1}}_{=\mathfrak{g}}\cdot
\underbrace{\mathfrak{g}^{\otimes\left(  q-1\right)  }\cdot C}_{=D}%
=\mathfrak{g}\cdot D$.

We are now in the following situation: We know that $D$ is an $\mathfrak{h}%
$-submodule of $\otimes\mathfrak{g}$ satisfying $\varphi\left(  D\right)  =0$.
We want to prove that $\varphi\left(  \mathfrak{g}\cdot D\right)  =0$.

In order to show this, it is obviously enough to show that $\varphi\left(
u\cdot U\right)  =0$ for every $u\in\mathfrak{g}$ and every $U\in D$ (since
$\varphi$ is $k$-linear). This, however, is very easy, because%
\[
\varphi\left(  u\cdot U\right)  =t\left(  u\right)  \cdot\varphi\left(
\underbrace{U}_{\in D}\right)  +\varphi\left(  \underbrace{s\left(  u\right)
\rightharpoonup U}_{\substack{\in D\text{ (since }s\left(  u\right)
\in\mathfrak{h}\text{, since }U\in D\\\text{and since }D\text{ is an
}\mathfrak{h}\text{-module)}}}\right)  \in t\left(  u\right)  \cdot
\underbrace{\varphi\left(  D\right)  }_{=0}+\underbrace{\varphi\left(
D\right)  }_{=0}=t\left(  U\right)  \cdot0+0=0.
\]


Thus we have shown that $\varphi\left(  \mathfrak{g}\cdot D\right)  =0$. Since
$\mathfrak{g}^{\otimes q}\cdot C=\mathfrak{g}\cdot D$, this becomes
$\varphi\left(  \mathfrak{g}^{\otimes q}\cdot C\right)  =0$. In other words,
(\ref{pf.l25.C.1}) holds for $p=q$. This completes the induction step, and
thus the induction proof of (\ref{pf.l25.C.1}) is done.
\end{vershort}

\begin{verlong}
\textit{Induction step:} Let $q\in\mathbb{N}_{+}$. Assume that
(\ref{pf.l25.C.1}) holds for $p=q-1$. Now we must prove that (\ref{pf.l25.C.1}%
) holds for $p=q$.

We have assumed that (\ref{pf.l25.C.1}) holds for $p=q-1$. In other words,
$\varphi\left(  \mathfrak{g}^{\otimes\left(  q-1\right)  }\cdot C\right)  =0$.

Let $D=\mathfrak{g}^{\otimes\left(  q-1\right)  }\cdot C$. Corollary
\ref{cor.biideal} \textbf{(c)} (applied to $\mathfrak{g}$, $q-1$ and $C$
instead of $V$, $p$ and $R$) yields that $\mathfrak{g}^{\otimes\left(
q-1\right)  }\cdot C$ is an $\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$
(since $C$ is an $\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$). Thus, $D$
is an $\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$ (since $D=\mathfrak{g}%
^{\otimes\left(  q-1\right)  }\cdot C$). Also, $D=\mathfrak{g}^{\otimes\left(
q-1\right)  }\cdot C$ yields $\varphi\left(  D\right)  =\varphi\left(
\mathfrak{g}^{\otimes\left(  q-1\right)  }\cdot C\right)  =0$.

On the other hand, Proposition \ref{prop.tensor.filt-is-alg-filt} \textbf{(a)}
(applied to $V=\mathfrak{g}$, $i=1$ and $j=q-1$) yields $\mathfrak{g}%
^{\otimes1}\cdot\mathfrak{g}^{\otimes\left(  q-1\right)  }=\mathfrak{g}%
^{\otimes\left(  1+\left(  q-1\right)  \right)  }=\mathfrak{g}^{\otimes q}$.
Thus,%
\[
\underbrace{\mathfrak{g}^{\otimes q}}_{=\mathfrak{g}^{\otimes1}\cdot
\mathfrak{g}^{\otimes\left(  q-1\right)  }}\cdot C=\underbrace{\mathfrak{g}%
^{\otimes1}}_{=\mathfrak{g}}\cdot\underbrace{\mathfrak{g}^{\otimes\left(
q-1\right)  }\cdot C}_{=D}=\mathfrak{g}\cdot D.
\]


We are now in the following situation: We know that $D$ is an $\mathfrak{h}%
$-submodule of $\otimes\mathfrak{g}$ satisfying $\varphi\left(  D\right)  =0$.
We want to prove that $\varphi\left(  \mathfrak{g}\cdot D\right)  =0$.

Let $E$ be the subset $\left\{  u\cdot U\ \mid\ \left(  u,U\right)
\in\mathfrak{g}\times D\right\}  $ of $\otimes\mathfrak{g}$. Then,%
\[
\mathfrak{g}\cdot D=\left\langle u\cdot U\ \mid\ \left(  u,U\right)
\in\mathfrak{g}\times D\right\rangle =\left\langle \underbrace{\left\{  u\cdot
U\ \mid\ \left(  u,U\right)  \in\mathfrak{g}\times D\right\}  }_{=E}%
\right\rangle =\left\langle E\right\rangle .
\]
Now we are going to show that $E\subseteq\operatorname*{Ker}\varphi$.

In fact, let $T\in E$ be arbitrary. Then, $T\in E=\left\{  u\cdot
U\ \mid\ \left(  u,U\right)  \in\mathfrak{g}\times D\right\}  $, so that there
exists some $\left(  u,U\right)  \in\mathfrak{g}\times D$ such that $T=u\cdot
U$. Consider this $\left(  u,U\right)  $. Clearly, $u\in\mathfrak{g}$ and
$U\in D$. Now, (\ref{prop.alpha.rec}) yields%
\begin{align*}
\varphi\left(  u\cdot U\right)   &  =t\left(  u\right)  \cdot\varphi\left(
\underbrace{U}_{\in D}\right)  +\varphi\left(  \underbrace{s\left(  u\right)
\rightharpoonup U}_{\substack{\in D\text{ (since }s\left(  u\right)
\in\mathfrak{h}\text{, since }U\in D\\\text{and since }D\text{ is an
}\mathfrak{h}\text{-module)}}}\right) \\
&  \in t\left(  u\right)  \cdot\underbrace{\varphi\left(  D\right)  }%
_{=0}+\underbrace{\varphi\left(  D\right)  }_{=0}=t\left(  u\right)
\cdot0+0=0,
\end{align*}
so that $\varphi\left(  u\cdot U\right)  =0$. Since $u\cdot U=T$, this
rewrites as $\varphi\left(  T\right)  =0$. Thus, $T\in\operatorname*{Ker}%
\varphi$.

We have therefore shown that every $T\in E$ satisfies $T\in\operatorname*{Ker}%
\varphi$. Thus, $E\subseteq\operatorname*{Ker}\varphi$.

Consequently, Proposition \ref{prop.<>} \textbf{(a)} (applied to
$\otimes\mathfrak{g}$, $E$ and $\operatorname*{Ker}\varphi$ instead of $M$,
$S$ and $Q$) yields $\left\langle E\right\rangle \subseteq\operatorname*{Ker}%
\varphi$ (since $\operatorname*{Ker}\varphi$ is a $k$-submodule of
$\otimes\mathfrak{g}$, since $\varphi$ is $k$-linear). Since $\left\langle
E\right\rangle =\mathfrak{g}\cdot D$, this rewrites as $\mathfrak{g}\cdot
D\subseteq\operatorname*{Ker}\varphi$, and thus $\varphi\left(  \mathfrak{g}%
\cdot D\right)  =0$. Since $\mathfrak{g}^{\otimes q}\cdot C=\mathfrak{g}\cdot
D$, this becomes $\varphi\left(  \mathfrak{g}^{\otimes q}\cdot C\right)  =0$.
In other words, (\ref{pf.l25.C.1}) holds for $p=q$. This completes the
induction step, and thus the induction proof of (\ref{pf.l25.C.1}) is done.
\end{verlong}

\begin{vershort}
Now, (\ref{pf.l25.C.1}) yields $\varphi\left(  \left(  \otimes\mathfrak{g}%
\right)  \cdot C\right)  =0$ (because $\otimes\mathfrak{g}=\bigoplus
\limits_{p\in\mathbb{N}}\mathfrak{g}^{\otimes p}$ and because $\varphi$ is
$k$-linear). This proves Lemma \ref{lem.l25.C}.
\end{vershort}

\begin{verlong}
Now, $\otimes\mathfrak{g}=\bigoplus\limits_{p\in\mathbb{N}}\mathfrak{g}%
^{\otimes p}=\sum\limits_{p\in\mathbb{N}}\mathfrak{g}^{\otimes p}$ (since
direct sums are sums) yields $\left(  \otimes\mathfrak{g}\right)  \cdot
C=\left(  \sum\limits_{p\in\mathbb{N}}\mathfrak{g}^{\otimes p}\right)  \cdot
C=\sum\limits_{p\in\mathbb{N}}\left(  \mathfrak{g}^{\otimes p}\cdot C\right)
$ and thus
\begin{align*}
\varphi\left(  \left(  \otimes\mathfrak{g}\right)  \cdot C\right)   &
=\varphi\left(  \sum\limits_{p\in\mathbb{N}}\left(  \mathfrak{g}^{\otimes
p}\cdot C\right)  \right)  =\sum\limits_{p\in\mathbb{N}}\underbrace{\varphi
\left(  \mathfrak{g}^{\otimes p}\cdot C\right)  }_{=0\text{ (due to
(\ref{pf.l25.C.1}))}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varphi\text{ is
}k\text{-linear}\right) \\
&  =\sum\limits_{p\in\mathbb{N}}0=0.
\end{align*}
This proves Lemma \ref{lem.l25.C}.
\end{verlong}

\subsection{$\varphi\left(  J\right)  =0$ and $\varphi\left(  \left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  =0$}

We are now ready to prove the following facts:

\begin{proposition}
\label{prop.l25.alpha(J)}We have $\varphi\left(  J\right)  =0$.
\end{proposition}

\begin{proposition}
\label{prop.l25.alpha(xgh)}We have $\varphi\left(  \left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  =0$.
\end{proposition}

\textit{Proof of Proposition \ref{prop.l25.alpha(J)}.} Consider the
$k$-submodule $J_{0}$ defined in Proposition \ref{prop.l25a} \textbf{(b)}.

We are going to prove that $\varphi\left(  J_{0}\cdot\left(  \otimes
\mathfrak{g}\right)  \right)  =0$ now (this will quickly yield $\varphi\left(
J\right)  =0$ then, due to $J=\left(  \otimes\mathfrak{g}\right)  \cdot
J_{0}\cdot\left(  \otimes\mathfrak{g}\right)  $ and Lemma \ref{lem.l25.C}).

\begin{vershort}
Let $S_{0}$ denote the subset $\left\{  v\otimes w-w\otimes v-\left[
v,w\right]  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}%
\right\}  $ of $\otimes\mathfrak{g}$. Then, $J_{0}=\left\langle S_{0}%
\right\rangle $ (this is how $J_{0}$ was defined).

Firstly, it is clear that in order to prove $\varphi\left(  J_{0}\cdot\left(
\otimes\mathfrak{g}\right)  \right)  =0$, it is enough to show that
$\varphi\left(  s\cdot T\right)  =0$ for every $s\in J_{0}$ and $T\in
\otimes\mathfrak{g}$ (because $\varphi$ is $k$-linear). Secondly, in order to
show that $\varphi\left(  s\cdot T\right)  =0$ for every $s\in J_{0}$ and
$T\in\otimes\mathfrak{g}$, it is enough to prove that $\varphi\left(  s\cdot
T\right)  =0$ for every $s\in S_{0}$ and $T\in\otimes\mathfrak{g}$ (because
$J_{0}=\left\langle S_{0}\right\rangle $, and because multiplication in
$\otimes\mathfrak{g}$ and the map $\varphi$ are $k$-linear). So let us prove
the latter assertion. Fix some $s\in S_{0}$ and $T\in\otimes\mathfrak{g}$.

By the very definition of $S_{0}$, the relation $s\in S_{0}$ means that there
exists some $\left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}$ such that
$s=v\otimes w-w\otimes v-\left[  v,w\right]  $. Since (\ref{*=(X)}) yields
$v\otimes w=v\cdot w$ and $w\otimes v=w\cdot v$, this rewrites as $s=v\cdot
w-w\cdot v-\left[  v,w\right]  $.

We are now going to show $\varphi\left(  s\cdot T\right)  =0$ through
straightforward computation:

We have $\varphi\left(  v\cdot T\right)  =t\left(  v\right)  \cdot
\varphi\left(  T\right)  +\varphi\left(  s\left(  v\right)  \rightharpoonup
T\right)  $ (according to (\ref{prop.alpha.rec}), applied to $u=v$ and $U=T$)
and $\varphi\left(  w\cdot T\right)  =\varphi\left(  w\rightharpoonup
T\right)  $ (according to (\ref{prop.alpha.rec.h}), applied to $u=w$ and $U=T$).

Now, (\ref{prop.alpha.rec}) (applied to $u=v$ and $U=w\cdot T$) yields
\begin{align*}
\varphi\left(  v\cdot w\cdot T\right)   &  =t\left(  v\right)  \cdot
\underbrace{\varphi\left(  w\cdot T\right)  }_{=\varphi\left(
w\rightharpoonup T\right)  }+\varphi\left(  \underbrace{s\left(  v\right)
\rightharpoonup\left(  w\cdot T\right)  }_{\substack{=\left(  s\left(
v\right)  \rightharpoonup w\right)  \cdot T+w\cdot\left(  s\left(  v\right)
\rightharpoonup T\right)  \\\text{(by (\ref{defs.g-alg.cond}) (applied to
}\otimes\mathfrak{g}\text{, }s\left(  v\right)  \text{, }w\text{ and
}T\\\text{instead of }A\text{, }a\text{, }u\text{ and }v\text{), because
}\otimes\mathfrak{g}\text{ is a }\mathfrak{g}\text{-algebra)}}}\right) \\
&  =t\left(  v\right)  \cdot\varphi\left(  w\rightharpoonup T\right)
+\underbrace{\varphi\left(  \left(  s\left(  v\right)  \rightharpoonup
w\right)  \cdot T+w\cdot\left(  s\left(  v\right)  \rightharpoonup T\right)
\right)  }_{\substack{=\varphi\left(  \left(  s\left(  v\right)
\rightharpoonup w\right)  \cdot T\right)  +\varphi\left(  w\cdot\left(
s\left(  v\right)  \rightharpoonup T\right)  \right)  \\\text{(since }%
\varphi\text{ is }k\text{-linear)}}}\\
&  =t\left(  v\right)  \cdot\varphi\left(  w\rightharpoonup T\right)
+\varphi\left(  \underbrace{\left(  s\left(  v\right)  \rightharpoonup
w\right)  }_{\substack{=\left[  s\left(  v\right)  ,w\right]  }}\cdot
T\right)  +\underbrace{\varphi\left(  w\cdot\left(  s\left(  v\right)
\rightharpoonup T\right)  \right)  }_{\substack{=\varphi\left(
w\rightharpoonup\left(  s\left(  v\right)  \rightharpoonup T\right)  \right)
\\\text{(by (\ref{prop.alpha.rec.h}), applied}\\\text{to }u=w\text{ and
}U=s\left(  v\right)  \rightharpoonup T\text{)}}}\\
&  =t\left(  v\right)  \cdot\varphi\left(  w\rightharpoonup T\right)
+\underbrace{\varphi\left(  \left[  s\left(  v\right)  ,w\right]  \cdot
T\right)  }_{\substack{=\varphi\left(  \left[  s\left(  v\right)  ,w\right]
\rightharpoonup T\right)  \\\text{(by (\ref{prop.alpha.rec.h}) (applied}%
\\\text{to }u=\left[  s\left(  v\right)  ,w\right]  \text{ and }U=T\text{),
because}\\\left[  s\left(  v\right)  ,w\right]  \in\mathfrak{h}\text{ (since
}s\left(  v\right)  \in\mathfrak{h}\text{ and }w\in\mathfrak{h}\text{
and}\\\text{since }\mathfrak{h}\text{ is a Lie subalgebra of }\mathfrak{g}%
\text{))}}}+\varphi\left(  w\rightharpoonup\left(  s\left(  v\right)
\rightharpoonup T\right)  \right) \\
&  =t\left(  v\right)  \cdot\varphi\left(  w\rightharpoonup T\right)
+\varphi\left(  \left[  s\left(  v\right)  ,w\right]  \rightharpoonup
T\right)  +\varphi\left(  w\rightharpoonup\left(  s\left(  v\right)
\rightharpoonup T\right)  \right)  .
\end{align*}
On the other hand, (\ref{prop.alpha.rec.h}) (applied to $u=w$ and $U=v\cdot
T$) yields%
\begin{align*}
\varphi\left(  w\cdot v\cdot T\right)   &  =\varphi\left(
\underbrace{w\rightharpoonup\left(  v\cdot T\right)  }_{\substack{=\left(
w\rightharpoonup v\right)  \cdot T+v\cdot\left(  w\rightharpoonup T\right)
\\\text{(by (\ref{defs.g-alg.cond}) (applied to }w\text{, }v\text{ and
}T\\\text{instead of }a\text{, }u\text{ and }v\text{), because }%
\otimes\mathfrak{g}\text{ is a }\mathfrak{g}\text{-algebra)}}}\right) \\
&  =\varphi\left(  \left(  w\rightharpoonup v\right)  \cdot T+v\cdot\left(
w\rightharpoonup T\right)  \right) \\
&  =\varphi\left(  \underbrace{\left(  w\rightharpoonup v\right)
}_{\substack{=\left[  w,v\right]  }}\cdot T\right)  +\underbrace{\varphi
\left(  v\cdot\left(  w\rightharpoonup T\right)  \right)  }%
_{\substack{=t\left(  v\right)  \cdot\varphi\left(  w\rightharpoonup T\right)
+\varphi\left(  s\left(  v\right)  \rightharpoonup\left(  w\rightharpoonup
T\right)  \right)  \\\text{(according to (\ref{prop.alpha.rec}),}%
\\\text{applied to }u=v\text{ and }U=w\rightharpoonup T\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varphi\text{ is }k\text{-linear}%
\right) \\
&  =\varphi\left(  \left[  w,v\right]  \cdot T\right)  +t\left(  v\right)
\cdot\varphi\left(  w\rightharpoonup T\right)  +\varphi\left(  s\left(
v\right)  \rightharpoonup\left(  w\rightharpoonup T\right)  \right)  .
\end{align*}
Finally,%
\[
\varphi\left(  \underbrace{\left[  v,w\right]  }_{=-\left[  w,v\right]  }\cdot
T\right)  =-\varphi\left(  \left[  w,v\right]  \cdot T\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varphi\text{ is }k\text{-linear}%
\right)  .
\]
Now, $s=v\cdot w-w\cdot v-\left[  v,w\right]  $ yields%
\begin{align*}
&  \varphi\left(  s\cdot T\right) \\
&  =\varphi\left(  \left(  v\cdot w-w\cdot v-\left[  v,w\right]  \right)
\cdot T\right) \\
&  =\underbrace{\varphi\left(  v\cdot w\cdot T\right)  }_{=t\left(  v\right)
\cdot\varphi\left(  w\rightharpoonup T\right)  +\varphi\left(  \left[
s\left(  v\right)  ,w\right]  \rightharpoonup T\right)  +\varphi\left(
w\rightharpoonup\left(  s\left(  v\right)  \rightharpoonup T\right)  \right)
}-\underbrace{\varphi\left(  w\cdot v\cdot T\right)  }_{=\varphi\left(
\left[  w,v\right]  \cdot T\right)  +t\left(  v\right)  \cdot\varphi\left(
w\rightharpoonup T\right)  +\varphi\left(  s\left(  v\right)  \rightharpoonup
\left(  w\rightharpoonup T\right)  \right)  }-\underbrace{\varphi\left(
\left[  v,w\right]  \cdot T\right)  }_{=-\varphi\left(  \left[  w,v\right]
\cdot T\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varphi\text{ is }k\text{-linear}%
\right) \\
&  =\varphi\left(  \left[  s\left(  v\right)  ,w\right]  \rightharpoonup
T\right)  +\varphi\left(  w\rightharpoonup\left(  s\left(  v\right)
\rightharpoonup T\right)  \right)  -\varphi\left(  s\left(  v\right)
\rightharpoonup\left(  w\rightharpoonup T\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{after some obvious cancellations}\right)
\\
&  =\varphi\left(  \underbrace{\left[  s\left(  v\right)  ,w\right]
\rightharpoonup T}_{\substack{=s\left(  v\right)  \rightharpoonup\left(
w\rightharpoonup T\right)  -w\rightharpoonup\left(  s\left(  v\right)
\rightharpoonup T\right)  \\\text{(according to (\ref{[L-mod]}) (applied
to}\\\otimes\mathfrak{g}\text{, }s\left(  v\right)  \text{, }w\text{ and
}T\text{ instead of }V\text{, }a\text{, }b\text{ and }v\text{),}\\\text{since
}\otimes\mathfrak{g}\text{ is a }\mathfrak{g}\text{-module)}}%
}+w\rightharpoonup\left(  s\left(  v\right)  \rightharpoonup T\right)
-s\left(  v\right)  \rightharpoonup\left(  w\rightharpoonup T\right)  \right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varphi\text{ is }k\text{-linear}%
\right) \\
&  =\varphi\left(  \underbrace{s\left(  v\right)  \rightharpoonup\left(
w\rightharpoonup T\right)  -w\rightharpoonup\left(  s\left(  v\right)
\rightharpoonup T\right)  +w\rightharpoonup\left(  s\left(  v\right)
\rightharpoonup T\right)  -s\left(  v\right)  \rightharpoonup\left(
w\rightharpoonup T\right)  }_{=0}\right) \\
&  =\varphi\left(  0\right)  =0,
\end{align*}
just as we wanted to show.

The proof of $\varphi\left(  J_{0}\cdot\left(  \otimes\mathfrak{g}\right)
\right)  =0$ is thus complete. Now, $J_{0}\cdot\left(  \otimes\mathfrak{g}%
\right)  $ is an $\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$ (according
to Proposition \ref{prop.l25a} \textbf{(b)}). Thus, Lemma \ref{lem.l25.C}
(applied to $C=J_{0}\cdot\left(  \otimes\mathfrak{g}\right)  $) yields that
$\varphi\left(  \left(  \otimes\mathfrak{g}\right)  \cdot\left(  J_{0}%
\cdot\left(  \otimes\mathfrak{g}\right)  \right)  \right)  =0$ (because
$\varphi\left(  J_{0}\cdot\left(  \otimes\mathfrak{g}\right)  \right)  =0$).
Since $\left(  \otimes\mathfrak{g}\right)  \cdot\left(  J_{0}\cdot\left(
\otimes\mathfrak{g}\right)  \right)  =\left(  \otimes\mathfrak{g}\right)
\cdot J_{0}\cdot\left(  \otimes\mathfrak{g}\right)  =J$ (according to
Proposition \ref{prop.l25a} \textbf{(b)} again), this rewrites as
$\varphi\left(  J\right)  =0$. Thus, Proposition \ref{prop.l25.alpha(J)} is proven.
\end{vershort}

\begin{verlong}
Let $S_{0}$ denote the subset $\left\{  v\otimes w-w\otimes v-\left[
v,w\right]  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}%
\right\}  $ of $\otimes\mathfrak{g}$. Then, we know from the definition of
$J_{0}$ that
\begin{align*}
J_{0}  &  =\left\langle v\otimes w-w\otimes v-\left[  v,w\right]
\ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}\right\rangle \\
&  =\left\langle \underbrace{\left\{  v\otimes w-w\otimes v-\left[
v,w\right]  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}%
\right\}  }_{=S_{0}}\right\rangle =\left\langle S_{0}\right\rangle .
\end{align*}


Now, let $T\in\otimes\mathfrak{g}$ be arbitrary. Let $\varrho_{T}%
:\otimes\mathfrak{g}\rightarrow\otimes N$ be the map defined by%
\[
\left(  \varrho_{T}\left(  U\right)  =\varphi\left(  U\cdot T\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }U\in\otimes\mathfrak{g}\right)  .
\]
This map $\varrho_{T}$ is $k$-linear (since it is the composition of the two
$k$-linear maps $\varphi$ and $\otimes\mathfrak{g}\rightarrow\otimes
\mathfrak{g},$ $U\mapsto U\cdot T$). Thus, $\operatorname*{Ker}\varrho_{T}$ is
a $k$-submodule of $\otimes\mathfrak{g}$.

We are now going to prove that every $s\in S_{0}$ satisfy $s\in
\operatorname*{Ker}\varrho_{T}$.

In fact, let $s\in S_{0}$ be arbitrary. Since $s\in S_{0}=\left\{  v\otimes
w-w\otimes v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)  \in
\mathfrak{g}\times\mathfrak{h}\right\}  $, there exists some $\left(
v,w\right)  \in\mathfrak{g}\times\mathfrak{h}$ such that $s=v\otimes
w-w\otimes v-\left[  v,w\right]  $. So we have $v\in\mathfrak{g}$ and
$w\in\mathfrak{h}$. Note that $v\in\mathfrak{g}=\mathfrak{g}^{\otimes1}$ and
$w\in\mathfrak{h}\subseteq\mathfrak{g}=\mathfrak{g}^{\otimes1}$ yield $v\cdot
w=v\otimes w$ (by (\ref{*=(X)}), applied to $1$ and $1$ instead of $n$ and
$m$), and similarly $w\cdot v=w\otimes v$. Using these observations, we see
that%
\[
s=\underbrace{v\otimes w}_{=v\cdot w}-\underbrace{w\otimes v}_{=w\cdot
v}-\left[  v,w\right]  =v\cdot w-w\cdot v-\left[  v,w\right]  .
\]


Now, we are going to prove that $\varrho_{T}\left(  s\right)  =0$. In order to
do this, we will compute $\varrho_{T}\left(  v\cdot w\right)  $, $\varrho
_{T}\left(  w\cdot v\right)  $ and $\varrho_{T}\left(  \left[  v,w\right]
\right)  $. First let us compute $\varphi\left(  v\cdot T\right)  $ and
$\varphi\left(  w\cdot T\right)  $.

We have $\varphi\left(  v\cdot T\right)  =t\left(  v\right)  \cdot
\varphi\left(  T\right)  +\varphi\left(  s\left(  v\right)  \rightharpoonup
T\right)  $ (according to (\ref{prop.alpha.rec}), applied to $u=v$ and $U=T$)
and $\varphi\left(  w\cdot T\right)  =\varphi\left(  w\rightharpoonup
T\right)  $ (according to (\ref{prop.alpha.rec.h}), applied to $u=w$ and $U=T$).

The definition of $\varrho_{T}$ yields
\begin{align}
\varrho_{T}\left(  v\cdot w\right)   &  =\varphi\left(  v\cdot w\cdot
T\right)  =t\left(  v\right)  \cdot\underbrace{\varphi\left(  w\cdot T\right)
}_{=\varphi\left(  w\rightharpoonup T\right)  }+\varphi\left(
\underbrace{s\left(  v\right)  \rightharpoonup\left(  w\cdot T\right)
}_{\substack{=\left(  s\left(  v\right)  \rightharpoonup w\right)  \cdot
T+w\cdot\left(  s\left(  v\right)  \rightharpoonup T\right)  \\\text{(by
(\ref{defs.g-alg.cond}) (applied to }\otimes\mathfrak{g}\text{, }s\left(
v\right)  \text{, }w\text{ and }T\\\text{instead of }A\text{, }a\text{,
}u\text{ and }v\text{), because }\otimes\mathfrak{g}\text{ is a }%
\mathfrak{g}\text{-algebra)}}}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{according to (\ref{prop.alpha.rec}),
applied to }u=v\text{ and }U=w\cdot T\right) \nonumber\\
&  =t\left(  v\right)  \cdot\varphi\left(  w\rightharpoonup T\right)
+\underbrace{\varphi\left(  \left(  s\left(  v\right)  \rightharpoonup
w\right)  \cdot T+w\cdot\left(  s\left(  v\right)  \rightharpoonup T\right)
\right)  }_{\substack{=\varphi\left(  \left(  s\left(  v\right)
\rightharpoonup w\right)  \cdot T\right)  +\varphi\left(  w\cdot\left(
s\left(  v\right)  \rightharpoonup T\right)  \right)  \\\text{(since }%
\varphi\text{ is }k\text{-linear)}}}\nonumber\\
&  =t\left(  v\right)  \cdot\varphi\left(  w\rightharpoonup T\right)
+\varphi\left(  \underbrace{\left(  s\left(  v\right)  \rightharpoonup
w\right)  }_{\substack{=\left[  s\left(  v\right)  ,w\right]
\\\text{(according to (\ref{prop.g-is-g-module.form}), applied}\\\text{to
}s\left(  v\right)  \text{ instead of }v\text{)}}}\cdot T\right)
+\underbrace{\varphi\left(  w\cdot\left(  s\left(  v\right)  \rightharpoonup
T\right)  \right)  }_{\substack{=\varphi\left(  w\rightharpoonup\left(
s\left(  v\right)  \rightharpoonup T\right)  \right)  \\\text{(by
(\ref{prop.alpha.rec.h}), applied}\\\text{to }u=w\text{ and }U=s\left(
v\right)  \rightharpoonup T\text{)}}}\nonumber\\
&  =t\left(  v\right)  \cdot\varphi\left(  w\rightharpoonup T\right)
+\underbrace{\varphi\left(  \left[  s\left(  v\right)  ,w\right]  \cdot
T\right)  }_{\substack{=\varphi\left(  \left[  s\left(  v\right)  ,w\right]
\rightharpoonup T\right)  \\\text{(by (\ref{prop.alpha.rec.h}) (applied}%
\\\text{to }u=\left[  s\left(  v\right)  ,w\right]  \text{ and }U=T\text{),
because}\\\left[  s\left(  v\right)  ,w\right]  \in\mathfrak{h}\text{ (since
}s\left(  v\right)  \in\mathfrak{h}\text{ and }w\in\mathfrak{h}\text{
and}\\\text{since }\mathfrak{h}\text{ is a Lie subalgebra of }\mathfrak{g}%
\text{))}}}+\varphi\left(  w\rightharpoonup\left(  s\left(  v\right)
\rightharpoonup T\right)  \right) \nonumber\\
&  =t\left(  v\right)  \cdot\varphi\left(  w\rightharpoonup T\right)
+\varphi\left(  \left[  s\left(  v\right)  ,w\right]  \rightharpoonup
T\right)  +\varphi\left(  w\rightharpoonup\left(  s\left(  v\right)
\rightharpoonup T\right)  \right)  . \label{pf.l25.alpha(J).vw}%
\end{align}


The definition of $\varrho_{T}$ yields%
\begin{align}
\varrho_{T}\left(  w\cdot v\right)   &  =\varphi\left(  w\cdot v\cdot
T\right)  =\varphi\left(  \underbrace{w\rightharpoonup\left(  v\cdot T\right)
}_{\substack{=\left(  w\rightharpoonup v\right)  \cdot T+v\cdot\left(
w\rightharpoonup T\right)  \\\text{(by (\ref{defs.g-alg.cond}) (applied to
}\otimes\mathfrak{g}\text{, }w\text{, }v\text{ and }T\\\text{instead of
}A\text{, }a\text{, }u\text{ and }v\text{), because }\otimes\mathfrak{g}\text{
is a }\mathfrak{g}\text{-algebra)}}}\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{prop.alpha.rec.h}), applied to
}u=w\text{ and }U=v\cdot T\right) \nonumber\\
&  =\varphi\left(  \left(  w\rightharpoonup v\right)  \cdot T+v\cdot\left(
w\rightharpoonup T\right)  \right) \nonumber\\
&  =\varphi\left(  \underbrace{\left(  w\rightharpoonup v\right)
}_{\substack{=\left[  w,v\right]  \\\text{(according to
(\ref{prop.g-is-g-module.form}), applied}\\\text{to }w\text{ and }v\text{
instead of }v\text{ and }w\text{)}}}\cdot T\right)  +\underbrace{\varphi
\left(  v\cdot\left(  w\rightharpoonup T\right)  \right)  }%
_{\substack{=t\left(  v\right)  \cdot\varphi\left(  w\rightharpoonup T\right)
+\varphi\left(  s\left(  v\right)  \rightharpoonup\left(  w\rightharpoonup
T\right)  \right)  \\\text{(according to (\ref{prop.alpha.rec}),}%
\\\text{applied to }u=v\text{ and }U=w\rightharpoonup T\text{)}}}\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varphi\text{ is }k\text{-linear}%
\right) \nonumber\\
&  =\varphi\left(  \left[  w,v\right]  \cdot T\right)  +t\left(  v\right)
\cdot\varphi\left(  w\rightharpoonup T\right)  +\varphi\left(  s\left(
v\right)  \rightharpoonup\left(  w\rightharpoonup T\right)  \right)  .
\label{pf.l25.alpha(J).wv}%
\end{align}
Finally, the definition of $\varrho_{T}$ yields%
\begin{equation}
\varrho_{T}\left(  \left[  v,w\right]  \right)  =\varphi\left(
\underbrace{\left[  v,w\right]  }_{\substack{=-\left[  w,v\right]
\\\text{(due to (\ref{[]-3}))}}}\cdot T\right)  =\varphi\left(  -\left[
w,v\right]  \cdot T\right)  =-\varphi\left(  \left[  w,v\right]  \cdot
T\right)  \label{pf.l25.alpha(J).[v,w]}%
\end{equation}
(since $\varphi$ is $k$-linear).

Now, $s=v\cdot w-w\cdot v-\left[  v,w\right]  $ yields%
\begin{align*}
&  \varrho_{T}\left(  s\right) \\
&  =\varrho_{T}\left(  v\cdot w-w\cdot v-\left[  v,w\right]  \right) \\
&  =\underbrace{\varrho_{T}\left(  v\cdot w\right)  }_{\substack{=t\left(
v\right)  \cdot\varphi\left(  w\rightharpoonup T\right)  +\varphi\left(
\left[  s\left(  v\right)  ,w\right]  \rightharpoonup T\right)  +\varphi
\left(  w\rightharpoonup\left(  s\left(  v\right)  \rightharpoonup T\right)
\right)  \\\text{(due to (\ref{pf.l25.alpha(J).vw}))}}}-\underbrace{\varrho
_{T}\left(  w\cdot v\right)  }_{\substack{=\varphi\left(  \left[  w,v\right]
\cdot T\right)  +t\left(  v\right)  \cdot\varphi\left(  w\rightharpoonup
T\right)  +\varphi\left(  s\left(  v\right)  \rightharpoonup\left(
w\rightharpoonup T\right)  \right)  \\\text{(due to (\ref{pf.l25.alpha(J).wv}%
))}}}-\underbrace{\varrho_{T}\left(  \left[  v,w\right]  \right)
}_{\substack{=-\varphi\left(  \left[  w,v\right]  \cdot T\right)  \\\text{(due
to (\ref{pf.l25.alpha(J).[v,w]}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varrho_{T}\text{ is
}k\text{-linear}\right) \\
&  =\left(  t\left(  v\right)  \cdot\varphi\left(  w\rightharpoonup T\right)
+\varphi\left(  \left[  s\left(  v\right)  ,w\right]  \rightharpoonup
T\right)  +\varphi\left(  w\rightharpoonup\left(  s\left(  v\right)
\rightharpoonup T\right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  \varphi\left(  \left[  w,v\right]  \cdot
T\right)  +t\left(  v\right)  \cdot\varphi\left(  w\rightharpoonup T\right)
+\varphi\left(  s\left(  v\right)  \rightharpoonup\left(  w\rightharpoonup
T\right)  \right)  \right)  -\left(  -\varphi\left(  \left[  w,v\right]  \cdot
T\right)  \right) \\
&  =\varphi\left(  \left[  s\left(  v\right)  ,w\right]  \rightharpoonup
T\right)  +\varphi\left(  w\rightharpoonup\left(  s\left(  v\right)
\rightharpoonup T\right)  \right)  -\varphi\left(  s\left(  v\right)
\rightharpoonup\left(  w\rightharpoonup T\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{after some obvious cancellations}\right)
\\
&  =\varphi\left(  \underbrace{\left[  s\left(  v\right)  ,w\right]
\rightharpoonup T}_{\substack{=s\left(  v\right)  \rightharpoonup\left(
w\rightharpoonup T\right)  -w\rightharpoonup\left(  s\left(  v\right)
\rightharpoonup T\right)  \\\text{(according to (\ref{[L-mod]}) (applied
to}\\\otimes\mathfrak{g}\text{, }s\left(  v\right)  \text{, }w\text{ and
}T\text{ instead of }V\text{, }a\text{, }b\text{ and }v\text{),}\\\text{since
}\otimes\mathfrak{g}\text{ is a }\mathfrak{g}\text{-module)}}%
}+w\rightharpoonup\left(  s\left(  v\right)  \rightharpoonup T\right)
-s\left(  v\right)  \rightharpoonup\left(  w\rightharpoonup T\right)  \right)
\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varphi\text{ is }k\text{-linear}%
\right) \\
&  =\varphi\left(  \underbrace{s\left(  v\right)  \rightharpoonup\left(
w\rightharpoonup T\right)  -w\rightharpoonup\left(  s\left(  v\right)
\rightharpoonup T\right)  +w\rightharpoonup\left(  s\left(  v\right)
\rightharpoonup T\right)  -s\left(  v\right)  \rightharpoonup\left(
w\rightharpoonup T\right)  }_{=0}\right) \\
&  =\varphi\left(  0\right)  =0,
\end{align*}
so that $s\in\operatorname*{Ker}\varrho_{T}$.

We have thus proven that every $s\in S_{0}$ satisfies $s\in\operatorname*{Ker}%
\varrho_{T}$. Thus, $S_{0}\subseteq\operatorname*{Ker}\varrho_{T}$. Since
$\operatorname*{Ker}\varrho_{T}$ is a $k$-module, Proposition \ref{prop.<>}
\textbf{(a)} (applied to $\otimes\mathfrak{g}$, $S_{0}$ and
$\operatorname*{Ker}\varrho_{T}$ instead of $M$, $S$ and $Q$) now yields
$\left\langle S_{0}\right\rangle \subseteq\operatorname*{Ker}\varrho_{T}$.
Since $\left\langle S_{0}\right\rangle =J_{0}$, this rewrites as
$J_{0}\subseteq\operatorname*{Ker}\varrho_{T}$. Thus, $\varrho_{T}\left(
J_{0}\right)  =0$. In other words, every $j\in J_{0}$ satisfies $\varrho
_{T}\left(  j\right)  =0$. Since $\varrho_{T}\left(  j\right)  =\varphi\left(
j\cdot T\right)  $ (by the definition of $\varrho_{T}$), this rewrites as
$\varphi\left(  j\cdot T\right)  =0$, and thus $j\cdot T\in\operatorname*{Ker}%
\varphi$ for every $j\in J_{0}$.

We have thus shown that every $j\in J_{0}$ and $T\in\otimes\mathfrak{g}$
satisfy $j\cdot T\in\operatorname*{Ker}\varphi$. In other words, every
$\left(  j,T\right)  \in J_{0}\times\left(  \otimes\mathfrak{g}\right)  $
satisfies $j\cdot T\in\operatorname*{Ker}\varphi$. In other words, $\left\{
j\cdot T\ \mid\ \left(  j,T\right)  \in J_{0}\times\left(  \otimes
\mathfrak{g}\right)  \right\}  \subseteq\operatorname*{Ker}\varphi$.

Now, applying Proposition \ref{prop.<>} \textbf{(a)} to $\otimes\mathfrak{g}$,
$\left\{  j\cdot T\ \mid\ \left(  j,T\right)  \in J_{0}\times\left(
\otimes\mathfrak{g}\right)  \right\}  $ and $\operatorname*{Ker}\varphi$
instead of $M$, $S$ and $Q$, we see that $\left\langle \left\{  j\cdot
T\ \mid\ \left(  j,T\right)  \in J_{0}\times\left(  \otimes\mathfrak{g}%
\right)  \right\}  \right\rangle \subseteq\operatorname*{Ker}\varphi$ (since
$\left\{  j\cdot T\ \mid\ \left(  j,T\right)  \in J_{0}\times\left(
\otimes\mathfrak{g}\right)  \right\}  \subseteq\operatorname*{Ker}\varphi$ and
since $\operatorname*{Ker}\varphi$ is a $k$-module). Now%
\[
J_{0}\cdot\left(  \otimes\mathfrak{g}\right)  =\left\langle j\cdot
T\ \mid\ \left(  j,T\right)  \in J_{0}\times\left(  \otimes\mathfrak{g}%
\right)  \right\rangle =\left\langle \left\{  j\cdot T\ \mid\ \left(
j,T\right)  \in J_{0}\times\left(  \otimes\mathfrak{g}\right)  \right\}
\right\rangle \subseteq\operatorname*{Ker}\varphi
\]
and thus $\varphi\left(  J_{0}\cdot\left(  \otimes\mathfrak{g}\right)
\right)  =0$.

Now, $J_{0}\cdot\left(  \otimes\mathfrak{g}\right)  $ is an $\mathfrak{h}%
$-submodule of $\otimes\mathfrak{g}$ (according to Proposition \ref{prop.l25a}
\textbf{(b)}). Thus, Lemma \ref{lem.l25.C} (applied to $C=J_{0}\cdot\left(
\otimes\mathfrak{g}\right)  $) yields that $\varphi\left(  \left(
\otimes\mathfrak{g}\right)  \cdot\left(  J_{0}\cdot\left(  \otimes
\mathfrak{g}\right)  \right)  \right)  =0$ (because $\varphi\left(  J_{0}%
\cdot\left(  \otimes\mathfrak{g}\right)  \right)  =0$). Since $\left(
\otimes\mathfrak{g}\right)  \cdot\left(  J_{0}\cdot\left(  \otimes
\mathfrak{g}\right)  \right)  =\left(  \otimes\mathfrak{g}\right)  \cdot
J_{0}\cdot\left(  \otimes\mathfrak{g}\right)  =J$ (according to Proposition
\ref{prop.l25a} \textbf{(b)} again), this rewrites as $\varphi\left(
J\right)  =0$. Thus, Proposition \ref{prop.l25.alpha(J)} is proven.
\end{verlong}

\textit{Proof of Proposition \ref{prop.l25.alpha(xgh)}.} Every $u\in
\mathfrak{h}$ satisfies%
\begin{align*}
\varphi\left(  u\right)   &  =t\left(  u\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{according to (\ref{rmk.alpha-1}), since }u\in\mathfrak{h}\subseteq
\mathfrak{g}\right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }u\in\mathfrak{h}\text{, while
}t\text{ is a projection with kernel }\mathfrak{h}\right)  .
\end{align*}
Thus, $\varphi\left(  \mathfrak{h}\right)  =0$. Since we know that
$\mathfrak{h}$ is an $\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$, we can
thus follow from Lemma \ref{lem.l25.C} (applied to $C=\mathfrak{h}$) that
$\varphi\left(  \left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
=0$ (because $\varphi\left(  \mathfrak{h}\right)  =0$). Thus, Proposition
\ref{prop.l25.alpha(xgh)} is proven.

\subsection{$\varphi$ induces a filtered $k$-module isomorphism}

We recall that, in order to prove Theorem \ref{thm.l25} \textbf{(b)}, we want
to construct an $\mathfrak{h}$-module isomorphism $F_{n}\diagup F_{n-1}%
\rightarrow\mathfrak{n}^{\otimes n}$. We will do this by constructing a
$k$-module isomorphism $\left(  \otimes\mathfrak{g}\right)  \diagup\left(
J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \rightarrow
\otimes\mathfrak{n}$ which respects the filtration. The associated graded
morphism of this isomorphism will then turn out to be an $\mathfrak{h}$-module isomorphism.

Here is how we construct our $k$-module isomorphism:

\begin{proposition}
\label{prop.l25.Ker-alpha}\textbf{(a)} The homomorphism $\varphi$ is
surjective and satisfies $\operatorname*{Ker}\varphi=J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}$. Thus, $\varphi$ induces a $k$-module
isomorphism $\overline{\varphi}:\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\rightarrow\otimes N$ which satisfies $\varphi=\overline{\varphi}\circ\zeta$
(where $\zeta$ denotes the canonical projection $\otimes\mathfrak{g}%
\rightarrow\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ as in Theorem
\ref{thm.l25}).\newline\textbf{(b)} The isomorphism $\overline{\varphi}$
respects the filtration. Here, the filtration on $\left(  \otimes
\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  $ is given by $\left(  F_{n}\right)  _{n\geq0}$, and
the filtration on $\otimes N$ is given by $\left(  N^{\otimes\leq n}\right)
_{n\geq0}$.\newline\textbf{(c)} The $k$-module homomorphism $\pi\mid
_{N}:N\rightarrow\mathfrak{n}$ is an isomorphism. Thus, the $k$-algebra
homomorphism $\otimes\left(  \pi\mid_{N}\right)  :\otimes N\rightarrow
\otimes\mathfrak{n}$ is an isomorphism as well. We are going to denote this
isomorphism $\otimes\left(  \pi\mid_{N}\right)  $ by $\eta$. The homomorphism
$\eta$ respects the filtration (where the filtrations on $\otimes N$ and
$\otimes\mathfrak{n}$ are the degree filtrations, as usual).\newline%
\textbf{(d)} Let $\iota:\otimes N\rightarrow\otimes\mathfrak{g}$ be the
canonical inclusion of $\otimes N$ in $\otimes\mathfrak{g}$. Then, the inverse
$\overline{\varphi}^{-1}$ of the isomorphism $\overline{\varphi}$ equals
$\zeta\circ\iota$. This inverse $\overline{\varphi}^{-1}$ also respects the
filtration.\newline\textbf{(e)} The composition $\eta\circ\overline{\varphi
}:\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \rightarrow\otimes\mathfrak{n}$
respects the filtration. For every $p\in\mathbb{N}$, the homomorphism
$\operatorname*{gr}\nolimits_{p}\left(  \eta\circ\overline{\varphi}\right)
:\operatorname*{gr}\nolimits_{p}\left(  \left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\right)  \rightarrow\operatorname*{gr}\nolimits_{p}\left(  \otimes
\mathfrak{n}\right)  $ is an $\mathfrak{h}$-module isomorphism.
\end{proposition}

In order to prove this, a very technical result:

\begin{lemma}
\label{lem.l25.sum}We have $\otimes\mathfrak{g}=J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(  \otimes N\right)  $ (where we,
as before, identify $\otimes N$ with a $k$-subalgebra of $\otimes\mathfrak{g}$).
\end{lemma}

There are two ways to prove this lemma. Let us sketch the first one (which is
hard to formalize, but rather straightforward) and detail the second one
(which is easier to formalize, but is not the simplest proof).

\textit{First proof of Lemma \ref{lem.l25.sum} (sketched).} It is enough to
prove that every pure tensor in $\otimes\mathfrak{g}$ lies in $J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(  \otimes N\right)  $
(because the pure tensors generate $\otimes\mathfrak{g}$ as a $k$-module). So
we must show that $g_{1}\otimes g_{2}\otimes...\otimes g_{n}\in J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(  \otimes N\right)  $ for
every $n\in\mathbb{N}$ and any elements $g_{1}$, $g_{2}$, $...$, $g_{n}$ of
$\mathfrak{g}$.

Since $g_{i}=t\left(  g_{i}\right)  +s\left(  g_{i}\right)  $ for every
$i\in\left\{  1,2,...,n\right\}  $, this rewrites as $\left(  t\left(
g_{1}\right)  +s\left(  g_{1}\right)  \right)  \otimes\left(  t\left(
g_{2}\right)  +s\left(  g_{2}\right)  \right)  \otimes...\otimes\left(
t\left(  g_{n}\right)  +s\left(  g_{n}\right)  \right)  \in J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(  \otimes N\right)  $.
Expanding the left hand side, it becomes a sum of $2^{n}$ addends. One of
these addends is $t\left(  g_{1}\right)  \otimes t\left(  g_{2}\right)
\otimes...\otimes t\left(  g_{n}\right)  $ and lies in $\otimes N$. Every of
the other $2^{n}-1$ addends is a pure tensor with at least one tensorand lying
in $\mathfrak{h}$. Some of these tensors have their rightmost tensorand lying
in $\mathfrak{h}$, which means that these tensors lie in $\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}$. The other ones still have a
tensorand lying in $\mathfrak{h}$, but it is not their rightmost tensorand.
Transform each of these latter tensors according to the following rewriting
rule:%
\begin{align*}
&  v_{1}\otimes v_{2}\otimes...\otimes v_{\ell-1}\otimes v_{\ell}\otimes
v_{\ell+1}\otimes v_{\ell+2}\otimes...\otimes v_{n}\\
&  \rightarrow v_{1}\otimes v_{2}\otimes...\otimes v_{\ell-1}\otimes
v_{\ell+1}\otimes v_{\ell}\otimes v_{\ell+2}\otimes...\otimes v_{n}\\
&  \ \ \ \ \ \ \ \ \ \ -v_{1}\otimes v_{2}\otimes...\otimes v_{\ell-1}%
\otimes\left[  v_{\ell+1},v_{\ell}\right]  \otimes v_{\ell+2}\otimes...\otimes
v_{n}\\
&  \ \ \ \ \ \ \ \ \ \ -v_{1}\otimes v_{2}\otimes...\otimes v_{\ell-1}%
\otimes\left(  v_{\ell+1}\otimes v_{\ell}-v_{\ell}\otimes v_{\ell+1}-\left[
v_{\ell+1},v_{\ell}\right]  \right)  \otimes v_{\ell+2}\otimes...\otimes
v_{n}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{whenever }v_{1},v_{2},...,v_{n}%
\in\mathfrak{g}\text{ and }v_{\ell}\in\mathfrak{h}\right)  .
\end{align*}
This rewriting rule splits the tensor $v_{1}\otimes v_{2}\otimes...\otimes
v_{\ell-1}\otimes v_{\ell}\otimes v_{\ell+1}\otimes v_{\ell+2}\otimes
...\otimes v_{n}$ into three tensors:

\begin{itemize}
\item a tensor $v_{1}\otimes v_{2}\otimes...\otimes v_{\ell-1}\otimes
v_{\ell+1}\otimes v_{\ell}\otimes v_{\ell+2}\otimes...\otimes v_{n}$ (we call
this tensor the "primary product") which still has a tensorand lying in
$\mathfrak{h}$ (namely, $v_{\ell}$), but now this tensorand has moved one step
to the right;

\item a tensor $-v_{1}\otimes v_{2}\otimes...\otimes v_{\ell-1}\otimes\left[
v_{\ell+1},v_{\ell}\right]  \otimes v_{\ell+2}\otimes...\otimes v_{n}$ (we
call this tensor a "fission product") which is of degree smaller than $n$;

\item a tensor $-v_{1}\otimes v_{2}\otimes...\otimes v_{\ell-1}\otimes\left(
v_{\ell+1}\otimes v_{\ell}-v_{\ell}\otimes v_{\ell+1}-\left[  v_{\ell
+1},v_{\ell}\right]  \right)  \otimes v_{\ell+2}\otimes...\otimes v_{n}$ (we
call this tensor a "J-product") which lies in $J$.
\end{itemize}

After applying this rewriting rule to all the tensors to which it can be
applied, let us apply it again to the primary products, then again to the
resulting primary products, etc. - until each of the primary products has its
rightmost tensorand lying in $\mathfrak{h}$ and thus cannot be rewritten
anymore. As a result, we obtain a sum of tensors whose rightmost tensorand
lies in $\mathfrak{h}$ (these tensors clearly lie in $\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}$), of "fission products" (these are
tensors of degree smaller than $n$) and of "J-products" (which lie in $J$).
And do not forget the tensor $t\left(  g_{1}\right)  \otimes t\left(
g_{2}\right)  \otimes...\otimes t\left(  g_{n}\right)  $ which lies in
$\otimes N$. As a result we know that%
\begin{align*}
g_{1}\otimes g_{2}\otimes...\otimes g_{n}  &  =\left(  \text{sum of some
tensors in }\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right) \\
&  +\left(  \text{sum of some tensors of degree smaller than }n\right) \\
&  +\left(  \text{sum of some tensors in }J\right)  +\left(  \text{a tensor in
}\otimes N\right) \\
&  \in\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\mathfrak{g}%
^{\otimes\leq\left(  n-1\right)  }+J+\left(  \otimes N\right)  .
\end{align*}
Since this holds for every pure tensor $g_{1}\otimes g_{2}\otimes...\otimes
g_{n}\in\mathfrak{g}^{\otimes n}$, this yields $\mathfrak{g}^{\otimes
n}\subseteq\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}%
+\mathfrak{g}^{\otimes\leq\left(  n-1\right)  }+J+\left(  \otimes N\right)  $.
But if we do an induction over $n$, we can assume that we already know that
$\mathfrak{g}^{\otimes\leq\left(  n-1\right)  }\subseteq J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(  \otimes N\right)  $, so
that this becomes
\[
\mathfrak{g}^{\otimes n}\subseteq\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}+\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot
\mathfrak{h}+\left(  \otimes N\right)  \right)  +J+\left(  \otimes N\right)
=J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(  \otimes
N\right)  ,
\]
completing the induction.

So much for the first proof of Lemma \ref{lem.l25.sum}.

\textit{Second proof of Lemma \ref{lem.l25.sum}.} Let us first show that%
\begin{equation}
J\cdot\left(  \otimes\mathfrak{g}\right)  =J\ \ \ \ \ \ \ \ \ \ \text{and}%
\ \ \ \ \ \ \ \ \ \ \left(  \otimes\mathfrak{g}\right)  \cdot J=J
\label{pf.l25.sum.-1}%
\end{equation}


\textit{Proof of (\ref{pf.l25.sum.-1}).} In fact, Proposition \ref{prop.l25a}
\textbf{(b)} yields $J=\left(  \otimes\mathfrak{g}\right)  \cdot J_{0}%
\cdot\left(  \otimes\mathfrak{g}\right)  $ (where $J_{0}$ was defined in that
proposition). This leads to $J\cdot\left(  \otimes\mathfrak{g}\right)
=\left(  \otimes\mathfrak{g}\right)  \cdot J_{0}\cdot\underbrace{\left(
\otimes\mathfrak{g}\right)  \cdot\left(  \otimes\mathfrak{g}\right)
}_{=\otimes\mathfrak{g}}=\left(  \otimes\mathfrak{g}\right)  \cdot J_{0}%
\cdot\left(  \otimes\mathfrak{g}\right)  =J$, but also leads to $\left(
\otimes\mathfrak{g}\right)  \cdot J=\underbrace{\left(  \otimes\mathfrak{g}%
\right)  \cdot\left(  \otimes\mathfrak{g}\right)  }_{=\otimes\mathfrak{g}%
}\cdot J_{0}\cdot\left(  \otimes\mathfrak{g}\right)  \subseteq\left(
\otimes\mathfrak{g}\right)  \cdot J_{0}\cdot\left(  \otimes\mathfrak{g}%
\right)  =J$. This proves (\ref{pf.l25.sum.-1}).

Next, we consider the $k$-submodule $J_{0}$ of $\otimes\mathfrak{g}$ defined
in Proposition \ref{prop.l25a} \textbf{(b)}. This $k$-submodule $J_{0}$ was
defined as $\left\langle v\otimes w-w\otimes v-\left[  v,w\right]
\ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}\right\rangle $.
Hence, obviously,
\begin{equation}
v\otimes w-w\otimes v-\left[  v,w\right]  \in J_{0}%
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  v,w\right)  \in\mathfrak{g}%
\times\mathfrak{h}. \label{pf.l25.sum.-2}%
\end{equation}
Now we are going to show that%
\begin{equation}
vw-wv-\left[  v,w\right]  \in J\ \ \ \ \ \ \ \ \ \ \text{for any }%
v\in\mathfrak{g}\text{ and }w\in\mathfrak{h} \label{pf.l25.sum.0}%
\end{equation}
(where $vw$ means the product of $v$ and $w$ in $\otimes\mathfrak{g}$, and
similarly $wv$ means the product of $w$ and $v$ in $\otimes\mathfrak{g}$).

\begin{vershort}
\textit{Proof of (\ref{pf.l25.sum.0}).} Clearly, $v\in\mathfrak{g}%
=\mathfrak{g}^{\otimes1}$ and $w\in\mathfrak{h}\subseteq\mathfrak{g}%
=\mathfrak{g}^{\otimes1}$ yield $v\cdot w=v\otimes w$ (by (\ref{*=(X)})) and
similarly $w\cdot v=w\otimes v$. Now,%
\begin{align*}
\underbrace{vw}_{=v\cdot w=v\otimes w}-\underbrace{wv}_{=w\cdot v=w\otimes
v}-\left[  v,w\right]   &  =v\otimes w-w\otimes v-\left[  v,w\right]
=\underbrace{1}_{\in\otimes\mathfrak{g}}\cdot\underbrace{\left(  v\otimes
w-w\otimes v-\left[  v,w\right]  \right)  }_{\in J_{0}\text{ (according to
(\ref{pf.l25.sum.-2}))}}\cdot\underbrace{1}_{\in\otimes\mathfrak{g}}\\
&  \in\left(  \otimes\mathfrak{g}\right)  \cdot J_{0}\cdot\left(
\otimes\mathfrak{g}\right)  =J\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition
\ref{prop.l25a} \textbf{(b)}}\right)  .
\end{align*}
This proves (\ref{pf.l25.sum.0}).
\end{vershort}

\begin{verlong}
\textit{Proof of (\ref{pf.l25.sum.0}).} Clearly, $v\in\mathfrak{g}%
=\mathfrak{g}^{\otimes1}$ and $w\in\mathfrak{h}\subseteq\mathfrak{g}%
=\mathfrak{g}^{\otimes1}$ yield $v\cdot w=v\otimes w$ (by (\ref{*=(X)}),
applied to $1$ and $1$ instead of $n$ and $m$), and similarly $w\cdot
v=w\otimes v$. On the other hand, $v\in\mathfrak{g}$ and $w\in\mathfrak{h}$
lead to $\left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}$. Now,%
\begin{align*}
\underbrace{vw}_{=v\cdot w=v\otimes w}-\underbrace{wv}_{=w\cdot v=w\otimes
v}-\left[  v,w\right]   &  =v\otimes w-w\otimes v-\left[  v,w\right]
=\underbrace{1}_{\in\otimes\mathfrak{g}}\cdot\underbrace{\left(  v\otimes
w-w\otimes v-\left[  v,w\right]  \right)  }_{\in J_{0}\text{ (according to
(\ref{pf.l25.sum.-2}))}}\cdot\underbrace{1}_{\in\otimes\mathfrak{g}}\\
&  \in\left(  \otimes\mathfrak{g}\right)  \cdot J_{0}\cdot\left(
\otimes\mathfrak{g}\right)  =J\ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition
\ref{prop.l25a} \textbf{(b)}}\right)  .
\end{align*}
This proves (\ref{pf.l25.sum.0}).
\end{verlong}

Next, let us show that%
\begin{equation}
\mathfrak{hg}\subseteq J+\mathfrak{gh}+\mathfrak{g}. \label{pf.l25.sum.1}%
\end{equation}


\begin{vershort}
\textit{Proof of (\ref{pf.l25.sum.1}).} In order to prove (\ref{pf.l25.sum.1}%
), it is enough to show that $wv\in J+\mathfrak{gh}+\mathfrak{g}$ for every
$w\in\mathfrak{h}$ and $v\in\mathfrak{g}$. But this is easy:%
\begin{align*}
wv  &  =-\underbrace{\left(  vw-wv-\left[  v,w\right]  \right)  }_{\in J\text{
(according to (\ref{pf.l25.sum.0}))}}+\underbrace{v}_{\in\mathfrak{g}%
}\underbrace{w}_{\in\mathfrak{h}}+\underbrace{\left(  -\left[  v,w\right]
\right)  }_{\in\mathfrak{g}}\in\underbrace{-J}_{\substack{=J\text{ (since
}J\text{ is}\\\text{a }k\text{-module)}}}+\mathfrak{gh}+\mathfrak{g}\\
&  =J+\mathfrak{gh}+\mathfrak{g}.
\end{align*}
Thus, (\ref{pf.l25.sum.1}) is proven.
\end{vershort}

\begin{verlong}
\textit{Proof of (\ref{pf.l25.sum.1}).} Let $G$ be the subset $\left\{
hg\ \mid\ \left(  h,g\right)  \in\mathfrak{h}\times\mathfrak{g}\right\}  $ of
$\otimes\mathfrak{g}$. We are going to prove that $G\subseteq J+\mathfrak{gh}%
+\mathfrak{g}$ now.

Indeed, let $U\in G$ be arbitrary. Then, $U\in G=\left\{  hg\ \mid\ \left(
h,g\right)  \in\mathfrak{h}\times\mathfrak{g}\right\}  $, so there exists some
$\left(  w,v\right)  \in\mathfrak{h}\times\mathfrak{g}$ such that $U=wv$.
Thus, $w\in\mathfrak{h}$ and $v\in\mathfrak{g}$, and $U=wv$. Now,%
\begin{align*}
U  &  =wv=-\underbrace{\left(  vw-wv-\left[  v,w\right]  \right)  }_{\in
J\text{ (according to (\ref{pf.l25.sum.0}))}}+\underbrace{v}_{\in\mathfrak{g}%
}\underbrace{w}_{\in\mathfrak{h}}+\underbrace{\left(  -\left[  v,w\right]
\right)  }_{\in\mathfrak{g}}\in\underbrace{-J}_{\substack{=J\text{ (since
}J\text{ is}\\\text{a }k\text{-module)}}}+\mathfrak{gh}+\mathfrak{g}\\
&  =J+\mathfrak{gh}+\mathfrak{g}.
\end{align*}
We have thus shown that every $U\in G$ satisfies $U\in J+\mathfrak{gh}%
+\mathfrak{g}$. In other words, $G\subseteq J+\mathfrak{gh}+\mathfrak{g}$.

Therefore, Proposition \ref{prop.<>} \textbf{(a)} (applied to $\otimes
\mathfrak{g}$, $G$ and $J+\mathfrak{gh}+\mathfrak{g}$ instead of $M$, $S$ and
$Q$) yields $\left\langle G\right\rangle \subseteq J+\mathfrak{gh}%
+\mathfrak{g}$. Since%
\begin{align*}
\left\langle G\right\rangle  &  =\left\langle \left\{  hg\ \mid\ \left(
h,g\right)  \in\mathfrak{h}\times\mathfrak{g}\right\}  \right\rangle
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }G=\left\{  hg\ \mid\ \left(
h,g\right)  \in\mathfrak{h}\times\mathfrak{g}\right\}  \right) \\
&  =\left\langle hg\ \mid\ \left(  h,g\right)  \in\mathfrak{h}\times
\mathfrak{g}\right\rangle =\mathfrak{hg},
\end{align*}
this rewrites as $\mathfrak{hg}\subseteq J+\mathfrak{gh}+\mathfrak{g}$. Thus,
(\ref{pf.l25.sum.1}) is proven.
\end{verlong}

Let us next conclude that%
\begin{equation}
\mathfrak{g}^{\otimes\left(  p-1\right)  }\mathfrak{hg}\subseteq
J+\mathfrak{g}^{\otimes p}\mathfrak{h}+\mathfrak{g}^{\otimes\leq
p}\ \ \ \ \ \ \ \ \ \ \text{for every }p\in\mathbb{N}_{+}.
\label{pf.l25.sum.2}%
\end{equation}


\begin{vershort}
\textit{Proof of (\ref{pf.l25.sum.2}).} Proposition
\ref{prop.tensor.filt-is-alg-filt} \textbf{(a)} yields $\mathfrak{g}%
^{\otimes\left(  p-1\right)  }\cdot\mathfrak{g}^{\otimes1}=\mathfrak{g}%
^{\otimes p}$. In other words, $\mathfrak{g}^{\otimes\left(  p-1\right)
}\mathfrak{g}=\mathfrak{g}^{\otimes p}$.

From (\ref{pf.l25.sum.1}), we have
\begin{align*}
\mathfrak{g}^{\otimes\left(  p-1\right)  }\underbrace{\mathfrak{hg}%
}_{\subseteq J+\mathfrak{gh}+\mathfrak{g}}  &  \subseteq\mathfrak{g}%
^{\otimes\left(  p-1\right)  }\left(  J+\mathfrak{gh}+\mathfrak{g}\right)
=\underbrace{\mathfrak{g}^{\otimes\left(  p-1\right)  }}_{\subseteq
\otimes\mathfrak{g}}J+\underbrace{\mathfrak{g}^{\otimes\left(  p-1\right)
}\mathfrak{g}}_{=\mathfrak{g}^{\otimes p}}\mathfrak{h}%
+\underbrace{\mathfrak{g}^{\otimes\left(  p-1\right)  }\mathfrak{g}%
}_{=\mathfrak{g}^{\otimes p}\subseteq\mathfrak{g}^{\otimes\leq p}}\\
&  \subseteq\underbrace{\left(  \otimes\mathfrak{g}\right)  \cdot J}_{=J\text{
(by (\ref{pf.l25.sum.-1}))}}+\mathfrak{g}^{\otimes p}\mathfrak{h}%
+\mathfrak{g}^{\otimes\leq p}\subseteq J+\mathfrak{g}^{\otimes p}%
\mathfrak{h}+\mathfrak{g}^{\otimes\leq p}.
\end{align*}
This proves (\ref{pf.l25.sum.2}).
\end{vershort}

\begin{verlong}
\textit{Proof of (\ref{pf.l25.sum.2}).} Proposition
\ref{prop.tensor.filt-is-alg-filt} \textbf{(a)} (applied to $\mathfrak{g}$,
$p-1$ and $1$ instead of $V$, $i$ and $j$) yields $\mathfrak{g}^{\otimes
\left(  p-1\right)  }\cdot\mathfrak{g}^{\otimes1}=\mathfrak{g}^{\otimes\left(
\left(  p-1\right)  +1\right)  }=\mathfrak{g}^{\otimes p}$. Since
$\mathfrak{g}^{\otimes1}=\mathfrak{g}$, this becomes $\mathfrak{g}%
^{\otimes\left(  p-1\right)  }\cdot\mathfrak{g}=\mathfrak{g}^{\otimes p}$. In
other words, $\mathfrak{g}^{\otimes\left(  p-1\right)  }\mathfrak{g}%
=\mathfrak{g}^{\otimes p}$.

From (\ref{pf.l25.sum.1}), we have
\begin{align*}
\mathfrak{g}^{\otimes\left(  p-1\right)  }\underbrace{\mathfrak{hg}%
}_{\subseteq J+\mathfrak{gh}+\mathfrak{g}}  &  \subseteq\mathfrak{g}%
^{\otimes\left(  p-1\right)  }\left(  J+\mathfrak{gh}+\mathfrak{g}\right)
=\underbrace{\mathfrak{g}^{\otimes\left(  p-1\right)  }}_{\subseteq
\otimes\mathfrak{g}}J+\underbrace{\mathfrak{g}^{\otimes\left(  p-1\right)
}\mathfrak{g}}_{=\mathfrak{g}^{\otimes p}}\mathfrak{h}%
+\underbrace{\mathfrak{g}^{\otimes\left(  p-1\right)  }\mathfrak{g}%
}_{=\mathfrak{g}^{\otimes p}\subseteq\mathfrak{g}^{\otimes\leq p}}\\
&  \subseteq\underbrace{\left(  \otimes\mathfrak{g}\right)  \cdot J}_{=J\text{
(by (\ref{pf.l25.sum.-1}))}}+\mathfrak{g}^{\otimes p}\mathfrak{h}%
+\mathfrak{g}^{\otimes\leq p}\subseteq J+\mathfrak{g}^{\otimes p}%
\mathfrak{h}+\mathfrak{g}^{\otimes\leq p}.
\end{align*}
This proves (\ref{pf.l25.sum.2}).
\end{verlong}

Our next step is to prove that%
\begin{equation}
\mathfrak{g}^{\otimes q}\subseteq\mathfrak{g}^{\otimes\leq\left(  q-1\right)
}+J+\mathfrak{g}^{\otimes\left(  q-1\right)  }\mathfrak{h}+N^{\otimes
q}\ \ \ \ \ \ \ \ \ \ \text{for every }q\in\mathbb{N}_{+}.
\label{pf.l25.sum.3}%
\end{equation}


\textit{Proof of (\ref{pf.l25.sum.3}).} We are going to prove
(\ref{pf.l25.sum.3}) by induction over $q$:

\textit{Induction base:} In the case $q=1$, we have $\mathfrak{g}^{\otimes
q}=\mathfrak{g}^{\otimes1}=\mathfrak{g}=\mathfrak{h}\oplus N=\mathfrak{h}+N$
and%
\[
\mathfrak{g}^{\otimes\leq\left(  q-1\right)  }+J+\mathfrak{g}^{\otimes\left(
q-1\right)  }\mathfrak{h}+N^{\otimes q}\supseteq\underbrace{\mathfrak{g}%
^{\otimes\left(  q-1\right)  }}_{=\mathfrak{g}^{\otimes\left(  1-1\right)
}=\mathfrak{g}^{\otimes0}=k}\mathfrak{h}+\underbrace{N^{\otimes q}%
}_{=N^{\otimes1}=N}=\underbrace{k\mathfrak{h}}_{\substack{=\mathfrak{h}\text{
(since }\mathfrak{h}\text{ is}\\\text{a }k\text{-module)}}}+N=\mathfrak{h}+N.
\]
Thus, in the case $q=1$, we have $\mathfrak{g}^{\otimes q}=\mathfrak{h}%
+N\subseteq\mathfrak{g}^{\otimes\leq\left(  q-1\right)  }+J+\mathfrak{g}%
^{\otimes\left(  q-1\right)  }\mathfrak{h}+N^{\otimes q}$. In other words,
(\ref{pf.l25.sum.3}) holds for $q=1$. This completes the induction base.

\textit{Induction step:} Let $p\in\mathbb{N}_{+}$ be arbitrary. Assume that
(\ref{pf.l25.sum.3}) holds for $q=p$. Now our task is to prove that
(\ref{pf.l25.sum.3}) holds for $q=p+1$ as well.

Since (\ref{pf.l25.sum.3}) holds for $q=p$, we have $\mathfrak{g}^{\otimes
p}\subseteq\mathfrak{g}^{\otimes\leq\left(  p-1\right)  }+J+\mathfrak{g}%
^{\otimes\left(  p-1\right)  }\mathfrak{h}+N^{\otimes p}$.

\begin{vershort}
Now, Proposition \ref{prop.tensor.filt-is-alg-filt} \textbf{(a)} yields
$\mathfrak{g}^{\otimes p}\cdot\mathfrak{g}^{\otimes1}=\mathfrak{g}%
^{\otimes\left(  p+1\right)  }$. In other words, $\mathfrak{g}^{\otimes
p}\cdot\mathfrak{g}=\mathfrak{g}^{\otimes\left(  p+1\right)  }$. Thus,%
\begin{align*}
\mathfrak{g}^{\otimes\left(  p+1\right)  }  &  =\underbrace{\mathfrak{g}%
^{\otimes p}}_{\subseteq\mathfrak{g}^{\otimes\leq\left(  p-1\right)
}+J+\mathfrak{g}^{\otimes\left(  p-1\right)  }\mathfrak{h}+N^{\otimes p}}%
\cdot\mathfrak{g}\subseteq\left(  \mathfrak{g}^{\otimes\leq\left(  p-1\right)
}+J+\mathfrak{g}^{\otimes\left(  p-1\right)  }\mathfrak{h}+N^{\otimes
p}\right)  \cdot\mathfrak{g}\\
&  =\mathfrak{g}^{\otimes\leq\left(  p-1\right)  }\cdot
\underbrace{\mathfrak{g}}_{=\mathfrak{g}^{\otimes1}\subseteq\mathfrak{g}%
^{\otimes\leq1}}+J\cdot\underbrace{\mathfrak{g}}_{\subseteq\otimes
\mathfrak{g}}+\underbrace{\mathfrak{g}^{\otimes\left(  p-1\right)
}\mathfrak{hg}}_{\substack{\subseteq J+\mathfrak{g}^{\otimes p}\mathfrak{h}%
+\mathfrak{g}^{\otimes\leq p}\\\text{(by (\ref{pf.l25.sum.2}))}}}+N^{\otimes
p}\cdot\underbrace{\mathfrak{g}}_{=\mathfrak{h}\oplus N=\mathfrak{h}+N}\\
&  \subseteq\underbrace{\mathfrak{g}^{\otimes\leq\left(  p-1\right)  }%
\cdot\mathfrak{g}^{\otimes\leq1}}_{\substack{\subseteq\mathfrak{g}%
^{\otimes\leq p}\\\text{(by Proposition \ref{prop.tensor.filt-is-alg-filt}
\textbf{(b)})}}}+\underbrace{J\cdot\left(  \otimes\mathfrak{g}\right)
}_{=J\text{ (by (\ref{pf.l25.sum.-1}))}}+J+\mathfrak{g}^{\otimes
p}\mathfrak{h}+\mathfrak{g}^{\otimes\leq p}+\underbrace{N^{\otimes p}%
\cdot\left(  \mathfrak{h}+N\right)  }_{=N^{\otimes p}\cdot\mathfrak{h}%
+N^{\otimes p}\cdot N}\\
&  \subseteq\mathfrak{g}^{\otimes\leq p}+\underbrace{J+J}_{\substack{=J\text{
(since }J\\\text{is a }k\text{-module)}}}+\mathfrak{g}^{\otimes p}%
\mathfrak{h}+\mathfrak{g}^{\otimes\leq p}+\underbrace{N^{\otimes p}%
}_{\subseteq\mathfrak{g}^{\otimes p}}\cdot\mathfrak{h}+N^{\otimes p}%
\cdot\underbrace{N}_{=N^{\otimes1}}\\
&  \subseteq\mathfrak{g}^{\otimes\leq p}+J+\mathfrak{g}^{\otimes
p}\mathfrak{h}+\mathfrak{g}^{\otimes\leq p}+\mathfrak{g}^{\otimes
p}\mathfrak{h}+\underbrace{N^{\otimes p}\cdot N^{\otimes1}}_{=N^{\otimes
\left(  p+1\right)  }\text{ (by Proposition \ref{prop.tensor.filt-is-alg-filt}
\textbf{(a)})}}\\
&  =\mathfrak{g}^{\otimes\leq p}+J+\mathfrak{g}^{\otimes p}\mathfrak{h}%
+\mathfrak{g}^{\otimes\leq p}+\mathfrak{g}^{\otimes p}\mathfrak{h}%
+N^{\otimes\left(  p+1\right)  }\\
&  =\underbrace{\mathfrak{g}^{\otimes\leq p}+\mathfrak{g}^{\otimes\leq p}%
}_{=\mathfrak{g}^{\otimes\leq p}}+J+\underbrace{\mathfrak{g}^{\otimes
p}\mathfrak{h}+\mathfrak{g}^{\otimes p}\mathfrak{h}}_{=\mathfrak{g}^{\otimes
p}\mathfrak{h}}+N^{\otimes\left(  p+1\right)  }=\mathfrak{g}^{\otimes\leq
p}+J+\mathfrak{g}^{\otimes p}\mathfrak{h}+N^{\otimes\left(  p+1\right)  }.
\end{align*}
In other words, (\ref{pf.l25.sum.3}) holds for $q=p+1$. This completes the
induction step, and thus (\ref{pf.l25.sum.3}) is proven.
\end{vershort}

\begin{verlong}
Now, Proposition \ref{prop.tensor.filt-is-alg-filt} \textbf{(a)} (applied to
$\mathfrak{g}$, $p$ and $1$ instead of $V$, $i$ and $j$) yields $\mathfrak{g}%
^{\otimes p}\cdot\mathfrak{g}^{\otimes1}=\mathfrak{g}^{\otimes\left(
p+1\right)  }$. Since $\mathfrak{g}^{\otimes1}=\mathfrak{g}$, this becomes
$\mathfrak{g}^{\otimes p}\cdot\mathfrak{g}=\mathfrak{g}^{\otimes\left(
p+1\right)  }$. Thus,%
\begin{align*}
\mathfrak{g}^{\otimes\left(  p+1\right)  }  &  =\underbrace{\mathfrak{g}%
^{\otimes p}}_{\subseteq\mathfrak{g}^{\otimes\leq\left(  p-1\right)
}+J+\mathfrak{g}^{\otimes\left(  p-1\right)  }\mathfrak{h}+N^{\otimes p}}%
\cdot\mathfrak{g}\subseteq\left(  \mathfrak{g}^{\otimes\leq\left(  p-1\right)
}+J+\mathfrak{g}^{\otimes\left(  p-1\right)  }\mathfrak{h}+N^{\otimes
p}\right)  \cdot\mathfrak{g}\\
&  =\mathfrak{g}^{\otimes\leq\left(  p-1\right)  }\cdot
\underbrace{\mathfrak{g}}_{=\mathfrak{g}^{\otimes1}\subseteq\mathfrak{g}%
^{\otimes\leq1}}+J\cdot\underbrace{\mathfrak{g}}_{\subseteq\otimes
\mathfrak{g}}+\underbrace{\mathfrak{g}^{\otimes\left(  p-1\right)
}\mathfrak{hg}}_{\substack{\subseteq J+\mathfrak{g}^{\otimes p}\mathfrak{h}%
+\mathfrak{g}^{\otimes\leq p}\\\text{(by (\ref{pf.l25.sum.2}))}}}+N^{\otimes
p}\cdot\underbrace{\mathfrak{g}}_{=\mathfrak{h}\oplus N=\mathfrak{h}+N}\\
&  \subseteq\underbrace{\mathfrak{g}^{\otimes\leq\left(  p-1\right)  }%
\cdot\mathfrak{g}^{\otimes\leq1}}_{\substack{\subseteq\mathfrak{g}%
^{\otimes\leq\left(  \left(  p-1\right)  +1\right)  }\\\text{(by Proposition
\ref{prop.tensor.filt-is-alg-filt} \textbf{(b)},}\\\text{applied to
}\mathfrak{g}\text{, }p-1\text{ and }1\text{ instead of }V\text{, }n\text{ and
}m\text{)}}}+\underbrace{J\cdot\left(  \otimes\mathfrak{g}\right)  }_{=J\text{
(by (\ref{pf.l25.sum.-1}))}}+J+\mathfrak{g}^{\otimes p}\mathfrak{h}%
+\mathfrak{g}^{\otimes\leq p}+\underbrace{N^{\otimes p}\cdot\left(
\mathfrak{h}+N\right)  }_{=N^{\otimes p}\cdot\mathfrak{h}+N^{\otimes p}\cdot
N}\\
&  \subseteq\underbrace{\mathfrak{g}^{\otimes\leq\left(  \left(  p-1\right)
+1\right)  }}_{=\mathfrak{g}^{\otimes\leq p}}+\underbrace{J+J}%
_{\substack{=J\text{ (since }J\\\text{is a }k\text{-module)}}}+\mathfrak{g}%
^{\otimes p}\mathfrak{h}+\mathfrak{g}^{\otimes\leq p}+\underbrace{N^{\otimes
p}}_{\subseteq\mathfrak{g}^{\otimes p}}\cdot\mathfrak{h}+N^{\otimes p}%
\cdot\underbrace{N}_{=N^{\otimes1}}\\
&  \subseteq\mathfrak{g}^{\otimes\leq p}+J+\mathfrak{g}^{\otimes
p}\mathfrak{h}+\mathfrak{g}^{\otimes\leq p}+\mathfrak{g}^{\otimes
p}\mathfrak{h}+\underbrace{N^{\otimes p}\cdot N^{\otimes1}}%
_{\substack{=N^{\otimes\left(  p+1\right)  }\text{ (by Proposition
\ref{prop.tensor.filt-is-alg-filt} \textbf{(a)},}\\\text{applied to }N\text{,
}p\text{ and }1\text{ instead of }V\text{, }i\text{ and }j\text{)}}}\\
&  =\mathfrak{g}^{\otimes\leq p}+J+\mathfrak{g}^{\otimes p}\mathfrak{h}%
+\mathfrak{g}^{\otimes\leq p}+\mathfrak{g}^{\otimes p}\mathfrak{h}%
+N^{\otimes\left(  p+1\right)  }\\
&  =\underbrace{\mathfrak{g}^{\otimes\leq p}+\mathfrak{g}^{\otimes\leq p}%
}_{\substack{=\mathfrak{g}^{\otimes\leq p}\text{ (since }\mathfrak{g}%
^{\otimes\leq p}\\\text{is a }k\text{-module)}}}+J+\underbrace{\mathfrak{g}%
^{\otimes p}\mathfrak{h}+\mathfrak{g}^{\otimes p}\mathfrak{h}}%
_{\substack{=\mathfrak{g}^{\otimes p}\mathfrak{h}\text{ (since }%
\mathfrak{g}^{\otimes p}\mathfrak{h}\\\text{is a }k\text{-module)}%
}}+N^{\otimes\left(  p+1\right)  }\\
&  =\underbrace{\mathfrak{g}^{\otimes\leq p}}_{=\mathfrak{g}^{\otimes
\leq\left(  \left(  p+1\right)  -1\right)  }}+J+\underbrace{\mathfrak{g}%
^{\otimes p}}_{=\mathfrak{g}^{\otimes\left(  \left(  p+1\right)  -1\right)  }%
}\mathfrak{h}+N^{\otimes\left(  p+1\right)  }=\mathfrak{g}^{\otimes\leq\left(
\left(  p+1\right)  -1\right)  }+J+\mathfrak{g}^{\otimes\left(  \left(
p+1\right)  -1\right)  }\mathfrak{h}+N^{\otimes\left(  p+1\right)  }.
\end{align*}
In other words, (\ref{pf.l25.sum.3}) holds for $q=p+1$. This completes the
induction step, and thus (\ref{pf.l25.sum.3}) is proven.
\end{verlong}

Now, our next (and, for this proof, our last) claim is:%
\begin{equation}
\mathfrak{g}^{\otimes\leq p}\subseteq J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}+\left(  \otimes N\right)  \ \ \ \ \ \ \ \ \ \ \text{for
every }p\in\mathbb{N}. \label{pf.l25.sum.4}%
\end{equation}


\textit{Proof of (\ref{pf.l25.sum.4}).} We are going to verify
(\ref{pf.l25.sum.4}) by induction over $p$:

\begin{vershort}
\textit{Induction base:} The induction base is clear, since
(\ref{pf.l25.sum.4}) holds for $p=0$ (because $\mathfrak{g}^{\otimes\leq
0}=k\subseteq\otimes N$).
\end{vershort}

\begin{verlong}
\textit{Induction base:} For $p=0$, we have $\mathfrak{g}^{\otimes\leq
p}=\mathfrak{g}^{\otimes\leq0}=\bigoplus\limits_{i=0}^{0}\mathfrak{g}^{\otimes
i}=\mathfrak{g}^{\otimes0}=k\subseteq\otimes N\subseteq J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(  \otimes N\right)  $.
Thus, (\ref{pf.l25.sum.4}) holds for $p=0$. This settles the induction base.
\end{verlong}

\textit{Induction step:} Let $q\in\mathbb{N}_{+}$. Assume that
(\ref{pf.l25.sum.4}) is already proven for $p=q-1$. We now must prove
(\ref{pf.l25.sum.4}) for $p=q$.

Since we assumed that (\ref{pf.l25.sum.4}) is already proven for $p=q-1$, we
know that $\mathfrak{g}^{\otimes\leq\left(  q-1\right)  }\subseteq J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(  \otimes N\right)  $.

\begin{vershort}
Now, (\ref{pf.l25.sum.3}) yields%
\begin{align*}
\mathfrak{g}^{\otimes q}  &  \subseteq\underbrace{\mathfrak{g}^{\otimes
\leq\left(  q-1\right)  }}_{\subseteq J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}+\left(  \otimes N\right)  }+J+\underbrace{\mathfrak{g}%
^{\otimes\left(  q-1\right)  }}_{\subseteq\otimes\mathfrak{g}}\mathfrak{h}%
+\underbrace{N^{\otimes q}}_{\subseteq\otimes N}\subseteq\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(  \otimes N\right)
\right)  +J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(
\otimes N\right) \\
&  =\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(
\otimes N\right)  \right)  +\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}+\left(  \otimes N\right)  \right)  =J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(  \otimes N\right)  .
\end{align*}
Now, the definitions of $\mathfrak{g}^{\otimes\leq q}$ and $\mathfrak{g}%
^{\otimes\leq\left(  q-1\right)  }$ lead to $\mathfrak{g}^{\otimes\leq
q}=\mathfrak{g}^{\otimes\leq\left(  q-1\right)  }+\mathfrak{g}^{\otimes q}$.
Since both $\mathfrak{g}^{\otimes\leq\left(  q-1\right)  }$ and $\mathfrak{g}%
^{\otimes q}$ are subsets of the $k$-module $J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}+\left(  \otimes N\right)  $, this yields
that$\mathfrak{\ g}^{\otimes\leq q}\subseteq J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}+\left(  \otimes N\right)  $. In other words,
(\ref{pf.l25.sum.4}) holds for $p=q$. This completes the induction step, and
thus the induction proof of (\ref{pf.l25.sum.4}) is done.
\end{vershort}

\begin{verlong}
Now, (\ref{pf.l25.sum.3}) yields%
\begin{align*}
\mathfrak{g}^{\otimes q}  &  \subseteq\underbrace{\mathfrak{g}^{\otimes
\leq\left(  q-1\right)  }}_{\subseteq J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}+\left(  \otimes N\right)  }+J+\underbrace{\mathfrak{g}%
^{\otimes\left(  q-1\right)  }}_{\subseteq\otimes\mathfrak{g}}\mathfrak{h}%
+\underbrace{N^{\otimes q}}_{\subseteq\otimes N}\subseteq\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(  \otimes N\right)
\right)  +J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(
\otimes N\right) \\
&  =\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(
\otimes N\right)  \right)  +\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}+\left(  \otimes N\right)  \right)  =J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(  \otimes N\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(
\otimes N\right)  \text{ is a }k\text{-module}\\
\text{(because }J\text{, }\left(  \otimes\mathfrak{g}\right)  \cdot
\mathfrak{h}\text{ and }\otimes N\text{ are }k\text{-modules)}%
\end{array}
\right)  .
\end{align*}
Now, the definition of $\mathfrak{g}^{\otimes\leq q}$ says $\mathfrak{g}%
^{\otimes\leq q}=\bigoplus\limits_{i=0}^{q}\mathfrak{g}^{\otimes i}$, and the
definition of $\mathfrak{g}^{\otimes\leq\left(  q-1\right)  }$ says
$\mathfrak{g}^{\otimes\leq\left(  q-1\right)  }=\bigoplus\limits_{i=0}%
^{q-1}\mathfrak{g}^{\otimes i}$. Therefore,
\begin{align*}
\mathfrak{g}^{\otimes\leq q}  &  =\bigoplus\limits_{i=0}^{q}\mathfrak{g}%
^{\otimes i}=\underbrace{\left(  \bigoplus\limits_{i=0}^{q-1}\mathfrak{g}%
^{\otimes i}\right)  }_{=\mathfrak{g}^{\otimes\leq\left(  q-1\right)  }}%
\oplus\mathfrak{g}^{\otimes q}=\mathfrak{g}^{\otimes\leq\left(  q-1\right)
}\oplus\mathfrak{g}^{\otimes q}=\underbrace{\mathfrak{g}^{\otimes\leq\left(
q-1\right)  }}_{\subseteq J+\left(  \otimes\mathfrak{g}\right)  \cdot
\mathfrak{h}+\left(  \otimes N\right)  }+\underbrace{\mathfrak{g}^{\otimes q}%
}_{\subseteq J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(
\otimes N\right)  }\\
&  \subseteq\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}%
+\left(  \otimes N\right)  \right)  +\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}+\left(  \otimes N\right)  \right)  \subseteq
J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(  \otimes
N\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(
\otimes N\right)  \text{ is a }k\text{-module}\\
\text{(because }J\text{, }\left(  \otimes\mathfrak{g}\right)  \cdot
\mathfrak{h}\text{ and }\otimes N\text{ are }k\text{-modules)}%
\end{array}
\right)  .
\end{align*}
In other words, (\ref{pf.l25.sum.4}) holds for $p=q$. This completes the
induction step, and thus the induction proof of (\ref{pf.l25.sum.4}) is done.
\end{verlong}

\begin{vershort}
Now Lemma \ref{lem.l25.sum} directly follows from (\ref{pf.l25.sum.4}),
because $\otimes\mathfrak{g}=\bigcup\limits_{p\in\mathbb{N}}\mathfrak{g}%
^{\otimes\leq p}$.
\end{vershort}

\begin{verlong}
Now we are ready to make short shrift of Lemma \ref{lem.l25.sum}: We have%
\begin{align*}
\otimes\mathfrak{g}  &  =\bigoplus\limits_{i\in\mathbb{N}}\mathfrak{g}%
^{\otimes i}=\bigoplus\limits_{p\in\mathbb{N}}\mathfrak{g}^{\otimes
p}\ \ \ \ \ \ \ \ \ \ \left(  \text{here we substituted }p\text{ for }i\text{
in the sum}\right) \\
&  =\sum\limits_{p\in\mathbb{N}}\underbrace{\mathfrak{g}^{\otimes p}%
}_{\substack{\subseteq\mathfrak{g}^{\otimes\leq p}\subseteq J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(  \otimes N\right)
\\\text{(according to (\ref{pf.l25.sum.4}))}}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since direct sums are sums}\right) \\
&  \subseteq\sum\limits_{p\in\mathbb{N}}\left(  J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(  \otimes N\right)  \right)
\subseteq J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(
\otimes N\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(
\otimes N\right)  \text{ is a }k\text{-module}\\
\text{(because }J\text{, }\left(  \otimes\mathfrak{g}\right)  \cdot
\mathfrak{h}\text{ and }\otimes N\text{ are }k\text{-modules)}%
\end{array}
\right)  .
\end{align*}
Together with $J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(
\otimes N\right)  \subseteq\otimes\mathfrak{g}$ (which is obvious), this
yields $\otimes\mathfrak{g}=J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}+\left(  \otimes N\right)  $. This proves Lemma
\ref{lem.l25.sum}.
\end{verlong}

\subsection{The factor map $\overline{\varphi}$}

\begin{vershort}
Before we start proving Proposition \ref{prop.l25.Ker-alpha}, let us notice
that $\left(  F_{n}\right)  _{n\geq0}$ is a filtration of $\left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  $. This follows trivially from the
definition of $F_{n}$.
\end{vershort}

\begin{verlong}
Before we start proving Proposition \ref{prop.l25.Ker-alpha}, let us show that
$\left(  F_{n}\right)  _{n\geq0}$ is a filtration of $\left(  \otimes
\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  $ (this is part of the claim of Theorem
\ref{thm.l25} \textbf{(b)}):

Every $n\in\mathbb{N}$ satisfies $\mathfrak{g}^{\otimes\leq n}\subseteq
\mathfrak{g}^{\otimes\leq\left(  n+1\right)  }$ (since $\left(  \mathfrak{g}%
^{\otimes\leq n}\right)  _{n\geq0}$ is a filtration of $\otimes\mathfrak{g}$),
and thus $\zeta\left(  \mathfrak{g}^{\otimes\leq n}\right)  \subseteq
\zeta\left(  \mathfrak{g}^{\otimes\leq\left(  n+1\right)  }\right)  $. Since
$F_{n}=\zeta\left(  \mathfrak{g}^{\otimes\leq n}\right)  $ (by the definition
of $F_{n}$) and $F_{n+1}=\zeta\left(  \mathfrak{g}^{\otimes\leq\left(
n+1\right)  }\right)  $ (by the definition of $F_{n+1}$), we thus have
$F_{n}=\zeta\left(  \mathfrak{g}^{\otimes\leq n}\right)  \subseteq\zeta\left(
\mathfrak{g}^{\otimes\leq\left(  n+1\right)  }\right)  =F_{n+1}$. Since this
holds for all $n\in\mathbb{N}$, we thus have $F_{0}\subseteq F_{1}\subseteq
F_{2}\subseteq...$. On the other hand, $\bigcup\limits_{n\in\mathbb{N}%
}\mathfrak{g}^{\otimes\leq n}=\otimes\mathfrak{g}$ (since $\left(
\mathfrak{g}^{\otimes\leq n}\right)  _{n\geq0}$ is a filtration of
$\otimes\mathfrak{g}$). Now,%
\[
\bigcup\limits_{n\in\mathbb{N}}\underbrace{F_{n}}_{=\zeta\left(
\mathfrak{g}^{\otimes\leq n}\right)  }=\bigcup\limits_{n\in\mathbb{N}}%
\zeta\left(  \mathfrak{g}^{\otimes\leq n}\right)  =\zeta\left(
\underbrace{\bigcup\limits_{n\in\mathbb{N}}\mathfrak{g}^{\otimes\leq n}%
}_{=\otimes\mathfrak{g}}\right)  =\zeta\left(  \otimes\mathfrak{g}\right)
=\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\]
(since $\zeta$ is the canonical projection on $\left(  \otimes\mathfrak{g}%
\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot
\mathfrak{h}\right)  $). Together with $F_{0}\subseteq F_{1}\subseteq
F_{2}\subseteq...$, this yields that $\left(  F_{n}\right)  _{n\geq0}$ is a
filtration of $\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $.
\end{verlong}

We are now able to show the first four parts of Proposition
\ref{prop.l25.Ker-alpha}:

\textit{Proof of Proposition \ref{prop.l25.Ker-alpha}.} \textbf{(a)} For every
$U\in\otimes N$, we have $U=\varphi\left(  U\right)  $ (because we know from
Proposition \ref{prop.l25.alpha.(X)N} that $\varphi\mid_{\otimes
N}=\operatorname*{id}\nolimits_{\otimes N}$, and now $U\in\otimes N$ yields
$\varphi\left(  U\right)  =\underbrace{\left(  \varphi\mid_{\otimes N}\right)
}_{=\operatorname*{id}\nolimits_{\otimes N}}\left(  U\right)
=\operatorname*{id}\nolimits_{\otimes N}\left(  U\right)  =U$). Thus, for
every $U\in\otimes N$, we have $U\in\varphi\left(  \otimes\mathfrak{g}\right)
$. In other words, $\varphi$ is surjective.

Next we must prove that $\operatorname*{Ker}\varphi=J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}$.

First of all, we know that
\begin{align*}
\varphi\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
&  =\underbrace{\varphi\left(  J\right)  }_{=0\text{ (by Proposition
\ref{prop.l25.alpha(J)})}}+\underbrace{\varphi\left(  \left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  }_{=0\text{ (by Proposition
\ref{prop.l25.alpha(xgh)})}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\varphi\text{ is }k\text{-linear}\right) \\
&  =0+0=0.
\end{align*}


Lemma \ref{lem.l25.sum} yields $\otimes\mathfrak{g}=J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}+\left(  \otimes N\right)  $. In other
words, $\otimes\mathfrak{g}=\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  +\left(  \otimes N\right)  $.

We are going to strengthen this to $\otimes\mathfrak{g}=\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \oplus\left(  \otimes
N\right)  $ now:

\begin{vershort}
Any element $U$ of $\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  \cap\left(  \otimes N\right)  $ satisfies
$\varphi\left(  U\right)  =0$ (because $U\in J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}$ leads to $\varphi\left(  U\right)  \in
\varphi\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
=0$) and, at the same time, $\varphi\left(  U\right)  =U$ (since $U\in\otimes
N$ yields $\varphi\left(  U\right)  =\underbrace{\left(  \varphi\mid_{\otimes
N}\right)  }_{=\operatorname*{id}\nolimits_{\otimes N}\text{ (by Proposition
\ref{prop.l25.alpha.(X)N})}}\left(  U\right)  =U$). Thus, any element $U$ of
$\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\cap\left(  \otimes N\right)  $ must satisfy $U=\varphi\left(  U\right)  =0$.
Hence, $\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}%
\right)  \cap\left(  \otimes N\right)  =0$, so that the sum $\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  +\left(  \otimes
N\right)  $ is a direct sum. That is, $\otimes\mathfrak{g}=\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  +\left(  \otimes
N\right)  $ becomes $\otimes\mathfrak{g}=\left(  J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \oplus\left(  \otimes N\right)
$.

Now, the map $\varphi$ is a projection from this direct sum $\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \oplus\left(  \otimes
N\right)  $ onto its addend $\otimes N$ (because it is a $k$-linear map which
acts as $0$ on the addend $J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}$ and acts as the identity on the addend $\otimes N$). Hence,
$\operatorname*{Ker}\varphi=J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}$.
\end{vershort}

\begin{verlong}
Any element $U$ of $\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  \cap\left(  \otimes N\right)  $ satisfies
$\varphi\left(  U\right)  =0$ (because $U\in\left(  J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \cap\left(  \otimes N\right)  $
yields $U\in J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}$ and thus
$\varphi\left(  U\right)  \in\varphi\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  =0$, so that $\varphi\left(  U\right)
=0$). But on the other hand, any element $U$ of $\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \cap\left(  \otimes
N\right)  $ satisfies $\varphi\left(  U\right)  =U$ (since $U\in\left(
J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \cap\left(
\otimes N\right)  $ yields $U\in\otimes N$, so that $\varphi\left(  U\right)
=\underbrace{\left(  \varphi\mid_{\otimes N}\right)  }_{=\operatorname*{id}%
\nolimits_{\otimes N}\text{ (by Proposition \ref{prop.l25.alpha.(X)N})}%
}\left(  U\right)  =\operatorname*{id}\nolimits_{\otimes N}\left(  U\right)
=U$). Thus, any element $U$ of $\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  \cap\left(  \otimes N\right)  $ satisfies
$U=\varphi\left(  U\right)  =0$. Hence, $\left(  J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \cap\left(  \otimes N\right)
=0$. Thus, the sum $\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  +\left(  \otimes N\right)  $ is a direct sum; hence,
$\otimes\mathfrak{g}=\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  +\left(  \otimes N\right)  $ becomes $\otimes
\mathfrak{g}=\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot
\mathfrak{h}\right)  \oplus\left(  \otimes N\right)  $. Thus, there exists a
$k$-linear projection $\varphi^{\prime}$ of $\otimes\mathfrak{g}$ on $\otimes
N$ with kernel $J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}$. We
are now going to show that $\varphi=\varphi^{\prime}$.

In fact,%
\[
\left(  \varphi-\varphi^{\prime}\right)  \left(  J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  =\underbrace{\varphi\left(
J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  }%
_{=0}-\underbrace{\varphi^{\prime}\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  }_{\substack{=0\text{ (since }%
\varphi^{\prime}\text{ is a projection}\\\text{with kernel }J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\text{)}}}=0-0=0
\]
and thus $J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}%
\in\operatorname*{Ker}\left(  \varphi-\varphi^{\prime}\right)  $. On the other
hand, every $U\in\otimes N$ satisfies%
\begin{align*}
\varphi\left(  U\right)   &  =\underbrace{\left(  \varphi\mid_{\otimes
N}\right)  }_{=\operatorname*{id}\nolimits_{\otimes N}\text{ (by Proposition
\ref{prop.l25.alpha.(X)N})}}\left(  U\right)  =\operatorname*{id}%
\nolimits_{\otimes N}\left(  U\right)  =U=\varphi^{\prime}\left(  U\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varphi^{\prime}\text{ is a
projection on }\otimes N\text{, whereas }U\text{ is an element of }\otimes
N\right)
\end{align*}
and thus $\left(  \varphi-\varphi^{\prime}\right)  \left(  U\right)
=\underbrace{\varphi\left(  U\right)  }_{=\varphi^{\prime}\left(  U\right)
}-\varphi^{\prime}\left(  U\right)  =0$, so that $U\in\operatorname*{Ker}%
\left(  \varphi-\varphi^{\prime}\right)  $. So we have shown that every
$U\in\otimes N$ satisfies $U\in\operatorname*{Ker}\left(  \varphi
-\varphi^{\prime}\right)  $. Thus, $\otimes N\subseteq\operatorname*{Ker}%
\left(  \varphi-\varphi^{\prime}\right)  $. Now,%
\begin{align*}
\otimes\mathfrak{g}  &  =\underbrace{J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}}_{\subseteq\operatorname*{Ker}\left(  \varphi-\varphi
^{\prime}\right)  }+\underbrace{\left(  \otimes N\right)  }_{\subseteq
\operatorname*{Ker}\left(  \varphi-\varphi^{\prime}\right)  }\subseteq
\operatorname*{Ker}\left(  \varphi-\varphi^{\prime}\right)
+\operatorname*{Ker}\left(  \varphi-\varphi^{\prime}\right) \\
&  =\operatorname*{Ker}\left(  \varphi-\varphi^{\prime}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname*{Ker}\left(
\varphi-\varphi^{\prime}\right)  \text{ is a }k\text{-module (since }%
\varphi-\varphi^{\prime}\text{ is }k\text{-linear)}\right)  ,
\end{align*}
so that $\varphi-\varphi^{\prime}=0$. Thus, $\varphi=\varphi^{\prime}$. Thus,
$\operatorname*{Ker}\varphi^{\prime}=J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}$ (since $\varphi^{\prime}$ is a projection with kernel
$J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}$) rewrites as
$\operatorname*{Ker}\varphi=J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}$.
\end{verlong}

By the isomorphism theorem, the $k$-module homomorphism $\varphi
:\otimes\mathfrak{g}\rightarrow\otimes N$ induces a $k$-module isomorphism
$\overline{\varphi}:\left(  \otimes\mathfrak{g}\right)  \diagup
\operatorname*{Ker}\varphi\rightarrow\varphi\left(  \otimes\mathfrak{g}%
\right)  $ which satisfies $\varphi=\overline{\varphi}\circ\zeta^{\prime}$,
where $\zeta^{\prime}$ is the canonical projection of $\otimes\mathfrak{g}$
onto $\left(  \otimes\mathfrak{g}\right)  \diagup\operatorname*{Ker}\varphi$.
Since $\operatorname*{Ker}\varphi=J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}$ and $\varphi\left(  \otimes\mathfrak{g}\right)  =\otimes N$
(because $\varphi$ is surjective), this is therefore a $k$-module isomorphism
$\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \rightarrow\otimes N$.
Moreover, $\zeta^{\prime}$ is the canonical projection of $\otimes
\mathfrak{g}$ onto $\left(  \otimes\mathfrak{g}\right)  \diagup
\operatorname*{Ker}\varphi$, and therefore equal to the canonical projection
of $\otimes\mathfrak{g}$ onto $\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
$ (since $\operatorname*{Ker}\varphi=J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}$). Since the canonical projection of $\otimes\mathfrak{g}$
onto $\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ is $\zeta$, this
yields that $\zeta^{\prime}$ is equal to $\zeta$. Therefore, $\varphi
=\overline{\varphi}\circ\zeta^{\prime}$ rewrites as $\varphi=\overline
{\varphi}\circ\zeta$. This completes the proof of Proposition
\ref{prop.l25.Ker-alpha} \textbf{(a)}.

\textbf{(b)} Let $n\in\mathbb{N}$ be arbitrary. Since $F_{n}=\zeta\left(
\mathfrak{g}^{\otimes\leq n}\right)  $, we have%
\begin{align*}
\overline{\varphi}\left(  F_{n}\right)   &  =\overline{\varphi}\left(
\zeta\left(  \mathfrak{g}^{\otimes\leq n}\right)  \right)
=\underbrace{\left(  \overline{\varphi}\circ\zeta\right)  }_{=\varphi}\left(
\mathfrak{g}^{\otimes\leq n}\right)  =\varphi\left(  \mathfrak{g}^{\otimes\leq
n}\right)  \subseteq N^{\otimes\leq n}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{according to Proposition
\ref{prop.l25.alpha.filt}}\right)  .
\end{align*}
In other words, the isomorphism $\overline{\varphi}$ respects the filtration.
This proves Proposition \ref{prop.l25.Ker-alpha} \textbf{(b)}.

\textbf{(c)} The homomorphism $\eta$ respects the filtration (because
$\eta=\otimes\left(  \pi\mid_{N}\right)  $, and because Proposition
\ref{prop.tensor.filt.(X)f} (applied to $\pi\mid_{N}$, $N$ and $\mathfrak{n}$
instead of $f$, $V$ and $W$) yields that $\otimes\left(  \pi\mid_{N}\right)  $
respects the filtration). Everything else claimed in Proposition
\ref{prop.l25.Ker-alpha} \textbf{(c)} is trivial. Thus, Proposition
\ref{prop.l25.Ker-alpha} \textbf{(c)} is proven.

\textbf{(d)} We have defined $\iota:\otimes N\rightarrow\otimes\mathfrak{g}$
as the canonical inclusion of $\otimes N$ in $\otimes\mathfrak{g}$. Therefore,
$\iota$ clearly respects the filtration. Also, we know that $\zeta$ respects
the filtration, because the filtration on $\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
$ was defined to be $\left(  \underbrace{F_{n}}_{=\zeta\left(  \mathfrak{g}%
^{\otimes\leq n}\right)  }\right)  _{n\geq0}=\left(  \zeta\left(
\mathfrak{g}^{\otimes\leq n}\right)  \right)  _{n\geq0}$.

Since $\zeta$ and $\iota$ respect the filtration, the composition $\zeta
\circ\iota$ respects the filtration as well (by Proposition
\ref{prop.gr.functor} \textbf{(b)}).

Besides, $\overline{\varphi}\circ\left(  \zeta\circ\iota\right)
=\underbrace{\overline{\varphi}\circ\zeta}_{=\varphi}\circ\iota=\varphi
\circ\iota=\operatorname*{id}\nolimits_{\otimes N}$ (because every
$U\in\otimes N$ satisfies%
\begin{align*}
\left(  \varphi\circ\iota\right)  \left(  U\right)   &  =\varphi\left(
\underbrace{\iota\left(  U\right)  }_{\substack{=U\text{ (since }\iota\text{
is the}\\\text{ inclusion map)}}}\right)  =\varphi\left(  U\right)
=\underbrace{\left(  \varphi\mid_{\otimes N}\right)  }%
_{\substack{=\operatorname*{id}\nolimits_{\otimes N}\\\text{(due to
Proposition \ref{prop.l25.alpha.(X)N})}}}\left(  U\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }U\in\otimes N\right) \\
&  =\operatorname*{id}\nolimits_{\otimes N}\left(  U\right)
\end{align*}
). Hence, $\zeta\circ\iota=\overline{\varphi}^{-1}$ (here we know that
$\overline{\varphi}^{-1}$ exists, because $\overline{\varphi}$ is a $k$-module
isomorphism). Thus, $\overline{\varphi}^{-1}$ respects the filtration (since
$\zeta\circ\iota$ respects the filtration). This proves Proposition
\ref{prop.l25.Ker-alpha} \textbf{(d)}.

\subsection{$\varphi$ and $\overline{\varphi}$ on the associated graded
objects}

Before we can verify Proposition \ref{prop.l25.Ker-alpha} \textbf{(e)}, we are
going to show the next result, which will also help us in explicitly
describing the associated graded morphism of $\overline{\varphi}$ later:

\begin{proposition}
\label{prop.l25.alpha.mod}Consider the $k$-module homomorphism $\otimes
t:\otimes\mathfrak{g}\rightarrow\otimes N$ induced by the $k$-module
homomorphism $t:\mathfrak{g}\rightarrow N$. This homomorphism $\otimes t$
respects the filtration and satisfies%
\begin{equation}
\left(  \varphi-\left(  \otimes t\right)  \right)  \left(  \mathfrak{g}%
^{\otimes\leq n}\right)  \subseteq N^{\otimes\leq\left(  n-1\right)
}\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}\text{.}
\label{prop.l25.alpha.mod.in}%
\end{equation}

\end{proposition}

\textit{Proof of Proposition \ref{prop.l25.alpha.mod}.} Proposition
\ref{prop.tensor.filt.(X)f} (applied to $t$, $\mathfrak{g}$ and $N$ instead of
$f$, $V$ and $W$) yields that $\otimes t$ respects the filtration. Thus, the
only thing that remains to be done for the proof of Proposition
\ref{prop.l25.alpha.mod} is proving the relation (\ref{prop.l25.alpha.mod.in}).

We are going to prove (\ref{prop.l25.alpha.mod.in}) by induction over $n$:

\begin{vershort}
\textit{Induction base:} Let us prove that (\ref{prop.l25.alpha.mod.in}) holds
for $n=0$. This means showing that $\left(  \varphi-\left(  \otimes t\right)
\right)  \left(  k\right)  \subseteq0$ (because $\mathfrak{g}^{\otimes\leq
0}=k$ and $N^{\otimes\leq\left(  -1\right)  }=0$). This, however, is clear,
because every $\lambda\in k$ satisfies $\varphi\left(  \lambda\right)
=\lambda$ (according to (\ref{rmk.alpha-0})) and $\left(  \otimes t\right)
\left(  \lambda\right)  =\lambda$ (by the definition of $\otimes t$). Thus,
the induction base belongs to us.
\end{vershort}

\begin{verlong}
\textit{Induction base:} Every $\lambda\in\mathfrak{g}^{\otimes\leq0}$
satisfies $\lambda\in k$ (since $\mathfrak{g}^{\otimes\leq0}=\bigoplus
\limits_{i=0}^{0}\mathfrak{g}^{\otimes i}=\mathfrak{g}^{\otimes0}=k$).
Therefore, every $\lambda\in\mathfrak{g}^{\otimes\leq0}$ satisfies
$\varphi\left(  \lambda\right)  =\lambda$ (according to (\ref{rmk.alpha-0}),
since $\lambda\in k$) and $\left(  \otimes t\right)  \left(  \lambda\right)
=\lambda$ (by the definition of $\otimes t$, since $\lambda\in k$), so that
$\left(  \varphi-\left(  \otimes t\right)  \right)  \left(  \lambda\right)
=\underbrace{\varphi\left(  \lambda\right)  }_{=\lambda}-\underbrace{\left(
\otimes t\right)  \left(  \lambda\right)  }_{=\lambda}=0$ and thus $\lambda
\in\operatorname*{Ker}\left(  \varphi-\left(  \otimes t\right)  \right)  $. We
have thus shown that every $\lambda\in\mathfrak{g}^{\otimes\leq0}$ satisfies
$\lambda\in\operatorname*{Ker}\left(  \varphi-\left(  \otimes t\right)
\right)  $. Hence, $\mathfrak{g}^{\otimes\leq0}\subseteq\operatorname*{Ker}%
\left(  \varphi-\left(  \otimes t\right)  \right)  $, so that $\left(
\varphi-\left(  \otimes t\right)  \right)  \left(  \mathfrak{g}^{\otimes\leq
0}\right)  =0\subseteq N^{\otimes\leq\left(  0-1\right)  }$. In other words,
(\ref{prop.l25.alpha.mod.in}) holds for $n=0$. This completes the induction base.
\end{verlong}

\textit{Induction step:} Let $p\in\mathbb{N}$. Assume that
(\ref{prop.l25.alpha.mod.in}) holds for $n=p$. We now must show that
(\ref{prop.l25.alpha.mod.in}) also holds for $n=p+1$.

Since (\ref{prop.l25.alpha.mod.in}) holds for $n=p$, we have $\left(
\varphi-\left(  \otimes t\right)  \right)  \left(  \mathfrak{g}^{\otimes\leq
p}\right)  \subseteq N^{\otimes\leq\left(  p-1\right)  }$.

We are now going to show that $\left(  \varphi-\left(  \otimes t\right)
\right)  \left(  \mathfrak{g}^{\otimes\left(  p+1\right)  }\right)  \subseteq
N^{\otimes\leq p}$. This will quickly yield $\left(  \varphi-\left(  \otimes
t\right)  \right)  \left(  \mathfrak{g}^{\otimes\leq\left(  p+1\right)
}\right)  \subseteq N^{\otimes\leq p}$, which will bring us to the end of the
induction step.

\begin{vershort}
Proposition \ref{prop.left-ind2} yields $\mathfrak{g}^{\otimes\left(
p+1\right)  }=\left\langle \mathfrak{g}_{\operatorname*{lind}}^{\otimes\left(
p+1\right)  }\right\rangle $.
\end{vershort}

\begin{verlong}
Proposition \ref{prop.left-ind2} (applied to $p+1$ and $\mathfrak{g}$ instead
of $p$ and $V$) yields $\mathfrak{g}^{\otimes\left(  p+1\right)
}=\left\langle \mathfrak{g}_{\operatorname*{lind}}^{\otimes\left(  p+1\right)
}\right\rangle $.
\end{verlong}

Note that $\varphi-\left(  \otimes t\right)  $ is a $k$-linear map (since
$\varphi$ and $\otimes t$ are $k$-linear), and thus $\operatorname*{Ker}%
\left(  \varphi-\left(  \otimes t\right)  \right)  $ is a $k$-module.

\begin{vershort}
Now we are going to prove that $\mathfrak{g}_{\operatorname*{lind}}%
^{\otimes\left(  p+1\right)  }\subseteq\left(  \varphi-\left(  \otimes
t\right)  \right)  ^{-1}\left(  N^{\otimes\leq p}\right)  $. Indeed, let
$V\in\mathfrak{g}_{\operatorname*{lind}}^{\otimes\left(  p+1\right)  }$ be
arbitrary. Then, $V$ is a left-induced tensor in $\mathfrak{g}^{\otimes\left(
p+1\right)  }$ (since $V\in\mathfrak{g}_{\operatorname*{lind}}^{\otimes\left(
p+1\right)  }$), and thus there exist $u\in\mathfrak{g}$ and $U\in
\mathfrak{g}^{\otimes p}$ such that $V=u\otimes U$. Consider these $u$ and
$U$. They satisfy $u\cdot U=u\otimes U$ (due to (\ref{*=(X)})).
\end{vershort}

\begin{verlong}
Now we are going to prove that $\mathfrak{g}_{\operatorname*{lind}}%
^{\otimes\left(  p+1\right)  }\subseteq\left(  \varphi-\left(  \otimes
t\right)  \right)  ^{-1}\left(  N^{\otimes\leq p}\right)  $. Indeed, let
$V\in\mathfrak{g}_{\operatorname*{lind}}^{\otimes\left(  p+1\right)  }$ be
arbitrary. Then, $V$ is a left-induced tensor in $\mathfrak{g}^{\otimes\left(
p+1\right)  }$ (since $V\in\mathfrak{g}_{\operatorname*{lind}}^{\otimes\left(
p+1\right)  }$), and thus there exist $u\in\mathfrak{g}$ and $U\in
\mathfrak{g}^{\otimes p}$ such that $V=u\otimes U$. Consider these $u$ and
$U$. Since $u\in\mathfrak{g}=\mathfrak{g}^{\otimes1}$ and $U\in\mathfrak{g}%
^{\otimes p}$, we have $u\cdot U=u\otimes U$ (due to (\ref{*=(X)}), applied to
$u$, $U$, $1$ and $p$ instead of $a$, $b$, $n$ and $m$).
\end{verlong}

Since $U\in\mathfrak{g}^{\otimes p}$ and since $\mathfrak{g}^{\otimes p}$ is a
$\mathfrak{g}$-module, we have $s\left(  u\right)  \rightharpoonup
U\in\mathfrak{g}^{\otimes p}\subseteq\mathfrak{g}^{\otimes\leq p}$. Thus
$\varphi\left(  s\left(  u\right)  \rightharpoonup U\right)  \in\varphi\left(
\mathfrak{g}^{\otimes\leq p}\right)  \subseteq N^{\otimes\leq p}$ (according
to Proposition \ref{prop.l25.alpha.filt}), so that $\varphi\left(  s\left(
u\right)  \rightharpoonup U\right)  \equiv0\operatorname{mod}N^{\otimes\leq
p}$.

\begin{vershort}
On the other hand, $U\in\mathfrak{g}^{\otimes p}\subseteq\mathfrak{g}%
^{\otimes\leq p}$ yields $\left(  \varphi-\left(  \otimes t\right)  \right)
\left(  U\right)  \in\left(  \varphi-\left(  \otimes t\right)  \right)
\left(  \mathfrak{g}^{\otimes\leq p}\right)  \subseteq N^{\otimes\leq\left(
p-1\right)  }$. Since $\left(  \varphi-\left(  \otimes t\right)  \right)
\left(  U\right)  =\varphi\left(  U\right)  -\left(  \otimes t\right)  \left(
U\right)  $, this rewrites as $\varphi\left(  U\right)  -\left(  \otimes
t\right)  \left(  U\right)  \in N^{\otimes\leq\left(  p-1\right)  }$. On the
other hand, $\otimes t$ is a $k$-algebra homomorphism, so that%
\[
\left(  \otimes t\right)  \left(  u\cdot U\right)  =\underbrace{\left(
\otimes t\right)  \left(  u\right)  }_{\substack{=t\left(  u\right)  \text{
(by the definition}\\\text{of }\otimes t\text{, since }u\in\mathfrak{g}%
=\mathfrak{g}^{\otimes1}\text{)}}}\cdot\left(  \otimes t\right)  \left(
U\right)  =t\left(  u\right)  \cdot\left(  \otimes t\right)  \left(  U\right)
.
\]
Now,%
\begin{align*}
&  t\left(  u\right)  \cdot\varphi\left(  U\right)  -\left(  \otimes t\right)
\left(  \underbrace{V}_{=u\otimes U=u\cdot U}\right) \\
&  =t\left(  u\right)  \cdot\varphi\left(  U\right)  -\underbrace{\left(
\otimes t\right)  \left(  u\cdot U\right)  }_{=t\left(  u\right)  \cdot\left(
\otimes t\right)  \left(  U\right)  }=t\left(  u\right)  \cdot\varphi\left(
U\right)  -t\left(  u\right)  \cdot\left(  \otimes t\right)  \left(  U\right)
\\
&  =\underbrace{t\left(  u\right)  }_{\in N=N^{\otimes1}\subseteq
N^{\otimes\leq1}}\cdot\underbrace{\left(  \varphi\left(  U\right)  -\left(
\otimes t\right)  \left(  U\right)  \right)  }_{\in N^{\otimes\leq\left(
p-1\right)  }}\in N^{\otimes\leq1}\cdot N^{\otimes\leq\left(  p-1\right)  }\\
&  \subseteq N^{\otimes\leq p}\ \ \ \ \ \ \ \ \ \ \left(  \text{according to
Proposition \ref{prop.tensor.filt-is-alg-filt} \textbf{(b)}}\right)  .
\end{align*}
In other words, $t\left(  u\right)  \cdot\varphi\left(  U\right)
\equiv\left(  \otimes t\right)  \left(  V\right)  \operatorname{mod}%
N^{\otimes\leq p}$.
\end{vershort}

\begin{verlong}
On the other hand, $U\in\mathfrak{g}^{\otimes p}\subseteq\mathfrak{g}%
^{\otimes\leq p}$ yields $\left(  \varphi-\left(  \otimes t\right)  \right)
\left(  U\right)  \in\left(  \varphi-\left(  \otimes t\right)  \right)
\left(  \mathfrak{g}^{\otimes\leq p}\right)  \subseteq N^{\otimes\leq\left(
p-1\right)  }$. Since $\left(  \varphi-\left(  \otimes t\right)  \right)
\left(  U\right)  =\varphi\left(  U\right)  -\left(  \otimes t\right)  \left(
U\right)  $, this rewrites as $\varphi\left(  U\right)  -\left(  \otimes
t\right)  \left(  U\right)  \in N^{\otimes\leq\left(  p-1\right)  }$. On the
other hand, $\otimes t$ is a $k$-algebra homomorphism, so that%
\[
\left(  \otimes t\right)  \left(  u\cdot U\right)  =\underbrace{\left(
\otimes t\right)  \left(  u\right)  }_{\substack{=t\left(  u\right)  \text{
(by the definition}\\\text{of }\otimes t\text{, since }u\in\mathfrak{g}%
=\mathfrak{g}^{\otimes1}\text{)}}}\cdot\left(  \otimes t\right)  \left(
U\right)  =t\left(  u\right)  \cdot\left(  \otimes t\right)  \left(  U\right)
.
\]
Now,%
\begin{align*}
&  t\left(  u\right)  \cdot\varphi\left(  U\right)  -\left(  \otimes t\right)
\left(  \underbrace{V}_{=u\otimes U=u\cdot U}\right) \\
&  =t\left(  u\right)  \cdot\varphi\left(  U\right)  -\underbrace{\left(
\otimes t\right)  \left(  u\cdot U\right)  }_{=t\left(  u\right)  \cdot\left(
\otimes t\right)  \left(  U\right)  }=t\left(  u\right)  \cdot\varphi\left(
U\right)  -t\left(  u\right)  \cdot\left(  \otimes t\right)  \left(  U\right)
\\
&  =\underbrace{t\left(  u\right)  }_{\in N=N^{\otimes1}\subseteq
N^{\otimes\leq1}}\cdot\underbrace{\left(  \varphi\left(  U\right)  -\left(
\otimes t\right)  \left(  U\right)  \right)  }_{\in N^{\otimes\leq\left(
p-1\right)  }}\in N^{\otimes\leq1}\cdot N^{\otimes\leq\left(  p-1\right)  }\\
&  \subseteq N^{\otimes\leq\left(  1+\left(  p-1\right)  \right)
}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{according to Proposition \ref{prop.tensor.filt-is-alg-filt}
\textbf{(b)}, applied}\\
\text{to }N\text{, }1\text{ and }p-1\text{ instead of }V\text{, }n\text{ and
}m
\end{array}
\right) \\
&  =N^{\otimes\leq p}.
\end{align*}
In other words, $t\left(  u\right)  \cdot\varphi\left(  U\right)
\equiv\left(  \otimes t\right)  \left(  V\right)  \operatorname{mod}%
N^{\otimes\leq p}$.
\end{verlong}

Now, $V=u\otimes U=u\cdot U$ yields%
\begin{align*}
\varphi\left(  V\right)   &  =\varphi\left(  u\cdot U\right)
=\underbrace{t\left(  u\right)  \cdot\varphi\left(  U\right)  }_{\equiv\left(
\otimes t\right)  \left(  V\right)  \operatorname{mod}N^{\otimes\leq p}%
}+\underbrace{\varphi\left(  s\left(  u\right)  \rightharpoonup U\right)
}_{\equiv0\operatorname{mod}N^{\otimes\leq p}}\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{prop.alpha.rec})}\right) \\
&  \equiv\left(  \otimes t\right)  \left(  V\right)  +0=\left(  \otimes
t\right)  \left(  V\right)  \operatorname{mod}N^{\otimes\leq p}.
\end{align*}
In other words, $\varphi\left(  V\right)  -\left(  \otimes t\right)  \left(
V\right)  \in N^{\otimes\leq p}$. Since $\varphi\left(  V\right)  -\left(
\otimes t\right)  \left(  V\right)  =\left(  \varphi-\left(  \otimes t\right)
\right)  \left(  V\right)  $, this rewrites as $\left(  \varphi-\left(
\otimes t\right)  \right)  \left(  V\right)  \in N^{\otimes\leq p}$. In other
words, $V\in\left(  \varphi-\left(  \otimes t\right)  \right)  ^{-1}\left(
N^{\otimes\leq p}\right)  $.

\begin{vershort}
So we have proven that every $V\in\mathfrak{g}_{\operatorname*{lind}}%
^{\otimes\left(  p+1\right)  }$ satisfies $V\in\left(  \varphi-\left(  \otimes
t\right)  \right)  ^{-1}\left(  N^{\otimes\leq p}\right)  $. In other words,
$\mathfrak{g}_{\operatorname*{lind}}^{\otimes\left(  p+1\right)  }%
\subseteq\left(  \varphi-\left(  \otimes t\right)  \right)  ^{-1}\left(
N^{\otimes\leq p}\right)  $. Therefore, $\left\langle \mathfrak{g}%
_{\operatorname*{lind}}^{\otimes\left(  p+1\right)  }\right\rangle
\subseteq\left(  \varphi-\left(  \otimes t\right)  \right)  ^{-1}\left(
N^{\otimes\leq p}\right)  $. Since $\left\langle \mathfrak{g}%
_{\operatorname*{lind}}^{\otimes\left(  p+1\right)  }\right\rangle
=\mathfrak{g}^{\otimes\left(  p+1\right)  }$, this becomes $\mathfrak{g}%
^{\otimes\left(  p+1\right)  }\subseteq\left(  \varphi-\left(  \otimes
t\right)  \right)  ^{-1}\left(  N^{\otimes\leq p}\right)  $, so that $\left(
\varphi-\left(  \otimes t\right)  \right)  \left(  \mathfrak{g}^{\otimes
\left(  p+1\right)  }\right)  \subseteq N^{\otimes\leq p}$.
\end{vershort}

\begin{verlong}
So we have proven that every $V\in\mathfrak{g}_{\operatorname*{lind}}%
^{\otimes\left(  p+1\right)  }$ satisfies $V\in\left(  \varphi-\left(  \otimes
t\right)  \right)  ^{-1}\left(  N^{\otimes\leq p}\right)  $. In other words,
$\mathfrak{g}_{\operatorname*{lind}}^{\otimes\left(  p+1\right)  }%
\subseteq\left(  \varphi-\left(  \otimes t\right)  \right)  ^{-1}\left(
N^{\otimes\leq p}\right)  $. Therefore, Proposition \ref{prop.<>} \textbf{(a)}
(applied to $\mathfrak{g}^{\otimes\left(  p+1\right)  }$, $\mathfrak{g}%
_{\operatorname*{lind}}^{\otimes\left(  p+1\right)  }$ and $\left(
\varphi-\left(  \otimes t\right)  \right)  ^{-1}\left(  N^{\otimes\leq
p}\right)  $ instead of $M$, $S$ and $Q$) yields $\left\langle \mathfrak{g}%
_{\operatorname*{lind}}^{\otimes\left(  p+1\right)  }\right\rangle
\subseteq\left(  \varphi-\left(  \otimes t\right)  \right)  ^{-1}\left(
N^{\otimes\leq p}\right)  $ (since $\left(  \varphi-\left(  \otimes t\right)
\right)  ^{-1}\left(  N^{\otimes\leq p}\right)  $ is a $k$-module). Altogether
we now have $\mathfrak{g}^{\otimes\left(  p+1\right)  }=\left\langle
\mathfrak{g}_{\operatorname*{lind}}^{\otimes\left(  p+1\right)  }\right\rangle
\subseteq\left(  \varphi-\left(  \otimes t\right)  \right)  ^{-1}\left(
N^{\otimes\leq p}\right)  $, so that $\left(  \varphi-\left(  \otimes
t\right)  \right)  \left(  \mathfrak{g}^{\otimes\left(  p+1\right)  }\right)
\subseteq N^{\otimes\leq p}$.
\end{verlong}

\begin{vershort}
We are not done yet, because our goal is to show that $\left(  \varphi-\left(
\otimes t\right)  \right)  \left(  \mathfrak{g}^{\otimes\leq\left(
p+1\right)  }\right)  \subseteq N^{\otimes\leq p}$, but we are almost there.
The definitions of $\mathfrak{g}^{\otimes\leq\left(  p+1\right)  }$ and
$\mathfrak{g}^{\otimes\leq p}$ yield $\mathfrak{g}^{\otimes\leq\left(
p+1\right)  }=\mathfrak{g}^{\otimes\left(  p+1\right)  }+\mathfrak{g}%
^{\otimes\leq p}$, so that%
\begin{align*}
\left(  \varphi-\left(  \otimes t\right)  \right)  \left(  \mathfrak{g}%
^{\otimes\leq\left(  p+1\right)  }\right)   &  =\left(  \varphi-\left(
\otimes t\right)  \right)  \left(  \mathfrak{g}^{\otimes\left(  p+1\right)
}+\mathfrak{g}^{\otimes\leq p}\right) \\
&  =\underbrace{\left(  \varphi-\left(  \otimes t\right)  \right)  \left(
\mathfrak{g}^{\otimes\leq p}\right)  }_{\substack{\subseteq N^{\otimes
\leq\left(  p-1\right)  }\subseteq N^{\otimes\leq p}\\\text{(since }\left(
N^{\otimes\leq n}\right)  _{n\geq0}\text{ is a filtration)}}%
}+\underbrace{\left(  \varphi-\left(  \otimes t\right)  \right)  \left(
\mathfrak{g}^{\otimes\left(  p+1\right)  }\right)  }_{\subseteq N^{\otimes\leq
p}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varphi-\left(  \otimes t\right)
\text{ is }k\text{-linear}\right) \\
&  \subseteq N^{\otimes\leq p}+N^{\otimes\leq p}=N^{\otimes\leq p}.
\end{align*}
In other words, (\ref{prop.l25.alpha.mod.in}) holds for $n=p+1$. This
completes the induction step. Thus, (\ref{prop.l25.alpha.mod.in}) is proven
for all $n\in\mathbb{N}$. In other words, Proposition \ref{prop.l25.alpha.mod}
is proven.
\end{vershort}

\begin{verlong}
Now, the definition of $\mathfrak{g}^{\otimes\leq\left(  p+1\right)  }$ is
$\mathfrak{g}^{\otimes\leq\left(  p+1\right)  }=\bigoplus\limits_{i=0}%
^{p+1}\mathfrak{g}^{\otimes i}$, while the definition of $\mathfrak{g}%
^{\otimes\leq p}=\bigoplus\limits_{i=0}^{p}\mathfrak{g}^{\otimes i}$. Thus,%
\[
\mathfrak{g}^{\otimes\leq\left(  p+1\right)  }=\bigoplus\limits_{i=0}%
^{p+1}\mathfrak{g}^{\otimes i}=\underbrace{\left(  \bigoplus\limits_{i=0}%
^{p}\mathfrak{g}^{\otimes i}\right)  }_{=\mathfrak{g}^{\otimes\leq p}}%
\oplus\mathfrak{g}^{\otimes\left(  p+1\right)  }=\mathfrak{g}^{\otimes\leq
p}\oplus\mathfrak{g}^{\otimes\left(  p+1\right)  }=\mathfrak{g}^{\otimes\leq
p}+\mathfrak{g}^{\otimes\left(  p+1\right)  }%
\]
(since direct sums are sums). Thus,%
\begin{align*}
\left(  \varphi-\left(  \otimes t\right)  \right)  \left(  \mathfrak{g}%
^{\otimes\leq\left(  p+1\right)  }\right)   &  =\left(  \varphi-\left(
\otimes t\right)  \right)  \left(  \mathfrak{g}^{\otimes\leq p}+\mathfrak{g}%
^{\otimes\left(  p+1\right)  }\right) \\
&  =\underbrace{\left(  \varphi-\left(  \otimes t\right)  \right)  \left(
\mathfrak{g}^{\otimes\leq p}\right)  }_{\substack{\subseteq N^{\otimes
\leq\left(  p-1\right)  }\subseteq N^{\otimes\leq p}\\\text{(since }\left(
N^{\otimes\leq n}\right)  _{n\geq0}\text{ is a filtration)}}%
}+\underbrace{\left(  \varphi-\left(  \otimes t\right)  \right)  \left(
\mathfrak{g}^{\otimes\left(  p+1\right)  }\right)  }_{\subseteq N^{\otimes\leq
p}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varphi-\left(  \otimes t\right)
\text{ is }k\text{-linear}\right) \\
&  \subseteq N^{\otimes\leq p}+N^{\otimes\leq p}=N^{\otimes\leq p}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }N^{\otimes\leq p}\text{ is a
}k\text{-module}\right) \\
&  =N^{\otimes\leq\left(  \left(  p+1\right)  -1\right)  }.
\end{align*}
In other words, (\ref{prop.l25.alpha.mod.in}) holds for $n=p+1$. This
completes the induction step. Thus, (\ref{prop.l25.alpha.mod.in}) is proven
for all $n\in\mathbb{N}$. In other words, Proposition \ref{prop.l25.alpha.mod}
is proven.
\end{verlong}

Now let us finally prove Proposition \ref{prop.l25.Ker-alpha} \textbf{(e)}:

Both maps $\overline{\varphi}$ and $\eta$ respect the filtration. Hence, their
composition $\eta\circ\overline{\varphi}:\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\rightarrow\otimes\mathfrak{n}$ respects the filtration as well (by
Proposition \ref{prop.gr.functor} \textbf{(b)}). In order to prove Proposition
\ref{prop.l25.Ker-alpha} \textbf{(e)}, it thus only remains to show that the
homomorphism $\operatorname*{gr}\nolimits_{p}\left(  \eta\circ\overline
{\varphi}\right)  :\operatorname*{gr}\nolimits_{p}\left(  \left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  \right)  \rightarrow\operatorname*{gr}%
\nolimits_{p}\left(  \otimes\mathfrak{n}\right)  $ is an $\mathfrak{h}$-module
isomorphism for every $p\in\mathbb{N}$. In order to do this, we must show two
things: we must show that it is a $k$-module isomorphism, and that it is an
$\mathfrak{h}$-module homomorphism.

First, we notice $\eta\circ\overline{\varphi}$ is a $k$-module isomorphism,
since both $\eta$ and $\overline{\varphi}$ are $k$-module isomorphisms. But
this alone is not enough to conclude that $\operatorname*{gr}\nolimits_{p}%
\left(  \eta\circ\overline{\varphi}\right)  $ is a$\mathfrak{\ }k$-module
isomorphism (see Warning \ref{warn.gr.iso}). However, we can get around this
as follows:

We also know that $\eta^{-1}=\left(  \otimes\left(  \pi\mid_{N}\right)
\right)  ^{-1}=\otimes\left(  \left(  \pi\mid_{N}\right)  ^{-1}\right)  $, so
that $\eta^{-1}$ respects the filtration (by Proposition
\ref{prop.tensor.filt.(X)f}). Since $\overline{\varphi}^{-1}$ and $\eta^{-1}$
respect the filtration, the composition $\overline{\varphi}^{-1}\circ\eta
^{-1}$ respects the filtration as well (by Proposition \ref{prop.gr.functor}
\textbf{(b)}). Since $\overline{\varphi}^{-1}\circ\eta^{-1}=\left(  \eta
\circ\overline{\varphi}\right)  ^{-1}$, this means that $\left(  \eta
\circ\overline{\varphi}\right)  ^{-1}$ respects the filtration.

\begin{vershort}
Now, Proposition \ref{prop.gr.iso} yields that $\operatorname*{gr}%
\nolimits_{p}\left(  \eta\circ\overline{\varphi}\right)  :\operatorname*{gr}%
\nolimits_{p}\left(  \left(  \otimes\mathfrak{g}\right)  \diagup\left(
J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \right)
\rightarrow\operatorname*{gr}\nolimits_{p}\left(  \otimes\mathfrak{n}\right)
$ is a $k$-module isomorphism for every $p\in\mathbb{N}$.
\end{vershort}

\begin{verlong}
Now, Proposition \ref{prop.gr.iso} (applied to $\left(  \otimes\mathfrak{g}%
\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot
\mathfrak{h}\right)  $, $\otimes\mathfrak{n}$, $\left(  F_{n}\right)
_{n\geq0}$, $\left(  N^{\otimes\leq n}\right)  _{n\geq0}$ and $\eta
\circ\overline{\varphi}$ instead of $V$, $W$, $\left(  V_{n}\right)  _{n\geq
0}$, $\left(  W_{n}\right)  _{n\geq0}$ and $f$) yields that
$\operatorname*{gr}\nolimits_{p}\left(  \eta\circ\overline{\varphi}\right)
:\operatorname*{gr}\nolimits_{p}\left(  \left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\right)  \rightarrow\operatorname*{gr}\nolimits_{p}\left(  \otimes
\mathfrak{n}\right)  $ is a $k$-module isomorphism for every $p\in\mathbb{N}$.
\end{verlong}

It remains now to prove that it is an $\mathfrak{h}$-module homomorphism. In
order to do this, we make some preparations:

We have $\left(  \pi\mid_{N}\right)  \circ t=\pi$ \ \ \ {\footnotemark}%
\footnotetext{\textit{Proof.} Let $v\in\mathfrak{g}$ be arbitrary. Then,
$v\in\mathfrak{g}=\mathfrak{h}\oplus N$. Thus, $v=h+n$ for some $h\in
\mathfrak{h}$ and $n\in N$. Consider these $h$ and $n$. Then, $v=h+n$ yields%
\begin{align*}
t\left(  v\right)   &  =t\left(  h+n\right)  =\underbrace{t\left(  h\right)
}_{\substack{=0\text{ (since }h\in\mathfrak{h}\text{, while }t\text{ is
a}\\\text{projection with kernel }\mathfrak{h}\text{)}}}+\underbrace{t\left(
n\right)  }_{\substack{=n\text{ (since }n\in N\text{, while }t\text{ is
a}\\\text{projection on }N\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}t\text{ is }k\text{-linear}\right) \\
&  =n
\end{align*}
But%
\[
\left(  \left(  \pi\mid_{N}\right)  \circ t\right)  \left(  v\right)  =\left(
\pi\mid_{N}\right)  \left(  t\left(  v\right)  \right)  =\pi\left(
\underbrace{t\left(  v\right)  }_{=n}\right)  =\pi\left(  n\right)  .
\]
Compared with%
\begin{align*}
\pi\left(  \underbrace{v}_{=h+n}\right)   &  =\pi\left(  h+n\right)
=\underbrace{\pi\left(  h\right)  }_{\substack{=0\text{ (since }%
h\in\mathfrak{h}\text{, while }\pi\text{ is a}\\\text{projection with kernel
}\mathfrak{h}\text{)}}}+\pi\left(  n\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }\pi\text{ is }k\text{-linear}\right) \\
&  =\pi\left(  n\right)  ,
\end{align*}
this yields $\left(  \left(  \pi\mid_{N}\right)  \circ t\right)  \left(
v\right)  =\pi\left(  v\right)  $. We have thus shown that every
$v\in\mathfrak{g}$ satisfies $\left(  \left(  \pi\mid_{N}\right)  \circ
t\right)  \left(  v\right)  =\pi\left(  v\right)  $. Thus, $\left(  \pi
\mid_{N}\right)  \circ t=\pi$.}. Thus, $\otimes\left(  \left(  \pi\mid
_{N}\right)  \circ t\right)  =\otimes\pi$. Since $\otimes\left(  \left(
\pi\mid_{N}\right)  \circ t\right)  =\underbrace{\left(  \otimes\left(
\pi\mid_{N}\right)  \right)  }_{=\eta}\circ\left(  \otimes t\right)
=\eta\circ\left(  \otimes t\right)  $, this becomes $\eta\circ\left(  \otimes
t\right)  =\otimes\pi$.

Thus,%
\[
\eta\circ\varphi-\underbrace{\left(  \otimes\pi\right)  }_{=\eta\circ\left(
\otimes t\right)  }=\eta\circ\varphi-\eta\circ\left(  \otimes t\right)
=\eta\circ\left(  \varphi-\left(  \otimes t\right)  \right)  .
\]


Both maps $\varphi$ and $\otimes t$ respect the filtration. Thus, their
difference $\varphi-\left(  \otimes t\right)  $ also respects the filtration
(by Proposition \ref{prop.gr.functor} \textbf{(c)}). Together with the fact
that $\eta$ respects the filtration, this yields that the composition
$\eta\circ\left(  \varphi-\left(  \otimes t\right)  \right)  $ also respects
the filtration (by Proposition \ref{prop.gr.functor} \textbf{(b)}). Since
$\eta\circ\left(  \varphi-\left(  \otimes t\right)  \right)  =\eta\circ
\varphi-\left(  \otimes\pi\right)  $, we have thus proven that the map
$\eta\circ\varphi-\left(  \otimes\pi\right)  $ respects the filtration.

Now let $p\in\mathbb{N}$. Proposition \ref{prop.l25.alpha.mod} (applied to
$n=p$) yields $\left(  \varphi-\left(  \otimes t\right)  \right)  \left(
\mathfrak{g}^{\otimes\leq p}\right)  \subseteq N^{\otimes\leq\left(
p-1\right)  }$. Now,%
\begin{align}
\left(  \underbrace{\eta\circ\varphi-\left(  \otimes\pi\right)  }_{=\eta
\circ\left(  \varphi-\left(  \otimes t\right)  \right)  }\right)  \left(
\mathfrak{g}^{\otimes\leq p}\right)   &  =\left(  \eta\circ\left(
\varphi-\left(  \otimes t\right)  \right)  \right)  \left(  \mathfrak{g}%
^{\otimes\leq p}\right)  =\eta\left(  \underbrace{\left(  \varphi-\left(
\otimes t\right)  \right)  \left(  \mathfrak{g}^{\otimes\leq p}\right)
}_{\subseteq N^{\otimes\leq\left(  p-1\right)  }}\right) \nonumber\\
&  \subseteq\eta\left(  N^{\otimes\leq\left(  p-1\right)  }\right)
\subseteq\mathfrak{n}^{\otimes\leq\left(  p-1\right)  }
\label{pf.l25.ha-Xp.lowers}%
\end{align}
(since $\eta$ respects the filtration). Using this relation, Proposition
\ref{prop.gr.0} (applied to $\otimes\mathfrak{g}$, $\otimes\mathfrak{n}$,
$\left(  \mathfrak{g}^{\otimes\leq n}\right)  _{n\geq0}$, $\left(
\mathfrak{n}^{\otimes\leq n}\right)  _{n\geq0}$ and $\eta\circ\left(
\varphi-\left(  \otimes t\right)  \right)  $ instead of $V$, $W$, $\left(
V_{n}\right)  _{n\geq0}$, $\left(  W_{n}\right)  _{n\geq0}$ and $f$) yields
that $\operatorname*{gr}\nolimits_{p}\left(  \eta\circ\varphi-\left(
\otimes\pi\right)  \right)  =0$. Thus, $0=\operatorname*{gr}\nolimits_{p}%
\left(  \eta\circ\varphi-\left(  \otimes\pi\right)  \right)
=\operatorname*{gr}\nolimits_{p}\left(  \eta\circ\varphi\right)
-\operatorname*{gr}\nolimits_{p}\left(  \otimes\pi\right)  $ (by Proposition
\ref{prop.gr.functor} \textbf{(c)}), so that $\operatorname*{gr}%
\nolimits_{p}\left(  \eta\circ\varphi\right)  =\operatorname*{gr}%
\nolimits_{p}\left(  \otimes\pi\right)  $.

Now, $F_{p}$ was defined by $F_{p}=\zeta\left(  \mathfrak{g}^{\otimes\leq
p}\right)  $. Now, $\operatorname*{gr}\nolimits_{p}\zeta:\operatorname*{gr}%
\nolimits_{p}\left(  \otimes\mathfrak{g}\right)  \rightarrow\operatorname*{gr}%
\nolimits_{p}\left(  \left(  \otimes\mathfrak{g}\right)  \diagup\left(
J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \right)  $ is
an $\mathfrak{h}$-module homomorphism (because $\zeta$ is an $\mathfrak{h}%
$-module homomorphism) and is surjective (according to Proposition
\ref{prop.gr.0} (applied to $\otimes\mathfrak{g}$, $\left(  \otimes
\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  $, $\left(  \mathfrak{g}^{\otimes\leq n}\right)
_{n\geq0}$, $\left(  F_{n}\right)  _{n\geq0}$ and $\zeta$ instead of $V$, $W$,
$\left(  V_{n}\right)  _{n\geq0}$, $\left(  W_{n}\right)  _{n\geq0}$ and $f$),
because $\zeta\left(  \mathfrak{g}^{\otimes\leq p}\right)  =F_{p}$ (by the
definition of $F_{p}$)).

Now, Proposition \ref{prop.gr.functor} \textbf{(b)} yields $\operatorname*{gr}%
\nolimits_{p}\left(  \left(  \eta\circ\overline{\varphi}\right)  \circ
\zeta\right)  =\operatorname*{gr}\nolimits_{p}\left(  \eta\circ\overline
{\varphi}\right)  \circ\operatorname*{gr}\nolimits_{p}\zeta$. Thus,%
\begin{equation}
\operatorname*{gr}\nolimits_{p}\left(  \eta\circ\overline{\varphi}\right)
\circ\operatorname*{gr}\nolimits_{p}\zeta=\operatorname*{gr}\nolimits_{p}%
\left(  \left(  \eta\circ\overline{\varphi}\right)  \circ\zeta\right)
=\operatorname*{gr}\nolimits_{p}\left(  \eta\circ\underbrace{\overline
{\varphi}\circ\zeta}_{=\varphi}\right)  =\operatorname*{gr}\nolimits_{p}%
\left(  \eta\circ\varphi\right)  =\operatorname*{gr}\nolimits_{p}\left(
\otimes\pi\right)  . \label{pf.l25.Ker-alpha.gr}%
\end{equation}
Since $\operatorname*{gr}\nolimits_{p}\left(  \otimes\pi\right)  $ is an
$\mathfrak{h}$-module homomorphism (because $\pi$ is an $\mathfrak{h}$-module
homomorphism, and thus $\otimes\pi$ is an $\mathfrak{h}$-module homomorphism),
this yields that $\operatorname*{gr}\nolimits_{p}\left(  \eta\circ
\overline{\varphi}\right)  \circ\operatorname*{gr}\nolimits_{p}\zeta$ is an
$\mathfrak{h}$-module homomorphism. Thus, Lemma \ref{lem.surj} (applied to
$\operatorname*{gr}\nolimits_{p}\left(  \otimes\mathfrak{g}\right)  $,
$\operatorname*{gr}\nolimits_{p}\left(  \left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\right)  $, $\operatorname*{gr}\nolimits_{p}\left(  \otimes\mathfrak{n}%
\right)  $, $\operatorname*{gr}\nolimits_{p}\zeta$ and $\operatorname*{gr}%
\nolimits_{p}\left(  \eta\circ\overline{\varphi}\right)  $ instead of $A$,
$B$, $C$, $f$ and $g$) yields that $\operatorname*{gr}\nolimits_{p}\left(
\eta\circ\overline{\varphi}\right)  $ is an $\mathfrak{h}$-module
homomorphism. Since we know that $\operatorname*{gr}\nolimits_{p}\left(
\eta\circ\overline{\varphi}\right)  $ is a $k$-module isomorphism, we can thus
conclude that $\operatorname*{gr}\nolimits_{p}\left(  \eta\circ\overline
{\varphi}\right)  $ is an $\mathfrak{h}$-module isomorphism (due to
Proposition \ref{prop.liemodiso}). This completes the proof of Proposition
\ref{prop.l25.Ker-alpha} \textbf{(e)}. Thus, Proposition
\ref{prop.l25.Ker-alpha} is finished.

Now we can finish off Theorem \ref{thm.l25}:

\textit{Proof of Theorem \ref{thm.l25}.} \textbf{(a)} We have already proven
Theorem \ref{thm.l25} \textbf{(a)}.

\textbf{(b)} We have already proven farther above that $\left(  F_{n}\right)
_{n\geq0}$ is an $\mathfrak{h}$-module filtration of $\left(  \otimes
\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  $.

Let $p\in\mathbb{N}$ be arbitrary. According to Proposition
\ref{prop.l25.Ker-alpha} \textbf{(e)}, there exists an $\mathfrak{h}$-module
isomorphism $\operatorname*{gr}\nolimits_{p}\left(  \eta\circ\overline
{\varphi}\right)  :\operatorname*{gr}\nolimits_{p}\left(  \left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  \right)  \rightarrow\operatorname*{gr}%
\nolimits_{p}\left(  \otimes\mathfrak{n}\right)  $. Thus, $\operatorname*{gr}%
\nolimits_{p}\left(  \left(  \otimes\mathfrak{g}\right)  \diagup\left(
J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \right)
\cong\operatorname*{gr}\nolimits_{p}\left(  \otimes\mathfrak{n}\right)  $ as
$\mathfrak{h}$-modules. Since $\operatorname*{gr}\nolimits_{p}\left(  \left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  \right)  =F_{p}\diagup F_{p-1}$ (because
the filtration on $\left(  \otimes\mathfrak{g}\right)  \diagup\left(
J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ is given by
$\left(  F_{n}\right)  _{n\geq0}$) and
\begin{align*}
&  \operatorname*{gr}\nolimits_{p}\left(  \otimes\mathfrak{n}\right) \\
&  =\mathfrak{n}^{\otimes\leq p}\diagup\mathfrak{n}^{\otimes\leq\left(
p-1\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since the filtration on
}\otimes\mathfrak{n}\text{ is }\left(  \mathfrak{n}^{\otimes\leq n}\right)
_{n\geq0}\right) \\
&  =\left(  \bigoplus_{i=0}^{p}\mathfrak{n}^{\otimes i}\right)  \diagup\left(
\bigoplus_{i=0}^{p-1}\mathfrak{n}^{\otimes i}\right)
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\mathfrak{n}^{\otimes\leq p}=\bigoplus\limits_{i=0}%
^{p}\mathfrak{n}^{\otimes i}\text{ by the definition of }\mathfrak{n}%
^{\otimes\leq p}\\
\text{and since }\mathfrak{n}^{\otimes\leq\left(  p-1\right)  }=\bigoplus
\limits_{i=0}^{p-1}\mathfrak{n}^{\otimes i}\text{ by the definition of
}\mathfrak{n}^{\otimes\leq\left(  p-1\right)  }%
\end{array}
\right) \\
&  \cong\mathfrak{n}^{\otimes p},
\end{align*}
this becomes $F_{p}\diagup F_{p-1}\cong\mathfrak{n}^{\otimes p}$ as
$\mathfrak{h}$-modules. Renaming $p$ as $n$, we conclude that $F_{n}\diagup
F_{n-1}\cong\mathfrak{n}^{\otimes n}$ as $\mathfrak{h}$-modules for every
$n\in\mathbb{N}$. This completes the proof of Theorem \ref{thm.l25}
\textbf{(b)}.

\textbf{(c)} Every $n\in\mathbb{N}$ satisfies $\operatorname*{gr}%
\nolimits_{n}\left(  \left(  \otimes\mathfrak{g}\right)  \diagup\left(
J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \right)
=F_{n}\diagup F_{n-1}$ (because the filtration on $\left(  \otimes
\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  $ is given by $\left(  F_{n}\right)  _{n\geq0}$).

Now let $n\in\mathbb{N}$ be arbitrary.

Proposition \ref{prop.l25.Ker-alpha} \textbf{(e)} yields that
$\operatorname*{gr}\nolimits_{p}\left(  \eta\circ\overline{\varphi}\right)  $
is an $\mathfrak{h}$-module isomorphism for every $p\in\mathbb{N}$. Applying
this to $p=n$, we conclude that $\operatorname*{gr}\nolimits_{n}\left(
\eta\circ\overline{\varphi}\right)  $ is an $\mathfrak{h}$-module isomorphism.

During the proof of Proposition \ref{prop.l25.Ker-alpha} \textbf{(e)}, we have
showed that every $p\in\mathbb{N}$ satisfies $\operatorname*{gr}%
\nolimits_{p}\left(  \eta\circ\overline{\varphi}\right)  \circ
\operatorname*{gr}\nolimits_{p}\zeta=\operatorname*{gr}\nolimits_{p}\left(
\otimes\pi\right)  $ (due to (\ref{pf.l25.Ker-alpha.gr})). Applying this to
$p=n$, we obtain $\operatorname*{gr}\nolimits_{n}\left(  \eta\circ
\overline{\varphi}\right)  \circ\operatorname*{gr}\nolimits_{n}\zeta
=\operatorname*{gr}\nolimits_{n}\left(  \otimes\pi\right)  $. In other words,%
\begin{equation}
\left(  \text{the diagram (\ref{thm.l25.c.diag}) commutes if }\Omega
_{n}=\operatorname*{gr}\nolimits_{n}\left(  \eta\circ\overline{\varphi
}\right)  \right)  . \label{pf.l25.c.0}%
\end{equation}


Now, let us prove that
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{if }\Omega_{n}:F_{n}\diagup F_{n-1}\rightarrow\operatorname*{gr}%
\nolimits_{n}\left(  \otimes\mathfrak{n}\right)  \text{ is a }k\text{-module
homomorphism}\\
\text{for which the diagram (\ref{thm.l25.c.diag}) commutes, then }\Omega
_{n}=\operatorname*{gr}\nolimits_{n}\left(  \eta\circ\overline{\varphi
}\right)
\end{array}
\right)  \label{pf.l25.c.1}%
\end{equation}


\textit{Proof of (\ref{pf.l25.c.1}).} Let $\Omega_{n}:F_{n}\diagup
F_{n-1}\rightarrow\operatorname*{gr}\nolimits_{n}\left(  \otimes
\mathfrak{n}\right)  $ be a $k$-module homomorphism for which the diagram
(\ref{thm.l25.c.diag}) commutes. Then, the diagram (\ref{thm.l25.c.diag})
commutes, so that $\Omega_{n}\circ\operatorname*{gr}\nolimits_{n}%
\zeta=\operatorname*{gr}\nolimits_{n}\left(  \otimes\pi\right)  $. Combining
this with $\operatorname*{gr}\nolimits_{n}\left(  \eta\circ\overline{\varphi
}\right)  \circ\operatorname*{gr}\nolimits_{n}\zeta=\operatorname*{gr}%
\nolimits_{n}\left(  \otimes\pi\right)  $, we obtain $\Omega_{n}%
\circ\operatorname*{gr}\nolimits_{n}\zeta=\operatorname*{gr}\nolimits_{n}%
\left(  \eta\circ\overline{\varphi}\right)  \circ\operatorname*{gr}%
\nolimits_{n}\zeta$.

So now we know that $\Omega_{n}\circ\operatorname*{gr}\nolimits_{n}%
\zeta=\operatorname*{gr}\nolimits_{n}\left(  \eta\circ\overline{\varphi
}\right)  \circ\operatorname*{gr}\nolimits_{n}\zeta$, and we want to show that
$\Omega_{n}=\operatorname*{gr}\nolimits_{n}\left(  \eta\circ\overline{\varphi
}\right)  $.

During the proof of Proposition \ref{prop.l25.Ker-alpha} \textbf{(e)}, we have
showed that $\operatorname*{gr}\nolimits_{p}\zeta$ is surjective for every
$p\in\mathbb{N}$. Applying this to $p=n$, we obtain that $\operatorname*{gr}%
\nolimits_{n}\zeta$ is surjective. Now let us show that $\Omega_{n}%
=\operatorname*{gr}\nolimits_{n}\left(  \eta\circ\overline{\varphi}\right)  $:

Let $v\in F_{n}\diagup F_{n-1}$ be arbitrary. Since $\operatorname*{gr}%
\nolimits_{n}\zeta$ is surjective, there exists some $w\in\operatorname*{gr}%
\nolimits_{n}\left(  \otimes\mathfrak{g}\right)  $ such that $v=\left(
\operatorname*{gr}\nolimits_{n}\zeta\right)  \left(  w\right)  $. Consider
this $w$. Then,
\begin{align*}
\Omega_{n}\left(  \underbrace{v}_{=\left(  \operatorname*{gr}\nolimits_{n}%
\zeta\right)  \left(  w\right)  }\right)   &  =\Omega_{n}\left(  \left(
\operatorname*{gr}\nolimits_{n}\zeta\right)  \left(  w\right)  \right)
=\left(  \underbrace{\Omega_{n}\circ\operatorname*{gr}\nolimits_{n}\zeta
}_{=\operatorname*{gr}\nolimits_{n}\left(  \eta\circ\overline{\varphi}\right)
\circ\operatorname*{gr}\nolimits_{n}\zeta}\right)  \left(  w\right)  =\left(
\operatorname*{gr}\nolimits_{n}\left(  \eta\circ\overline{\varphi}\right)
\circ\operatorname*{gr}\nolimits_{n}\zeta\right)  \left(  w\right) \\
&  =\left(  \operatorname*{gr}\nolimits_{n}\left(  \eta\circ\overline{\varphi
}\right)  \right)  \left(  \underbrace{\left(  \operatorname*{gr}%
\nolimits_{n}\zeta\right)  \left(  w\right)  }_{=v}\right)  =\left(
\operatorname*{gr}\nolimits_{n}\left(  \eta\circ\overline{\varphi}\right)
\right)  \left(  v\right)  .
\end{align*}
Since this holds for every $v\in F_{n}\diagup F_{n-1}$, we thus have proven
that $\Omega_{n}=\operatorname*{gr}\nolimits_{n}\left(  \eta\circ
\overline{\varphi}\right)  $. This proves (\ref{pf.l25.c.1}).

From (\ref{pf.l25.c.1}), we see that every $k$-module homomorphism $\Omega
_{n}:F_{n}\diagup F_{n-1}\rightarrow\operatorname*{gr}\nolimits_{n}\left(
\otimes\mathfrak{n}\right)  $ for which the diagram (\ref{thm.l25.c.diag})
commutes must be equal to $\operatorname*{gr}\nolimits_{n}\left(  \eta
\circ\overline{\varphi}\right)  $. Hence, there exists at most one $k$-module
homomorphism $\Omega_{n}:F_{n}\diagup F_{n-1}\rightarrow\operatorname*{gr}%
\nolimits_{n}\left(  \otimes\mathfrak{n}\right)  $ for which the diagram
(\ref{thm.l25.c.diag}) commutes. But since we also know that there exists at
least one $k$-module homomorphism $\Omega_{n}:F_{n}\diagup F_{n-1}%
\rightarrow\operatorname*{gr}\nolimits_{n}\left(  \otimes\mathfrak{n}\right)
$ for which the diagram (\ref{thm.l25.c.diag}) commutes (namely, the
homomorphism $\operatorname*{gr}\nolimits_{n}\left(  \eta\circ\overline
{\varphi}\right)  $, because of (\ref{pf.l25.c.0})), we thus conclude that
there exists one and only one $k$-module homomorphism $\Omega_{n}:F_{n}\diagup
F_{n-1}\rightarrow\operatorname*{gr}\nolimits_{n}\left(  \otimes
\mathfrak{n}\right)  $ for which the diagram (\ref{thm.l25.c.diag}) commutes.
This proves part of Theorem \ref{thm.l25} \textbf{(c)}.

According to Theorem \ref{thm.l25} \textbf{(c)}, we define $\omega_{n}$ as the
$k$-module homomorphism $\Omega_{n}:F_{n}\diagup F_{n-1}\rightarrow
\operatorname*{gr}\nolimits_{n}\left(  \otimes\mathfrak{n}\right)  $ for which
the diagram (\ref{thm.l25.c.diag}) commutes (the existence and uniqueness of
this homomorphism $\Omega_{n}$ was already proven above). This definition
immediately yields that the diagram (\ref{thm.l25.c.diag'}) commutes. In other
words, the diagram (\ref{thm.l25.c.diag}) commutes if $\Omega_{n}=\omega_{n}$.
Thus, (\ref{pf.l25.c.1}) (applied to $\Omega_{n}=\omega_{n}$) yields that
$\omega_{n}=\operatorname*{gr}\nolimits_{n}\left(  \eta\circ\overline{\varphi
}\right)  $. Thus, $\omega_{n}$ is an $\mathfrak{h}$-module isomorphism
(because we know that $\operatorname*{gr}\nolimits_{n}\left(  \eta
\circ\overline{\varphi}\right)  $ is an $\mathfrak{h}$-module isomorphism).

Now, all nontrivial statements in Theorem \ref{thm.l25} \textbf{(c)} are
proven. This completes the proof of Theorem \ref{thm.l25}.

\subsection{Independency of the splitting}

As a bonus from the above proof of Theorem \ref{thm.l25}, we obtain the
following strengthening of this theorem:

\begin{proposition}
\label{prop.l25.ind}In the context of Theorem \ref{thm.l25} \textbf{(b)}, for
every $n\in\mathbb{N}$, there exists an $\mathfrak{h}$-module isomorphism
$F_{n}\diagup F_{n-1}\rightarrow\mathfrak{n}^{\otimes n}$ which is independent
of the choice of $N$.
\end{proposition}

\begin{vershort}
\textit{Proof of Proposition \ref{prop.l25.ind}.} According to Theorem
\ref{thm.l25} \textbf{(c)}, the map $\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}^{-1}\circ\omega_{n}$ (where $\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}$ and $\omega_{n}$ are defined in Theorem
\ref{thm.l25} \textbf{(c)}) is an $\mathfrak{h}$-module isomorphism
$F_{n}\diagup F_{n-1}\rightarrow\mathfrak{n}^{\otimes n}$. This isomorphism is
clearly independent of the choice of $N$ (since the definitions of
$\operatorname*{grad}\nolimits_{\mathfrak{n},n}$ and $\omega_{n}$ are
independent of the choice of $N$). This proves Proposition \ref{prop.l25.ind}.
\end{vershort}

\begin{verlong}
\textit{First proof of Proposition \ref{prop.l25.ind}.} According to Theorem
\ref{thm.l25} \textbf{(c)}, the map $\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}^{-1}\circ\omega_{n}$ (where $\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}$ and $\omega_{n}$ are defined in Theorem
\ref{thm.l25} \textbf{(c)}) is an $\mathfrak{h}$-module isomorphism
$F_{n}\diagup F_{n-1}\rightarrow\mathfrak{n}^{\otimes n}$. This isomorphism is
clearly independent of the choice of $N$ (since the definitions of
$\operatorname*{grad}\nolimits_{\mathfrak{n},n}$ and $\omega_{n}$ are
independent of the choice of $N$). This proves Proposition \ref{prop.l25.ind}.

We can also show Proposition \ref{prop.l25.ind} without reference to Theorem
\ref{thm.l25} \textbf{(c)}:

\textit{Second proof of Proposition \ref{prop.l25.ind}.} Let $p\in\mathbb{N}$
be arbitrary.

In our above proof of Theorem \ref{thm.l25} \textbf{(b)}, we have constructed
an $\mathfrak{h}$-module isomorphism $\operatorname*{gr}\nolimits_{p}\left(
\eta\circ\overline{\varphi}\right)  :\operatorname*{gr}\nolimits_{p}\left(
\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \right)  \rightarrow
\operatorname*{gr}\nolimits_{p}\left(  \otimes\mathfrak{n}\right)  $. Let us
now prove that this isomorphism $\operatorname*{gr}\nolimits_{p}\left(
\eta\circ\overline{\varphi}\right)  $ is independent of the choice of $N$.
(Note that this is absolutely not trivial from its definition, because
$\overline{\varphi}$ \textit{does} depend on $N$.)

In fact, let us show that
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{for every }v\in\operatorname*{gr}\nolimits_{p}\left(  \left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  \right)  \text{, the image}\\
\left(  \operatorname*{gr}\nolimits_{p}\left(  \eta\circ\overline{\varphi
}\right)  \right)  \left(  v\right)  \text{ is independent of the choice of }N
\end{array}
\right)  . \label{pf.l25.ind}%
\end{equation}


\textit{Proof of (\ref{pf.l25.ind}).} Let $v\in\operatorname*{gr}%
\nolimits_{p}\left(  \left(  \otimes\mathfrak{g}\right)  \diagup\left(
J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \right)  $ be arbitrary.

In the proof of Proposition \ref{prop.l25.Ker-alpha} \textbf{(e)}, we showed
that $\operatorname*{gr}\nolimits_{p}\zeta$ is surjective. Hence, there exists
some $w\in\operatorname*{gr}\nolimits_{p}\left(  \otimes\mathfrak{g}\right)  $
such that $v=\left(  \operatorname*{gr}\nolimits_{p}\zeta\right)  \left(
w\right)  $. Thus%
\[
\left(  \operatorname*{gr}\nolimits_{p}\left(  \eta\circ\overline{\varphi
}\right)  \right)  \left(  v\right)  =\left(  \operatorname*{gr}%
\nolimits_{p}\left(  \eta\circ\overline{\varphi}\right)  \right)  \left(
\left(  \operatorname*{gr}\nolimits_{p}\zeta\right)  \left(  w\right)
\right)  =\underbrace{\left(  \operatorname*{gr}\nolimits_{p}\left(  \eta
\circ\overline{\varphi}\right)  \circ\operatorname*{gr}\nolimits_{p}%
\zeta\right)  }_{=\operatorname*{gr}\nolimits_{p}\left(  \otimes\pi\right)
\text{ (due to (\ref{pf.l25.Ker-alpha.gr}))}}\left(  w\right)  =\left(
\operatorname*{gr}\nolimits_{p}\left(  \otimes\pi\right)  \right)  \left(
w\right)  .
\]
Since $\left(  \operatorname*{gr}\nolimits_{p}\left(  \otimes\pi\right)
\right)  \left(  w\right)  $ is obviously independent of the choice of $N$, it
thus follows that $\left(  \operatorname*{gr}\nolimits_{p}\left(  \eta
\circ\overline{\varphi}\right)  \right)  \left(  v\right)  $ is independent of
the choice of $N$. This proves (\ref{pf.l25.ind}).

Now, (\ref{pf.l25.ind}) yields that the map $\operatorname*{gr}\nolimits_{p}%
\left(  \eta\circ\overline{\varphi}\right)  $ is independent of the choice of
$N$. Since $\operatorname*{gr}\nolimits_{p}\left(  \eta\circ\overline{\varphi
}\right)  $ is an $\mathfrak{h}$-module isomorphism $\operatorname*{gr}%
\nolimits_{p}\left(  \left(  \otimes\mathfrak{g}\right)  \diagup\left(
J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \right)
\rightarrow\operatorname*{gr}\nolimits_{p}\left(  \otimes\mathfrak{n}\right)
$, we have thus found an $\mathfrak{h}$-module isomorphism $\operatorname*{gr}%
\nolimits_{p}\left(  \left(  \otimes\mathfrak{g}\right)  \diagup\left(
J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \right)
\rightarrow\operatorname*{gr}\nolimits_{p}\left(  \otimes\mathfrak{n}\right)
$ independent of the choice of $N$. Since $\operatorname*{gr}\nolimits_{p}%
\left(  \left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \right)  =F_{p}\diagup
F_{p-1}$ (because the filtration on $\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
$ is given by $\left(  F_{n}\right)  _{n\geq0}$) and $\operatorname*{gr}%
\nolimits_{p}\left(  \otimes\mathfrak{n}\right)  =\mathfrak{n}^{\otimes\leq
p}\diagup\mathfrak{n}^{\otimes\leq\left(  p-1\right)  }$ (since the filtration
on $\otimes\mathfrak{n}$ is $\left(  \mathfrak{n}^{\otimes\leq n}\right)
_{n\geq0}$), this can be rewritten as follows: We have found an $\mathfrak{h}%
$-module isomorphism $F_{p}\diagup F_{p-1}\rightarrow\mathfrak{n}^{\otimes\leq
p}\diagup\mathfrak{n}^{\otimes\leq\left(  p-1\right)  }$ independent of the
choice of $N$. Composing it with the canonical $\mathfrak{h}$-module
isomorphism $\mathfrak{n}^{\otimes\leq p}\diagup\mathfrak{n}^{\otimes
\leq\left(  p-1\right)  }\rightarrow\mathfrak{n}^{\otimes p}$%
\ \ \ \ {\footnotemark}\footnotetext{This isomorphism is constructed as
follows:%
\begin{align*}
&  \mathfrak{n}^{\otimes\leq p}\diagup\mathfrak{n}^{\otimes\leq\left(
p-1\right)  }\\
&  =\left(  \bigoplus_{i=0}^{p}\mathfrak{n}^{\otimes i}\right)  \diagup\left(
\bigoplus_{i=0}^{p-1}\mathfrak{n}^{\otimes i}\right)
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\mathfrak{n}^{\otimes\leq p}=\bigoplus\limits_{i=0}%
^{p}\mathfrak{n}^{\otimes i}\text{ by the definition of }\mathfrak{n}%
^{\otimes\leq p}\\
\text{and since }\mathfrak{n}^{\otimes\leq\left(  p-1\right)  }=\bigoplus
\limits_{i=0}^{p-1}\mathfrak{n}^{\otimes i}\text{ by the definition of
}\mathfrak{n}^{\otimes\leq\left(  p-1\right)  }%
\end{array}
\right) \\
&  \cong\mathfrak{n}^{\otimes p}.
\end{align*}
}, which is also independent of the choice of $N$, we thus obtain an
$\mathfrak{h}$-module isomorphism $F_{p}\diagup F_{p-1}\rightarrow
\mathfrak{n}^{\otimes p}$ which is independent of the choice of $N$.

Thus, we have shown that for every $p\in\mathbb{N}$, there exists an
$\mathfrak{h}$-module isomorphism $F_{p}\diagup F_{p-1}\rightarrow
\mathfrak{n}^{\otimes p}$ which is independent of the choice of $N$. Renaming
$p$ into $n$ here, we obtain the assertion of Proposition \ref{prop.l25.ind}.
Thus, Proposition \ref{prop.l25.ind} is proven.
\end{verlong}

\section{\label{sect.(g,h)}$\left(  \mathfrak{g,h}\right)  $-semimodules}

Before we prove some more interesting results, we are going to introduce a
notion - that of a $\left(  \mathfrak{g},\mathfrak{h}\right)  $%
\textit{-semimodule}. This notion will be defined for every commutative ring
$k$, every $k$-Lie algebra $\mathfrak{g}$ and every Lie subalgebra
$\mathfrak{h}$ of $\mathfrak{g}$. It will be a kind of intermediate link
between the notion of a $\mathfrak{g}$-module and that of an $\mathfrak{h}%
$-module. Here is the definition:

\subsection{$\left(  \mathfrak{g,h}\right)  $-semimodules: the definition}

\begin{definition}
\label{defs.semimod}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let
$V$ be a $k$-module. Let $\mu:\mathfrak{g}\times V\rightarrow V$ be a
$k$-bilinear map. We say that $\left(  V,\mu\right)  $ is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $\textit{-semimodule} 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{h}\text{, }b\in\mathfrak{g}\text{ and }v\in V\right)  .
\label{defs.semimod.mu}%
\end{equation}
If $\left(  V,\mu\right)  $ is a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule, then the $k$-bilinear map $\mu:\mathfrak{g}\times V\rightarrow
V$ is called the \textit{Lie action} of the $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule $V$.\newline 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{defs.semimod.mu}) 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{h}\text{, }b\in\mathfrak{g}\text{ and }v\in
V\right)  . \label{defs.semimod.->}%
\end{equation}
\newline Also, an abuse of notation allows us to write "$V$ is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule" instead of "$\left(
V,\mu\right)  $ is a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule"
if the map $\mu$ is clear from the context or has not been introduced
yet.\newline Besides, when $\left(  V,\mu\right)  $ is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule, we will say that $\mu$ is a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-\textit{semimodule structure on
}$V$. In other words, if $V$ is a $k$-module, then a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $\textit{-semimodule structure on }$V$ means a map
$\mu:\mathfrak{g}\times V\rightarrow V$ such that $\left(  V,\mu\right)  $ is
a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule. (Thus, in order to
make a $k$-module into a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule, we must define a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule structure on it.)
\end{definition}

This definition is very similar to the Definition \ref{defs.liemod}. We will
see that this similarity is not just superficial, and that most properties of
$\mathfrak{g}$-modules have their analogues concerning $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules.

But first let us notice that:

\begin{proposition}
\label{prop.semimod.mod}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$.
Then, every $\mathfrak{g}$-module is a $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule.
\end{proposition}

In fact, this proposition follows trivially from comparing Definition
\ref{defs.liemod} with Definition \ref{defs.semimod}. The converse of this
proposition does not hold. However, a $\mathfrak{g}$-module is exactly the
same as a $\left(  \mathfrak{g},\mathfrak{g}\right)  $-semimodule, i. e., we have:

\begin{proposition}
\label{prop.semimod.(g,g)}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a $k$-Lie algebra. Let $V$ be a $k$-module. Let $\mu:\mathfrak{g}\times
V\rightarrow V$ be a map. Then, $\left(  V,\mu\right)  $ is a $\mathfrak{g}%
$-module if and only if $\left(  V,\mu\right)  $ is a $\left(  \mathfrak{g}%
,\mathfrak{g}\right)  $-semimodule.
\end{proposition}

This is again clear from comparing Definition \ref{defs.liemod} with
Definition \ref{defs.semimod}.

\begin{vershort}
Proposition \ref{prop.semimod.(g,g)} shows that the notion of a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule is a generalization of the
notion of a $\mathfrak{g}$-module. Much of this Section \ref{sect.(g,h)} will
be devoted to formulating some properties of $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules which are analogous to the well-known
properties of $\mathfrak{g}$-modules which we collected in Section
\ref{sect.intro}. We are \textit{not} going to prove all of these properties
anew, because the proofs will be identical to the corresponding proofs for
$\mathfrak{g}$-modules done in Section \ref{sect.intro}, up to changing
"$\mathfrak{g}$-module" to "$\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule" (and, similarly, "$\mathfrak{g}$-algebra" to "$\left(
\mathfrak{g},\mathfrak{h}\right)  $-semialgebra") and changing the references
to results about $\mathfrak{g}$-modules to the corresponding results about
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules.
\end{vershort}

\begin{verlong}
Proposition \ref{prop.semimod.(g,g)} shows that the notion of a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule is a generalization of the
notion of a $\mathfrak{g}$-module. Much of this Section \ref{sect.(g,h)} will
be devoted to formulating some properties of $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules which are analogous to the well-known
properties of $\mathfrak{g}$-modules which we collected in Section
\ref{sect.intro}. We are \textit{not} going to prove all of these properties
anew, because the proofs will often be almost identical to the corresponding
proofs for $\mathfrak{g}$-modules done in Section \ref{sect.intro}. We will
only point out which changes must be made to those proofs in order to make
them apply to $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules rather
than just to $\mathfrak{g}$-modules.
\end{verlong}

\begin{vershort}
Along with the theorems, most definitions from the theory of $\mathfrak{g}%
$-modules can be generalized to $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules. For example, we can define the tensor product of two $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodules in the same way as we have
defined the tensor product of two $\mathfrak{g}$-modules in Definition
\ref{defs.lie(X)}. Of course, this new notion of the tensor product of two
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules will not conflict
with the old notion of the tensor product of two $\mathfrak{g}$-modules
(because the definition is the same, so that, when we have two $\mathfrak{g}%
$-modules, their tensor product does not depend on whether we consider them as
$\mathfrak{g}$-modules or as $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules).
\end{vershort}

\begin{verlong}
In this Section \ref{sect.(g,h)}, we are also going to define several notions
related to $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules, such as
the notion of a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-subsemimodule,
and the notion of the tensor product of two $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules. \textbf{All the definitions that we are
going to give will be analogous to the definitions of the corresponding
notions for }$\mathfrak{g}$\textbf{-modules given in Section \ref{sect.intro},
and therefore will not conflict with these latter notions.} For example,
Definition \ref{defs.semi(X)} (the definition of the tensor product of two
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules) will be analogous to
Definition \ref{defs.lie(X)} (the definition of the tensor product of two
$\mathfrak{g}$-modules). Therefore, if we have a commutative ring $k$, some
$k$-Lie algebra $\mathfrak{g}$, some Lie subalgebra $\mathfrak{h}$ of
$\mathfrak{g}$, and two $\mathfrak{g}$-modules $V$ and $W$, then the tensor
product of the $\mathfrak{g}$-modules $V$ and $W$ (as defined in Definition
\ref{defs.lie(X)}) will be the same as the tensor product of the $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodules $V$ and $W$ (as defined in
Definition \ref{defs.semi(X)}). Similarly, the direct sum of the
$\mathfrak{g}$-modules $V$ and $W$ (as defined in Definition \ref{defs.lie(+)}%
) will be the same as the direct sum of the $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules $V$ and $W$ (as defined in Definition
\ref{defs.semi(+)}).
\end{verlong}

We begin with a convention:

\begin{condition}
\label{conv.semimod.->}We are going to use the notation $a\rightharpoonup v$
as a universal notation for the Lie action of a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule. This means that whenever we have some Lie
algebra $\mathfrak{g}$, some Lie subalgebra $\mathfrak{h}$ of $\mathfrak{g}$
and some $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule $V$ (they
need not be actually called $\mathfrak{g}$, $\mathfrak{h}$ and $V$; I only
refer to them as $\mathfrak{g}$, $\mathfrak{h}$ and $V$ here in this
Convention), and we are given two elements $a\in\mathfrak{g}$ and $v\in V$
(they need not be actually called $a$ and $v$; I only refer to them by $a$ and
$v$ here in this Convention), 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).
\end{condition}

This convention is, of course, just the extension of Convention \ref{conv.->}
to $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules.

\begin{conclusion}
\label{warn.semimod.->}We know from Proposition \ref{prop.semimod.mod} that
every $\mathfrak{g}$-module is a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule. Thus, when $k$ is a commutative ring, $\mathfrak{g}$ is a
$k$-Lie algebra, $\mathfrak{h}$ is a Lie subalgebra of $\mathfrak{g}$, and $V$
is some $\mathfrak{g}$-module, then $V$ is a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule as well, and thus, the notation
$a\rightharpoonup v$ (where $a\in\mathfrak{g}$ and $v\in V$) is overloaded: It
can be interpreted according to Convention \ref{conv.->}, but it can also be
interpreted according to Convention \ref{conv.semimod.->}. However,
fortunately these two conventions give the same definition for
$a\rightharpoonup v$, and thus they do not conflict. So when we have a
$\mathfrak{g}$-module $V$, then we do not have to worry about Convention
\ref{conv.->} and Convention \ref{conv.semimod.->} leading to different
interpretations $a\rightharpoonup v$; they don't.\newline\textbf{However,}
Convention \ref{conv.semimod.->} can conflict with Convention \ref{conv.->} in
two other cases: The first case is when we have a $\mathfrak{g}$-module
structure and a \textbf{different} $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule structure defined on one and the same $k$-module; the second case
is when we have an $\mathfrak{h}$-module structure and a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule structure defined on one and
the same $k$-module. The first of these cases will not appear in our studies;
however, the second will appear. In such a case, we will \textit{not} be
allowed to use Convention \ref{conv.semimod.->} until we verify that, for
every $a\in\mathfrak{h}$ and $v\in V$, the meaning of $a\rightharpoonup v$
according to Convention \ref{conv.->} (that is, the Lie action of the
$\mathfrak{h}$-module $V$ applied to $\left(  a,v\right)  $) equals to the
meaning of $a\rightharpoonup v$ according to Convention \ref{conv.semimod.->}
(that is, the Lie action of the $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule $V$ applied to $\left(  a,v\right)  $), so that the value of
$a\rightharpoonup v$ does not depend on which of the two Conventions we are
using.\newline Fortunately, within this paper, this will always be fulfilled
and easy to verify. In fact, within this paper, each time when we have an
$\mathfrak{h}$-module structure and a $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule structure defined on one and the same $k$-module, the
$\mathfrak{h}$-module will be the restriction of the $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule to $\mathfrak{h}$ (see Definition
\ref{defs.semi.restrict} for the definition of "restriction"), and thus we
\textit{will} be allowed to use Convention \ref{conv.semimod.->} (according to
Remark \ref{rmk.semi.restrict}).
\end{conclusion}

Now that we have defined a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule, let us do the next logical step and define a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule homomorphism:

\begin{definition}
\label{defs.semimodhom}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let
$V$ and $W$ be two $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules.
Let $f:V\rightarrow W$ be a $k$-linear map. Then, $f$ is said to be a $\left(
\mathfrak{g},\mathfrak{h}\right)  $\textit{-semimodule homomorphism} if and
only if%
\[
\left(  f\left(  a\rightharpoonup v\right)  =a\rightharpoonup\left(  f\left(
v\right)  \right)  \ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{
and }v\in V\right)  .
\]
\newline Often, we will use the words "$\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule map" or the words "homomorphism of $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodules" or the
words "$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semilinear map" as
synonyms for
"$\left(  \mathfrak{g},\mathfrak{h}\right)$-semimodule homomorphism".
\end{definition}

This Definition \ref{defs.semimodhom} is the analogue of Definition
\ref{defs.liemodhom} for $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules. Therefore, we have:

\begin{proposition}
Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie algebra. Let
$\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $V$ and $W$ be two
$\mathfrak{g}$-modules. Let $f:V\rightarrow W$ be a map. Then, $f$ is a
$\mathfrak{g}$-module homomorphism if and only if $f$ is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule homomorphism. (Here, it makes
sense to say that "$f$ is a $\mathfrak{g}$-module homomorphism" since $V$ and
$W$ are $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules (which is
because Proposition \ref{prop.semimod.mod} yields that every $\mathfrak{g}%
$-module is a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule).)
\end{proposition}

\begin{vershort}
It is easy to see that for every commutative ring $k$, for every $k$-Lie
algebra $\mathfrak{g}$, and for every Lie subalgebra $\mathfrak{h}$ of
$\mathfrak{g}$, there is a category whose objects are $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules and whose morphisms are $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule homomorphisms. We define the
notion of a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule
isomorphism as an isomorphism in this category; in other words, we define the
notion of a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule
isomorphism by analogy to the notion of a $\mathfrak{g}$-module isomorphism
(which we defined in Definition \ref{defs.liemod}). The obvious analogue of
Proposition \ref{prop.liemodiso} holds.
\end{vershort}

\begin{verlong}
It is easy to see that for every commutative ring $k$, for every $k$-Lie
algebra $\mathfrak{g}$, and for every Lie subalgebra $\mathfrak{h}$ of
$\mathfrak{g}$, there is a category whose objects are $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules and whose morphisms are $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule homomorphisms. We further
define a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule isomorphism
as an isomorphism in this category; this is equivalent to the following definition:

\begin{definition}
\label{defs.semimodiso}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let
$V$ and $W$ be two $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules.
Let $f:V\rightarrow W$ be a $k$-linear map. Then, $f$ is said to be a $\left(
\mathfrak{g},\mathfrak{h}\right)  $\textit{-semimodule isomorphism} if and
only if $f$ is an invertible $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule homomorphism whose inverse $f^{-1}$ is also a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule homomorphism.
\end{definition}

We can easily prove that this definition is somewhat redundant, viz., the
condition that $f^{-1}$ be also a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule homomorphism can be omitted:

\begin{proposition}
\label{prop.semimodiso}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let
$V$ and $W$ be two $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules.
Let $f:V\rightarrow W$ be a $k$-linear map. Then, $f$ is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule isomorphism if and only if $f$
is an invertible $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule
homomorphism. In other words, $f$ is a $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule isomorphism if and only if $f$ is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule homomorphism and a $k$-module
isomorphism at the same time.
\end{proposition}

The proof of this proposition is identical with the proof of Proposition
\ref{prop.liemodiso}.

Clearly, Definition \ref{defs.semimodiso} is the analogue of Definition
\ref{defs.liemodiso} for $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules. Thus:

\begin{proposition}
Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie algebra. Let
$\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $V$ and $W$ be two
$\mathfrak{g}$-modules. Let $f:V\rightarrow W$ be a map. Then, $f$ is a
$\mathfrak{g}$-module isomorphism if and only if $f$ is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule isomorphism. (Here, it makes
sense to say that "$f$ is a $\mathfrak{g}$-module isomorphism" since $V$ and
$W$ are $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules (which is
because Proposition \ref{prop.semimod.mod} yields that every $\mathfrak{g}%
$-module is a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule).)
\end{proposition}

Also, Proposition \ref{prop.semimodiso} is the analogue of Proposition
\ref{prop.liemodiso} for $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules.
\end{verlong}

\subsection{Restriction of $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules}

If $\mathfrak{h}$ is a Lie subalgebra of a $k$-Lie algebra $\mathfrak{g}$,
then we can canonically make every $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule into an $\mathfrak{h}$-module according to the following definition:

\begin{definition}
\label{defs.semi.restrict}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a $k$-Lie algebra, and let $\mathfrak{h}$ be a Lie subalgebra of
$\mathfrak{g}$. Then, every $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule $V$ canonically becomes an $\mathfrak{h}$-module (by restricting
its Lie action $\mu:\mathfrak{g}\times V\rightarrow V$ to $\mathfrak{h}\times
V$). This $\mathfrak{h}$-module is called the \textit{restriction of }$V$
\textit{to }$\mathfrak{h}$, and denoted by $\operatorname*{Res}%
\nolimits_{\mathfrak{h}}^{\mathfrak{g}}V$. However, when there is no
possibility of confusion, we will denote this $\mathfrak{h}$-module by $V$,
and we will distinguish it from the original $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule $V$ by means of referring to the former one
as "the $\mathfrak{h}$-module $V$" and referring to the latter one as "the
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule $V$".
\end{definition}

This Definition \ref{defs.semi.restrict} is the analogue of Definition
\ref{defs.g-as-h-module} for $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules. Therefore:

\begin{proposition}
Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie algebra, and
let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $V$ be a
$\mathfrak{g}$-module. Then, the restriction of $V$ to $\mathfrak{h}$ defined
in Definition \ref{defs.g-as-h-module} is the same $\mathfrak{h}$-module as
the restriction of $V$ to $\mathfrak{h}$ defined in Definition
\ref{defs.semi.restrict} (which is well-defined since $V$ is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule (which is because Proposition
\ref{prop.semimod.mod} says that every $\mathfrak{g}$-module is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule)). This allows us to speak of
"the restriction of $V$ to $\mathfrak{h}$" (or simply of "the $\mathfrak{h}%
$-module $V$") without having to worry whether it is understood according to
Definition \ref{defs.g-as-h-module} or according to Definition
\ref{defs.semi.restrict} (because it doesn't matter, as both definitions yield
the same result).
\end{proposition}

\begin{remark}
\label{rmk.semi.restrict}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a $k$-Lie algebra, and let $\mathfrak{h}$ be a Lie subalgebra of
$\mathfrak{g}$. Let $V$ be a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule. Then, Definition \ref{defs.semi.restrict} makes $V$ into an
$\mathfrak{h}$-module. Hence, $V$ is a $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule and an $\mathfrak{h}$-module at the same time. As we
know from Warning \ref{warn.semimod.->}, we are normally \textit{not} allowed
to use Convention \ref{conv.semimod.->} when we have a $k$-module $V$ which is
a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule and an
$\mathfrak{h}$-module at the same time, because in this case each of the two
Conventions \ref{conv.semimod.->} and \ref{conv.->} defines $a\rightharpoonup
v$ for $a\in\mathfrak{h}$ and $v\in V$, and these definitions might conflict.
However, in \textit{our} case (the case when the $\mathfrak{h}$-module $V$ is
obtained from the $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule $V$
according to Definition \ref{defs.semi.restrict}), these definitions cannot
conflict, because every $a\in\mathfrak{h}$ and $v\in V$ satisfy%
\begin{align*}
&  \left(  \text{the meaning of the term }a\rightharpoonup v\text{ according
to Convention \ref{conv.semimod.->}}\right) \\
&  =\left(  \text{the meaning of the term }a\rightharpoonup v\text{ according
to Convention \ref{conv.->}}\right)
\end{align*}
{\footnotemark}, so that both Conventions \ref{conv.semimod.->} and
\ref{conv.->} define $a\rightharpoonup v$ to mean one and the same value.
Therefore, we can use Convention \ref{conv.semimod.->} in our case (the case
when the $\mathfrak{h}$-module $V$ is obtained from the $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule $V$ according to Definition
\ref{defs.semi.restrict}) without worrying that it might conflict with
Convention \ref{conv.->}.
\end{remark}

\footnotetext{\textit{Proof.} Let us denote by $\mu$ the Lie action of the
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule $V$. Then, the Lie
action of the $\mathfrak{h}$-module $V$ is defined to be $\mu\mid
_{\mathfrak{h}\times V}$ (in fact, this is how the $\mathfrak{h}$-module $V$
was defined in Definition \ref{defs.semi.restrict}). Now,%
\begin{align*}
&  \left(  \text{the meaning of the term }a\rightharpoonup v\text{ according
to Convention \ref{conv.semimod.->}}\right) \\
&  =\left(  \text{the Lie action of the }\left(  \mathfrak{g},\mathfrak{h}%
\right)  \text{-semimodule }V\text{ applied to }\left(  a,v\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{according to Convention
\ref{conv.semimod.->}}\right) \\
&  =\mu\left(  a,v\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Lie
action of the }\left(  \mathfrak{g},\mathfrak{h}\right)  \text{-semimodule
}V\text{ is }\mu\right) \\
&  =\left(  \mu\mid_{\mathfrak{h}\times V}\right)  \left(  a,v\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }a\in\mathfrak{h}\text{ and }v\in
V\text{, so that }\left(  a,v\right)  \in\mathfrak{h}\times V\right) \\
&  =\left(  \text{the Lie action of the }\mathfrak{h}\text{-module }V\text{
applied to }\left(  a,v\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{because }\mu\mid_{\mathfrak{h}\times
V}\text{ is the Lie action of the }\mathfrak{h}\text{-module }V\right) \\
&  =\left(  \text{the meaning of the term }a\rightharpoonup v\text{ according
to Convention \ref{conv.->}}\right)  ,
\end{align*}
qed.}In Definition \ref{defs.semi.restrict}, we have defined the restriction
of a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule to an
$\mathfrak{h}$-module. We could also define a more general kind of
restriction, namely restriction of a $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule to a $\left(  \mathfrak{g}^{\prime},\mathfrak{h}%
^{\prime}\right)  $-semimodule (where $\mathfrak{g}^{\prime}$ is a Lie
subalgebra of $\mathfrak{g}$ and where $\mathfrak{h}^{\prime}$ is a Lie
subalgebra of $\mathfrak{h}$), but we will not find need for this form of
restriction in the following and therefore overlook its (trivial) definition.

\subsection{Subsemimodules, factors and direct sums of $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules}

From Proposition \ref{prop.semimod.mod}, we know that every $\mathfrak{g}%
$-module is a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule (where
$\mathfrak{g}$ is a $k$-Lie algebra, and $\mathfrak{h}$ is a Lie subalgebra of
$\mathfrak{g}$). This gives us plenty of examples of $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules: For instance, we know (from Definition
\ref{defs.g-mod-g}) that $\mathfrak{g}$ itself canonically is a $\mathfrak{g}%
$-module, and thus we conclude that $\mathfrak{g}$ is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule. Also we know (from Definition
\ref{defs.k-mod-g}) that $k$ canonically is a $\mathfrak{g}$-module, and thus
we conclude that $k$ is a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule.

\begin{vershort}
To obtain more $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule
structures, we can define the notion of a $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $\textit{-subsemimodule} of a $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule (by analogy with Definition \ref{defs.submod}
\textbf{(a)}). Then, the quotient $k$-module of any $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule by a $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-subsemimodule canonically becomes a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule (similarly to Definition \ref{defs.submod}
\textbf{(b)}). The obvious analogue of Proposition \ref{prop.lie(+)} for
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules holds; thus, we can
define the direct sum of two $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules (by analogy to Definition \ref{defs.lie(+)}). Also, we can
define the direct sum of arbitrarily many $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodules (by analogy to Definition \ref{defs.lie(++)}). The
obvious analogues of Proposition \ref{prop.lie(++).inj} and Proposition
\ref{prop.lie.(+)vsRes} for $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules are satisfied as well.
\end{vershort}

\begin{verlong}
To obtain more $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule
structures, we can factor existing $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules by submodules:

\begin{definition}
\label{defs.subsemimod}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $V$
be a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule.\newline%
\textbf{(a)} A $k$-submodule $W$ of $V$ is said to be a $\left(
\mathfrak{g},\mathfrak{h}\right)  $\textit{-subsemimodule} of $V$ if and only
if%
\[
\left(  a\rightharpoonup w\in W\text{ for every }a\in\mathfrak{g}\text{ and
}w\in W\right)  .
\]
In other words, a $k$-submodule $W$ of $V$ is said to be a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-subsemimodule of $V$ if and only if
$\mu\left(  \mathfrak{g}\times W\right)  \subseteq W$, where $\mu$ denotes the
Lie action of $V$. (We remind ourselves that the Lie action of $V$ means the
$k$-bilinear map $\mu:\mathfrak{g}\times V\rightarrow V$ from Definition
\ref{defs.semimod}.)\newline\textbf{(b)} If $W$ is a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-subsemimodule of $V$, then the quotient $k$-module
$V\diagup W$ becomes a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule
by setting%
\[
\left(  a\rightharpoonup\overline{v}=\overline{a\rightharpoonup v}\text{ for
every }a\in\mathfrak{g}\text{ and }v\in V\right)
\]
(where $\overline{u}$ denotes the residue class of $u$ modulo $W$ for every
$u\in V$). (This $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule
structure is indeed well-defined, as can be easily seen.)
\end{definition}

This Definition \ref{defs.subsemimod} is the analogue of Definition
\ref{defs.submod} for $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules. This immediately yields:

\begin{proposition}
Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a Lie algebra. Let
$\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$.\newline\textbf{(a)} Let
$V$ be a $\mathfrak{g}$-module. Let $W$ be a subset of $V$. Then, $W$ is a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-subsemimodule of $V$ if and only
if $W$ is a $\mathfrak{g}$-submodule of $V$. (Here, the words "$W$ is a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-subsemimodule" make sense because
$V$ is a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule (which is
because Proposition \ref{prop.semimod.mod} yields that every $\mathfrak{g}%
$-module is a $\left(  \mathfrak{g},\mathfrak{h}\right)  $%
-semimodule).)\newline\textbf{(b)} Let $V$ be a $\mathfrak{g}$-module, and let
$W$ be a $\mathfrak{g}$-submodule of $V$. Then, the quotient $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule $V\diagup W$ defined in
Definition \ref{defs.subsemimod} \textbf{(b)} is identic with the quotient
$\mathfrak{g}$-module $V\diagup W$ defined in Definition \ref{defs.submod}
\textbf{(b)}. This allows us to speak of the $\mathfrak{g}$-module $V\diagup
W$ without having to worry whether it is understood according to Definition
\ref{defs.subsemimod} or according to Definition \ref{defs.submod} (because it
doesn't matter, as both definitions yield the same result).
\end{proposition}

We can also add $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules via
the direct sum:

\begin{proposition}
\label{prop.semi(+)}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a Lie
algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $V$ and
$W$ be two $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules. Define a
map $\mu_{V\oplus W}:\mathfrak{g}\times\left(  V\oplus W\right)  \rightarrow
V\oplus W$ by%
\begin{equation}
\left(  \mu_{V\oplus W}\left(  a,\left(  v,w\right)  \right)  =\left(
a\rightharpoonup v,a\rightharpoonup w\right)  \ \ \ \ \ \ \ \ \ \ \text{for
every }a\in\mathfrak{g}\text{, }v\in V\text{ and }w\in W\right)  .
\label{prop.semi(+).cond}%
\end{equation}
Then, this map $\mu_{V\oplus W}$ is $k$-bilinear, and $\left(  V\oplus
W,\mu_{V\oplus W}\right)  $ is a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule satisfying%
\begin{equation}
a\rightharpoonup\left(  v,w\right)  =\left(  a\rightharpoonup
v,a\rightharpoonup w\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }%
a\in\mathfrak{g}\text{, }v\in V\text{ and }w\in W. \label{prop.semi(+).form}%
\end{equation}

\end{proposition}

This proposition is straightforward to prove, so we are not going to elaborate
on its proof. Anyway it allows a definition:

\begin{definition}
\label{defs.semi(+)}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a Lie
algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $V$ and
$W$ be two $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules.\newline
The $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule $\left(  V\oplus
W,\mu_{V\oplus W}\right)  $ constructed in Proposition \ref{prop.semi(+)} is
called the \textit{direct sum} of the $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodules $V$ and $W$. We are going to denote this $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule $\left(  V\oplus W,\mu_{V\oplus
W}\right)  $ simply by $V\oplus W$.
\end{definition}

Note that Proposition \ref{prop.semi(+)} is the analogue of Proposition
\ref{prop.lie(+)} for $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules. Also, Definition \ref{defs.semi(+)} is the analogue of
Definition \ref{defs.lie(+)} for $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules. This yields:

\begin{proposition}
Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie algebra, and
let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $V$ and $W$ be
two $\mathfrak{g}$-modules. Then, the $\mathfrak{g}$-module $V\oplus W$
defined in Definition \ref{defs.lie(+)} is identic with the $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule $V\oplus W$ defined in
Definition \ref{defs.semi(+)} (which is well-defined since $V$ and $W$ are
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules (which is because
Proposition \ref{prop.semimod.mod} says that every $\mathfrak{g}$-module is a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule)). This allows us to
speak of "the $\mathfrak{g}$-module $V\oplus W$" without having to worry
whether it is understood according to Definition \ref{defs.lie(+)} or
according to Definition \ref{defs.semi(+)} (because it doesn't matter, as both
definitions yield the same result).
\end{proposition}

We can similarly define the direct sum of several (not necessarily just two)
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules:

\begin{proposition}
\label{prop.semi(++)}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $S$
be a set. For every $s\in S$, let $V_{s}$ be a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule. Define a map $\mu_{\oplus}:\mathfrak{g}%
\times\left(  \bigoplus\limits_{s\in S}V_{s}\right)  \rightarrow
\bigoplus\limits_{s\in S}V_{s}$ by%
\begin{equation}
\left(  \mu_{\oplus}\left(  a,\left(  v_{s}\right)  _{s\in S}\right)  =\left(
a\rightharpoonup v_{s}\right)  _{s\in S}\ \ \ \ \ \ \ \ \ \ \text{for every
}a\in\mathfrak{g}\text{ and every family }\left(  v_{s}\right)  _{s\in S}%
\in\bigoplus\limits_{s\in S}V_{s}\right)  . \label{prop.semi(++).cond}%
\end{equation}
Then, this map $\mu_{\oplus}$ is $k$-bilinear, and $\left(  \bigoplus
\limits_{s\in S}V_{s},\mu_{\oplus}\right)  $ is a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule satisfying%
\begin{equation}
a\rightharpoonup\left(  v_{s}\right)  _{s\in S}=\left(  a\rightharpoonup
v_{s}\right)  _{s\in S}\ \ \ \ \ \ \ \ \ \ \text{for every }a\in
\mathfrak{g}\text{ and every family }\left(  v_{s}\right)  _{s\in S}%
\in\bigoplus\limits_{s\in S}V_{s}. \label{prop.semi(++).form}%
\end{equation}

\end{proposition}

\begin{definition}
\label{defs.semi(++)}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $S$
be a set. For every $s\in S$, let $V_{s}$ be a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule.\newline The $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule $\left(  \bigoplus\limits_{s\in S}V_{s}%
,\mu_{\oplus}\right)  $ constructed in Proposition \ref{prop.semi(++)} is
called the \textit{direct sum} of the $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodules $V_{s}$ over all $s\in S$. We are going to denote this
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule $\left(
\bigoplus\limits_{s\in S}V_{s},\mu_{\oplus}\right)  $ simply by $\bigoplus
\limits_{s\in S}V_{s}$.
\end{definition}

Again, there is nothing substantial to prove here. Notice that if
$S=\varnothing$, then $\bigoplus\limits_{s\in S}V_{s}$ is to be understood as
$0$.

Note that Proposition \ref{prop.semi(++)} is the analogue of Proposition
\ref{prop.lie(++)} for $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules, and that Definition \ref{defs.semi(++)} is the analogue of
Definition \ref{defs.lie(++)} for $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules. This yields:

\begin{proposition}
Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie algebra, and
let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $S$ be a set.
For every $s\in S$, let $V_{s}$ be a $\mathfrak{g}$-module.\newline Then, the
$\mathfrak{g}$-module $\bigoplus\limits_{s\in S}V_{s}$ defined in Definition
\ref{defs.lie(++)} is identic with the $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule $\bigoplus\limits_{s\in S}V_{s}$ defined in Definition
\ref{defs.semi(++)} (which is well-defined since $V_{s}$ is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule for every $s\in S$ (which is
because Proposition \ref{prop.semimod.mod} says that every $\mathfrak{g}%
$-module is a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule)). This
allows us to speak of "the $\mathfrak{g}$-module $\bigoplus\limits_{s\in
S}V_{s}$" without having to worry whether it is understood according to
Definition \ref{defs.lie(++)} or according to Definition \ref{defs.semi(++)}
(because it doesn't matter, as both definitions yield the same result).
\end{proposition}

Proposition \ref{prop.lie(++).inj} also has an analogue for $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodules:

\begin{proposition}
\label{prop.semi(++).inj}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let
$S$ be a set. For every $s\in S$, let $V_{s}$ be a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule. In Convention \ref{conv.(++).inj}, we have
identified the $k$-module $V_{t}$ with the image of $V_{t}$ under the
canonical injection $V_{t}\rightarrow\bigoplus\limits_{s\in S}V_{s}$ for every
$t\in S$. Thus, by means of this identification, $V_{t}$ becomes a
$k$-submodule of the direct sum $\bigoplus\limits_{s\in S}V_{s}$. But
actually, something stronger holds: By means of this identification, $V_{t}$
becomes a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-subsemimodule of the
direct sum $\bigoplus\limits_{s\in S}V_{s}$.
\end{proposition}

There is also a natural generalization of Proposition \ref{prop.lie.(+)vsRes}
to $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules:

\begin{proposition}
\label{prop.semi.(+)vsRes}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$.\newline\textbf{(a)} If $V$ and $W$ are two $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules, then $\operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}\left(  V\oplus W\right)  =\left(  \operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}V\right)  \oplus\left(  \operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}W\right)  $ as $\mathfrak{h}$-modules. This
allows us to speak of "the $\mathfrak{h}$-module $V\oplus W$" without having
to worry whether we mean $\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}%
}\left(  V\oplus W\right)  $ or $\left(  \operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}V\right)  \oplus\left(  \operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}W\right)  $ (because it does not matter, since
$\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}\left(  V\oplus W\right)
=\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V\right)
\oplus\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}W\right)
$).\newline\textbf{(b)} If $S$ is a set, and if $V_{t}$ is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule for every $t\in S$, then
$\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}\left(  \bigoplus
\limits_{s\in S}V_{s}\right)  =\bigoplus\limits_{s\in S}\left(
\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V_{s}\right)  $ as
$\mathfrak{h}$-modules. This allows us to speak of "the $\mathfrak{h}$-module
$\bigoplus\limits_{s\in S}V_{s}$" without having to worry whether we mean
$\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}\left(  \bigoplus
\limits_{s\in S}V_{s}\right)  $ or $\bigoplus\limits_{s\in S}\left(
\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V_{s}\right)  $ (because it
does not matter, since $\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}%
}\left(  \bigoplus\limits_{s\in S}V_{s}\right)  =\bigoplus\limits_{s\in
S}\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V_{s}\right)
$).\newline\textbf{(c)} The $\mathfrak{h}$-module $k$ is identical with the
restriction $\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}k$ of the
$\mathfrak{g}$-module $k$ to $\mathfrak{h}$.\newline\textbf{(d)} If $V$ is a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule, and if $W$ is a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-subsemimodule of $V$, then
$\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}W$ is an $\mathfrak{h}%
$-submodule of the $\mathfrak{h}$-module $\operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}V$ and satisfies $\operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}\left(  V\diagup W\right)  =\left(  \operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}V\right)  \diagup\left(  \operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}W\right)  $ as $\mathfrak{h}$-modules. This
allows us to speak of "the $\mathfrak{h}$-module $V\diagup W$" without having
to worry whether we mean $\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}%
}\left(  V\diagup W\right)  $ or $\left(  \operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}V\right)  \diagup\left(  \operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}W\right)  $ (because it does not matter, since
$\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}\left(  V\diagup W\right)
=\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V\right)
\diagup\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}W\right)  $).
\end{proposition}

Just as Proposition \ref{prop.lie.(+)vsRes}, this generalization will often be
tacitly used.
\end{verlong}

\subsection{Tensor products of two $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules}

\begin{vershort}
Proposition \ref{prop.lie(X)} can be generalized to $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules (with the same proof, up to obvious
modifications). Therefore, we can define the tensor product of two $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodules (in analogy with Definition
\ref{defs.lie(X)}). Generalizations of Proposition \ref{prop.tensor-assoc},
Proposition \ref{prop.lie(X)f(X)g}, and Proposition \ref{prop.lie.(X)vsRes} to
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules are just as easy to
obtain (the proofs work the same way as those of the original propositions).
\end{vershort}

\begin{verlong}
We can define tensor products of $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules:

\begin{proposition}
\label{prop.semi(X)}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a Lie
algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $V$ and
$W$ be two $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules. Then,
there exists one and only one $k$-bilinear map $m:\mathfrak{g}\times\left(
V\otimes W\right)  \rightarrow V\otimes W$ which satisfies%
\begin{equation}
\left(  m\left(  a,v\otimes w\right)  =\left(  a\rightharpoonup v\right)
\otimes w+v\otimes\left(  a\rightharpoonup w\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{, }v\in V\text{ and
}w\in W\right)  . \label{prop.semi(X).cond}%
\end{equation}
If we denote this map $m$ by $\mu_{V\otimes W}$, then $\left(  V\otimes
W,\mu_{V\otimes W}\right)  $ is a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule. This $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule
satisfies%
\begin{equation}
a\rightharpoonup\left(  v\otimes w\right)  =\left(  a\rightharpoonup v\right)
\otimes w+v\otimes\left(  a\rightharpoonup w\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{, }v\in V\text{ and
}w\in W. \label{prop.semi(X).form}%
\end{equation}

\end{proposition}

\begin{definition}
\label{defs.semi(X)}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a Lie
algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $V$ and
$W$ be two $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules.\newline
The $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule $\left(  V\otimes
W,\mu_{V\otimes W}\right)  $ constructed in Proposition \ref{prop.semi(X)} is
called the \textit{tensor product} of the $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodules $V$ and $W$. We are going to denote this $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule $\left(  V\otimes
W,\mu_{V\otimes W}\right)  $ simply by $V\otimes W$.
\end{definition}

Obviously, Proposition \ref{prop.semi(X)} is a generalization of Proposition
\ref{prop.lie(X)} to $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules,
and Definition \ref{defs.semi(X)} is a generalization of Definition
\ref{defs.lie(X)} to $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules.
This yields:

\begin{proposition}
Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie algebra, and
let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $V$ and $W$ be
two $\mathfrak{g}$-modules. Then, the $\mathfrak{g}$-module $V\otimes W$
defined in Definition \ref{defs.lie(X)} is identic with the $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule $V\otimes W$ defined in
Definition \ref{defs.semi(X)} (which is well-defined since $V$ and $W$ are
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules (which is because
Proposition \ref{prop.semimod.mod} says that every $\mathfrak{g}$-module is a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule)). This allows us to
speak of "the $\mathfrak{g}$-module $V\otimes W$" without having to worry
whether it is understood according to Definition \ref{defs.lie(X)} or
according to Definition \ref{defs.semi(X)} (because it doesn't matter, as both
definitions yield the same result).
\end{proposition}

\textit{Proof of Proposition \ref{prop.semi(X)}.} In order to obtain a proof
of Proposition \ref{prop.semi(X)}, it is enough to apply the following changes
to the proof of Proposition \ref{prop.lie(X)}:

\begin{itemize}
\item Replace "Proposition \ref{prop.lie(X)}" by "Proposition
\ref{prop.semi(X)}".

\item Replace "Definition \ref{defs.liemod}" by "Definition \ref{defs.semimod}".

\item Replace the words "$\mathfrak{g}$-module" by "$\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule".

\item Replace the references to (\ref{prop.lie(X).cond}) by references to
(\ref{prop.semi(X).cond}).

\item Replace the references to (\ref{prop.lie(X).form}) by references to
(\ref{prop.semi(X).form}).

\item In the proof of Assertion $\gamma$, replace "$a\in\mathfrak{g}$" by
"$a\in\mathfrak{h}$" (except in the formula (\ref{pf.lie(X).gamma.1})).

\item Replace the references to (\ref{[L-mod]}) by references to
(\ref{defs.semimod.->}).
\end{itemize}

This completes the proof of Proposition \ref{prop.semi(X)}.

The next proposition is a generalization of Proposition
\ref{prop.tensor-assoc} to $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules:

\begin{proposition}
\label{prop.semi.tensor-assoc}Let $k$ be a commutative ring. Let
$\mathfrak{g}$ be a $k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of
$\mathfrak{g}$.\newline\textbf{(a)} Let $V$ be a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule. Then, the $k$-linear map%
\[
V\rightarrow k\otimes V,\ \ \ \ \ \ \ \ \ \ v\mapsto1\otimes v
\]
is a canonical isomorphism of $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules. (Here, as usual, $k$ denotes the $\mathfrak{g}$-module $k$
defined in Definition \ref{defs.k-mod-g}.)\newline\textbf{(b)} Let $V$ be a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule. Then, the $k$-linear
map%
\[
V\rightarrow V\otimes k,\ \ \ \ \ \ \ \ \ \ v\mapsto v\otimes1
\]
is a canonical isomorphism of $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules. (Here, as usual, $k$ denotes the $\mathfrak{g}$-module $k$
defined in Definition \ref{defs.k-mod-g}.)\newline\textbf{(c)} Let $U$, $V$
and $W$ be $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules. Then, the
$k$-linear map%
\[
\left(  U\otimes V\right)  \otimes W\rightarrow U\otimes\left(  V\otimes
W\right)  ,\ \ \ \ \ \ \ \ \ \ \left(  u\otimes v\right)  \otimes w\mapsto
u\otimes\left(  v\otimes w\right)
\]
is a canonical isomorphism of $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules.
\end{proposition}

\textit{Proof of Proposition \ref{prop.semi.tensor-assoc}.} In order to obtain
a proof of Proposition \ref{prop.semi.tensor-assoc}, it is enough to apply the
following changes to the proof of Proposition \ref{prop.tensor-assoc}:

\begin{itemize}
\item Replace "Proposition \ref{prop.tensor-assoc}" by "Proposition
\ref{prop.semi.tensor-assoc}".

\item Replace the words "$\mathfrak{g}$-module" by "$\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule".

\item Replace the references to (\ref{prop.lie(X).form}) by references to
(\ref{prop.semi(X).form}).
\end{itemize}

This completes the proof of Proposition \ref{prop.semi.tensor-assoc}.

We also have:

\begin{corollary}
\label{cor.semi.k(X)V}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$.\newline Let $V$ be a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule. Then, the canonical $k$-module isomorphism $k\otimes
V\rightarrow V$ (this is the $k$-module homomorphism that sends $\lambda
\otimes v$ to $\lambda v$ for all $\lambda\in k$ and $v\in V$) is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule isomorphism.
\end{corollary}

This Corollary \ref{cor.semi.k(X)V} is the generalization of Corollary
\ref{cor.k(X)V} to $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules.

\textit{Proof of Corollary \ref{cor.semi.k(X)V}.} In order to obtain a proof
of Corollary \ref{cor.semi.k(X)V}, it is enough to apply the following changes
to the proof of Corollary \ref{cor.k(X)V}:

\begin{itemize}
\item Replace "Proposition \ref{prop.tensor-assoc}" by "Proposition
\ref{prop.semi.tensor-assoc}".

\item Replace "Corollary \ref{cor.k(X)V}" by "Corollary \ref{cor.semi.k(X)V}".

\item Replace the words "$\mathfrak{g}$-module" by "$\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule".
\end{itemize}

This completes the proof of Corollary \ref{cor.semi.k(X)V}.

Next we formulate a generalization of Proposition \ref{prop.lie(X)f(X)g} to
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules:

\begin{proposition}
\label{prop.semi(X)f(X)g}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a $k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$.\newline Let $V$, $W$, $V^{\prime}$ and $W^{\prime}$ be four $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodules, and let $f:V\rightarrow
V^{\prime}$ and $g:W\rightarrow W^{\prime}$ be two $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule homomorphisms. Then, $f\otimes g:V\otimes
W\rightarrow V^{\prime}\otimes W^{\prime}$ is a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule homomorphism.
\end{proposition}

\textit{Proof of Proposition \ref{prop.semi(X)f(X)g}.} In order to obtain a
proof of Proposition \ref{prop.semi(X)f(X)g}, it is enough to apply the
following changes to the proof of Proposition \ref{prop.lie(X)f(X)g}:

\begin{itemize}
\item Replace "Proposition \ref{prop.lie(X)f(X)g}" by "Proposition
\ref{prop.semi(X)f(X)g}".

\item Replace the references to (\ref{prop.lie(X).form}) by references to
(\ref{prop.semi(X).form}).

\item Replace the words "$\mathfrak{g}$-module" by "$\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule".
\end{itemize}

This completes the proof of Proposition \ref{prop.semi(X)f(X)g}.

We can also generalize Proposition \ref{prop.lie.(X)vsRes} to $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodules, obtaining the following proposition:

\begin{proposition}
\label{prop.semi.(X)vsRes}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$.\newline If $V$ and $W$ are two $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules, then $\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}\left(
V\otimes W\right)  =\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}%
}V\right)  \otimes\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}%
}W\right)  $ as $\mathfrak{h}$-modules. This allows us to speak of "the
$\mathfrak{h}$-module $V\otimes W$" without having to worry whether we mean
$\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}\left(  V\otimes W\right)  $
or $\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V\right)
\otimes\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}W\right)  $
(because it does not matter, since $\operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}\left(  V\otimes W\right)  =\left(  \operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}V\right)  \otimes\left(  \operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}W\right)  $).
\end{proposition}

This follows from the definitions.
\end{verlong}

\subsection{Tensor products of several $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodules}

\begin{vershort}
We can also define the tensor product of $n$ arbitrary $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules (in the same way as we did it for
$\mathfrak{g}$-modules in Definition \ref{defs.lie(X)mult}). The obvious
analogues of Propositions \ref{prop.lie(X).bracketing}, \ref{prop.lie(X)mor},
\ref{prop.lie.(X)multvsRes} and Remarks \ref{rmk.lie(X)mult} and
\ref{rmk.lie(X)mult.replace.short} still continue to hold for $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodules.
\end{vershort}

\begin{verlong}
We now define the tensor product of several $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules:

\begin{definition}
\label{defs.semi(X)mult}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let
$n\in\mathbb{N}$.\newline Now, by induction over $n$, we are going to define a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule $V_{1}\otimes
V_{2}\otimes...\otimes V_{n}$ for any $n$ arbitrary $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules $V_{1}$, $V_{2}$, $...$, $V_{n}$%
:\newline\textit{Induction base:} For $n=0$, we define $V_{1}\otimes
V_{2}\otimes...\otimes V_{n}$ as the $\mathfrak{g}$-module $k$ defined in
Definition \ref{defs.k-mod-g}.\newline\textit{Induction step:} Let
$p\in\mathbb{N}$. Assuming that we have defined a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule $V_{1}\otimes V_{2}\otimes...\otimes V_{p}$
for any $p$ arbitrary $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules
$V_{1}$, $V_{2}$, $...$, $V_{p}$, we now define a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule $V_{1}\otimes V_{2}\otimes...\otimes
V_{p+1}$ for any $p+1$ arbitrary $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules $V_{1}$, $V_{2}$, $...$, $V_{p+1}$ by the equation%
\begin{equation}
V_{1}\otimes V_{2}\otimes...\otimes V_{p+1}=V_{1}\otimes\left(  V_{2}\otimes
V_{3}\otimes...\otimes V_{p+1}\right)  . \label{defs.semi(X)mult.left}%
\end{equation}
Here, $V_{1}\otimes\left(  V_{2}\otimes V_{3}\otimes...\otimes V_{p+1}\right)
$ is to be understood as the tensor product of the $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule $V_{1}$ with the $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule $V_{2}\otimes V_{3}\otimes...\otimes
V_{p+1}$ (note that the $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule $V_{2}\otimes V_{3}\otimes...\otimes V_{p+1}$ is already defined
because we assumed that we have defined a $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule $V_{1}\otimes V_{2}\otimes...\otimes V_{p}$ for any $p$
arbitrary $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules $V_{1}$,
$V_{2}$, $...$, $V_{p}$). This completes the inductive definition.\newline
Thus we have defined a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule
$V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ for any $n$ arbitrary $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodules $V_{1}$, $V_{2}$, $...$,
$V_{n}$ for any $n\in\mathbb{N}$. This $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ is called
the \textit{tensor product} of the $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules $V_{1}$, $V_{2}$, $...$, $V_{n}$.
\end{definition}

This Definition \ref{defs.semi(X)mult} is the obvious generalization of
Definition \ref{defs.lie(X)mult} to $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules. Therefore:

\begin{proposition}
Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie algebra, and
let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $n\in\mathbb{N}%
$. Let $V_{1}$, $V_{2}$, $...$, $V_{n}$ be $n$ arbitrary $\mathfrak{g}%
$-modules. Then, the $\mathfrak{g}$-module $V_{1}\otimes V_{2}\otimes
...\otimes V_{n}$ defined in Definition \ref{defs.lie(X)mult} is identic with
the $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule $V_{1}\otimes
V_{2}\otimes...\otimes V_{n}$ defined in Definition \ref{defs.semi(X)mult}
(which is well-defined since $V_{1}$, $V_{2}$, $...$, $V_{n}$ are $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodules (which is because Proposition
\ref{prop.semimod.mod} says that every $\mathfrak{g}$-module is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule)). This allows us to speak of
"the $\mathfrak{g}$-module $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$"
without having to worry whether it is understood according to Definition
\ref{defs.lie(X)mult} or according to Definition \ref{defs.semi(X)mult}
(because it doesn't matter, as both definitions yield the same result).
\end{proposition}

\begin{remark}
\textbf{(a)} In Definition \ref{defs.semi(X)mult}, we could have replaced the
equation (\ref{defs.semi(X)mult.left}) by%
\[
V_{1}\otimes V_{2}\otimes...\otimes V_{p+1}=\left(  V_{1}\otimes V_{2}%
\otimes...\otimes V_{p}\right)  \otimes V_{p+1}.
\]
This would have given us a \textit{different} $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$
for any $n$ arbitrary $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules
$V_{1}$, $V_{2}$, $...$, $V_{n}$ for any $n\in\mathbb{N}$ than the one defined
in Definition \ref{defs.semi(X)mult}. However, this $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule would still be \textit{canonically
isomorphic} to the one defined in Definition \ref{defs.semi(X)mult} (we will
prove this and actually something more general in Proposition
\ref{prop.semi(X).bracketing}), and thus it is commonly considered to be "more
or less the same $\left(  \mathfrak{g},\mathfrak{h}\right)  $%
-semimodule".\newline\textbf{(b)} Definition \ref{defs.semi(X)mult}, applied
to $n=1$, defines the tensor product of \textit{one} $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule $V_{1}$ as $V_{1}\otimes k$. This takes
some getting used to, since it seems more natural to define the tensor product
of one $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule $V_{1}$ simply
as $V_{1}$. But this isn't really different because Proposition
\ref{prop.semi.tensor-assoc} \textbf{(b)} gives a canonical isomorphism of
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules $V_{1}\cong
V_{1}\otimes k$, so most people consider $V_{1}$ to be "more or less the same
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule" as $V_{1}\otimes
k$.\newline\textbf{(c)} Definition \ref{defs.semi(X)mult} does not conflict
with Definition \ref{defs.(X)mult}, because the underlying $k$-module of the
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule $V_{1}\otimes
V_{2}\otimes...\otimes V_{n}$ defined in Definition \ref{defs.semi(X)mult} is
indeed the $k$-module $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ defined in
Definition \ref{defs.(X)mult}. (This is trivial by induction.)
\end{remark}

Now we formulate the generalization of Proposition
\ref{prop.lie(X).bracketing} to $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules:

\begin{proposition}
\label{prop.semi(X).bracketing}Let $k$ be a commutative ring. Let
$\mathfrak{g}$ be a $k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of
$\mathfrak{g}$. Let $n\in\mathbb{N}$.\newline Then, for any $n$ arbitrary
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules $V_{1}$, $V_{2}$,
$...$, $V_{n}$ and every $i\in\left\{  0,1,...,n\right\}  $, the canonical
$k$-module isomorphism $\left(  V_{1}\otimes V_{2}\otimes...\otimes
V_{i}\right)  \otimes\left(  V_{i+1}\otimes V_{i+2}\otimes...\otimes
V_{n}\right)  \rightarrow V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ is a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule
isomorphism.{\footnotemark}
\end{proposition}

\footnotetext{Here and in the following, whenever I speak of "the canonical
$k$-module isomorphism $\left(  V_{1}\otimes V_{2}\otimes...\otimes
V_{i}\right)  \otimes\left(  V_{i+1}\otimes V_{i+2}\otimes...\otimes
V_{n}\right)  \rightarrow V_{1}\otimes V_{2}\otimes...\otimes V_{n}$", I mean
the $k$-module homomorphism $\left(  V_{1}\otimes V_{2}\otimes...\otimes
V_{i}\right)  \otimes\left(  V_{i+1}\otimes V_{i+2}\otimes...\otimes
V_{n}\right)  \rightarrow V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ which
sends $\left(  v_{1}\otimes v_{2}\otimes...\otimes v_{i}\right)
\otimes\left(  v_{i+1}\otimes v_{i+2}\otimes...\otimes v_{n}\right)  $ to
$v_{1}\otimes v_{2}\otimes...\otimes v_{n}$ for every $\left(  v_{1}%
,v_{2},...,v_{n}\right)  \in V_{1}\times V_{2}\times...\times V_{n}$. This
homomorphism is known (from linear algebra) to exist, be unique and be a
$k$-module isomorphism. (This is independent of the $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule structures on $V_{1}$, $V_{2}$, $...$,
$V_{n}$.)}

\textit{Proof of Proposition \ref{prop.semi(X).bracketing}.} In order to
obtain a proof of Proposition \ref{prop.semi(X).bracketing}, it is enough to
apply the following changes to the proof of Proposition
\ref{prop.lie(X).bracketing}:

\begin{itemize}
\item Replace "Proposition \ref{prop.lie(X).bracketing}" by "Proposition
\ref{prop.semi(X).bracketing}".

\item Replace "Proposition \ref{prop.tensor-assoc}" by "Proposition
\ref{prop.semi.tensor-assoc}".

\item Replace "Proposition \ref{prop.lie(X)f(X)g}" by "Proposition
\ref{prop.semi(X)f(X)g}".

\item Replace "Proposition \ref{prop.liemodiso}" by "Proposition
\ref{prop.semimodiso}".

\item Replace the words "$\mathfrak{g}$-module" by "$\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule".
\end{itemize}

This completes the proof of Proposition \ref{prop.semi(X).bracketing}.

\begin{vershort}
\begin{remark}
In Definition \ref{defs.semi(X)mult}, we defined the $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$
by induction over $n$. We can also describe its Lie action explicitly: Let $k$
be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie algebra. Let
$\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $n\in\mathbb{N}$.
Let $V_{1}$, $V_{2}$, $...$, $V_{n}$ be $n$ arbitrary $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules. Then, the $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$
defined in Definition \ref{defs.semi(X)mult} satisfies%
\[
a\rightharpoonup\left(  v_{1}\otimes v_{2}\otimes...\otimes v_{n}\right)
=\sum\limits_{i=1}^{n}\underbrace{v_{1}\otimes v_{2}\otimes...\otimes v_{i-1}%
}_{\substack{\text{tensor product of the}\\\text{first }i-1\text{ vectors
}v_{\ell}}}\otimes\left(  a\rightharpoonup v_{i}\right)  \otimes
\underbrace{v_{i+1}\otimes v_{i+2}\otimes...\otimes v_{n}}%
_{\substack{\text{tensor product of the}\\\text{last }n-i\text{ vectors
}v_{\ell}}}
\]
for every $a\in\mathfrak{g}$ and every $\left(  v_{1},v_{2},...,v_{n}\right)
\in V_{1}\times V_{2}\times...\times V_{n}$.
\end{remark}
\end{vershort}

In Definition \ref{defs.semi(X)mult}, we defined the $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$
by induction over $n$. It turns out that we can also easily describe the
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule structure on
$V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ explicitly:

\begin{proposition}
\label{prop.semi(X)mult.exp}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$. Let $n\in\mathbb{N}$. Let $V_{1}$, $V_{2}$, $...$, $V_{n}$ be $n$ arbitrary
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules. Then, the $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule $V_{1}\otimes V_{2}%
\otimes...\otimes V_{n}$ defined in Definition \ref{defs.semi(X)mult}
satisfies%
\[
a\rightharpoonup\left(  v_{1}\otimes v_{2}\otimes...\otimes v_{n}\right)
=\sum\limits_{i=1}^{n}v_{1}\otimes v_{2}\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup v_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
\ }\otimes...\otimes v_{n}%
\]
for every $a\in\mathfrak{g}$ and every $\left(  v_{1},v_{2},...,v_{n}\right)
\in V_{1}\times V_{2}\times...\times V_{n}$. Here, we are using Convention
\ref{conv.replace}.
\end{proposition}

This Proposition \ref{prop.semi(X)mult.exp} is the straightforward
generalization of Proposition \ref{prop.lie(X)mult.exp} to $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodules.

We are going to prove this later. First an obvious lemma, which generalizes
Proposition \ref{prop.lie(X)mult.left} to $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodules:

\begin{proposition}
\label{prop.semi(X)mult.left}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$. Let $n\in\mathbb{N}_{+}$. Let $V_{1}$, $V_{2}$, $...$, $V_{n}$ be $n$
arbitrary $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules. Then, the
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule $V_{1}\otimes
V_{2}\otimes...\otimes V_{n}$ defined in Definition \ref{defs.semi(X)mult}
satisfies%
\begin{equation}
a\rightharpoonup\left(  v\otimes T\right)  =\left(  a\rightharpoonup v\right)
\otimes T+v\otimes\left(  a\rightharpoonup T\right)
\label{prop.semi(X)mult.left.form}%
\end{equation}
for every $a\in\mathfrak{g}$, $v\in V_{1}$ and $T\in V_{2}\otimes V_{3}%
\otimes...\otimes V_{n}$.
\end{proposition}

\textit{Proof of Proposition \ref{prop.semi(X)mult.left}.} According to
Definition \ref{defs.semi(X)mult}, the $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule $V_{1}\otimes V_{2}\otimes...\otimes V_{n}$ is defined
as the tensor product $V_{1}\otimes\left(  V_{2}\otimes V_{3}\otimes...\otimes
V_{n}\right)  $. Hence, applying (\ref{prop.semi(X).form}) to $V=V_{1}$ and
$W=V_{2}\otimes V_{3}\otimes...\otimes V_{n}$, we see that%
\begin{align*}
a\rightharpoonup\left(  v\otimes w\right)   &  =\left(  a\rightharpoonup
v\right)  \otimes w+v\otimes\left(  a\rightharpoonup w\right) \\
&  \ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{, }v\in
V_{1}\text{ and }w\in V_{2}\otimes V_{3}\otimes...\otimes V_{n}%
\end{align*}
(because the tensor product $V\otimes W$ of two $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules $V$ and $W$ always satisfies
(\ref{prop.semi(X).form})). If we rename $w$ as $T$, this rewrites as follows:%
\begin{align*}
a\rightharpoonup\left(  v\otimes T\right)   &  =\left(  a\rightharpoonup
v\right)  \otimes T+v\otimes\left(  a\rightharpoonup T\right) \\
&  \ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{, }v\in
V_{1}\text{ and }T\in V_{2}\otimes V_{3}\otimes...\otimes V_{n}.
\end{align*}
This proves Proposition \ref{prop.semi(X)mult.left}.

\textit{Proof of Proposition \ref{prop.semi(X)mult.exp}.} In order to obtain a
proof of Proposition \ref{prop.semi(X)mult.exp}, it is enough to apply the
following changes to the proof of Proposition \ref{prop.lie(X)mult.exp}:

\begin{itemize}
\item Replace "Proposition \ref{prop.lie(X)mult.exp}" by "Proposition
\ref{prop.semi(X)mult.exp}".

\item Replace "Proposition \ref{prop.lie(X)mult.left}" by "Proposition
\ref{prop.semi(X)mult.left}".

\item Replace the words "$\mathfrak{g}$-module" by "$\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule".
\end{itemize}

This completes the proof of Proposition \ref{prop.semi(X)mult.exp}.

The next proposition is a multi-factor version of Proposition
\ref{prop.semi(X)f(X)g}, and generalizes Proposition \ref{prop.lie(X)mor} to
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules:

\begin{proposition}
\label{prop.semi(X)mor}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$.\newline Let $n\in\mathbb{N}$. Let $V_{1}$, $V_{2}$, $...$, $V_{n}$ be $n$
arbitrary $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules. Let
$V_{1}^{\prime}$, $V_{2}^{\prime}$, $...$, $V_{n}^{\prime}$ be $n$ arbitrary
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules. Let $f_{i}%
:V_{i}\rightarrow V_{i}^{\prime}$ be a $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule homomorphism for every $i\in\left\{  1,2,...,n\right\}
$. Then, $f_{1}\otimes f_{2}\otimes...\otimes f_{n}:V_{1}\otimes V_{2}%
\otimes...\otimes V_{n}\rightarrow V_{1}^{\prime}\otimes V_{2}^{\prime}%
\otimes...\otimes V_{n}^{\prime}$ is a $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule homomorphism.
\end{proposition}

The next proposition is a multi-factor version of Proposition
\ref{prop.semi.(X)vsRes}, and generalizes Proposition
\ref{prop.lie.(X)multvsRes} to $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules:

\begin{proposition}
\label{prop.semi.(X)multvsRes}Let $k$ be a commutative ring. Let
$\mathfrak{g}$ be a Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of
$\mathfrak{g}$. Let $n\in\mathbb{N}$.\newline If $V_{1}$, $V_{2}$, $...$,
$V_{n}$ are $n$ arbitrary $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules, then $\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}\left(
V_{1}\otimes V_{2}\otimes...\otimes V_{n}\right)  =\left(  \operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}V_{1}\right)  \otimes\left(  \operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}V_{2}\right)  \otimes...\otimes\left(
\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V_{n}\right)  $ as
$\mathfrak{h}$-modules. This allows us to speak of "the $\mathfrak{h}$-module
$V_{1}\otimes V_{2}\otimes...\otimes V_{n}$" without having to worry whether
we mean $\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}\left(  V_{1}\otimes
V_{2}\otimes...\otimes V_{n}\right)  $ or $\left(  \operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}V_{1}\right)  \otimes\left(  \operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}V_{2}\right)  \otimes...\otimes\left(
\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V_{n}\right)  $ (because it
does not matter, since $\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}%
}\left(  V_{1}\otimes V_{2}\otimes...\otimes V_{n}\right)  =\left(
\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V_{1}\right)  \otimes\left(
\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V_{2}\right)  \otimes
...\otimes\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}%
V_{n}\right)  $).
\end{proposition}
\end{verlong}

\subsection{Tensor powers of $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules}

\begin{vershort}
By analogy with Definition \ref{defs.lie(X)^n}, we can define the $n$-th
tensor power $V^{\otimes n}$ of a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule $V$. The obvious analogues of Remark \ref{rmk.lie(X)^0} and
Proposition \ref{prop.lie(X)^nmor}, \ref{prop.lie(X)^n.bracketing} and
\ref{prop.lie.(X)nvsRes} hold. Therefore, we are going to extend the reach of
Convention \ref{conv.lie(X)^n.ident} in that we are also going to use it when
$V$ is a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule.
\end{vershort}

\begin{verlong}
Next we define a particular case of tensor products of $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules, namely the tensor powers. Their
definition is analogous to the definition of the tensor powers of a $k$-module
(Definition \ref{defs.(X)^n}) and to the definition of the tensor powers of a
$\mathfrak{g}$-module (Definition \ref{defs.lie(X)^n}), but proceeds from a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule:

\begin{definition}
\label{defs.semi(X)^n}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let
$n\in\mathbb{N}$. For any $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule $V$, we define a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule $V^{\otimes n}$ by $V^{\otimes n}=\underbrace{V\otimes
V\otimes...\otimes V}_{n\text{ times}}$. This $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule $V^{\otimes n}$ is called the
$n$\textit{-th tensor power} of the $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule $V$.
\end{definition}

Since this Definition \ref{defs.semi(X)^n} is the analogue of Definition
\ref{defs.lie(X)^n} for $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules, we have:

\begin{proposition}
Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie algebra, and
let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $n\in\mathbb{N}%
$. Let $V$ be any $\mathfrak{g}$-module. Then, the $\mathfrak{g}$-module
$V^{\otimes n}$ defined in Definition \ref{defs.lie(X)^n} is identic with the
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule $V^{\otimes n}$
defined in Definition \ref{defs.semi(X)^n} (which is well-defined since $V$ is
a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule (which is because
Proposition \ref{prop.semimod.mod} says that every $\mathfrak{g}$-module is a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule)). This allows us to
speak of "the $\mathfrak{g}$-module $V^{\otimes n}$" without having to worry
whether it is understood according to Definition \ref{defs.lie(X)^n} or
according to Definition \ref{defs.semi(X)^n} (because it doesn't matter, as
both definitions yield the same result).
\end{proposition}

\begin{remark}
\label{rmk.semi(X)^0}Let $k$ be a commutative ring, let $\mathfrak{g}$ be a
$k$-Lie algebra, let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$, and
let $V$ be a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule. Then,
$V^{\otimes0}=k$ (as $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules)
and $V^{\otimes1}=V$ (as $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules), where we identify the $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule $V\otimes k$ with $V$. This is proven the same way as
Remark \ref{rmk.(X)^0}.
\end{remark}

As a consequence of Proposition \ref{prop.semi(X)mor}, we now have:

\begin{proposition}
\label{prop.semi(X)^nmor}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a $k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$.
Let $n\in\mathbb{N}$. Let $V$ and $V^{\prime}$ be $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules, and let $f:V\rightarrow V^{\prime}$ be a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule homomorphism. Then,
$f^{\otimes n}:V^{\otimes n}\rightarrow V^{\prime\otimes n}$ is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule homomorphism. (Here, we are
using Convention \ref{conv.f^(X)n}.)
\end{proposition}

This Proposition \ref{prop.semi(X)^nmor} is the analogue of Proposition
\ref{prop.lie(X)^nmor} for $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules.

The following proposition generalizes Proposition
\ref{prop.lie(X)^n.bracketing} to $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules:

\begin{proposition}
\label{prop.semi(X)^n.bracketing}Let $k$ be a commutative ring. Let
$\mathfrak{g}$ be a $k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of
$\mathfrak{g}$. Let $n\in\mathbb{N}$.\newline Then, for any $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule $V$ and every $i\in\left\{
0,1,...,n\right\}  $, the canonical $k$-module isomorphism $V^{\otimes
i}\otimes V^{\otimes\left(  n-i\right)  }\rightarrow V^{\otimes n}$ is a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule
isomorphism.{\footnotemark}
\end{proposition}

\footnotetext{Here and in the following, whenever I speak of "the canonical
$k$-module isomorphism $V^{\otimes i}\otimes V^{\otimes\left(  n-i\right)
}\rightarrow V^{\otimes n}$", I mean the $k$-module homomorphism $V^{\otimes
i}\otimes V^{\otimes\left(  n-i\right)  }\rightarrow V^{\otimes n}$ which
sends $\left(  v_{1}\otimes v_{2}\otimes...\otimes v_{i}\right)
\otimes\left(  v_{i+1}\otimes v_{i+2}\otimes...\otimes v_{n}\right)  $ to
$v_{1}\otimes v_{2}\otimes...\otimes v_{n}$ for every $\left(  v_{1}%
,v_{2},...,v_{n}\right)  \in V^{n}$. This homomorphism is known (from linear
algebra) to exist, be unique and be a $k$-module isomorphism. It is actually
the canonical $k$-module isomorphism $\underbrace{V\otimes V\otimes...\otimes
V}_{i\text{ times}}\otimes\underbrace{V\otimes V\otimes...\otimes
V}_{n-i\text{ times}}\rightarrow\underbrace{V\otimes V\otimes...\otimes
V}_{n\text{ times}}$.}

This proposition follows directly from applying Proposition
\ref{prop.semi(X).bracketing} to $V$, $V$, $...$, $V$ instead of $V_{1}$,
$V_{2}$, $...$, $V_{n}$.

The following convention extends Convention \ref{conv.lie(X)^n.ident} to
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules:

\begin{condition}
\label{conv.semi(X)^n.ident}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$. For every $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule $V$,
every $n\in\mathbb{N}$ and every $i\in\left\{  0,1,...,n\right\}  $, we are
going to identify the $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule
$V^{\otimes i}\otimes V^{\otimes\left(  n-i\right)  }$ with the $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule $V^{\otimes n}$ (this is
allowed because of Proposition \ref{prop.semi(X)^n.bracketing}). In other
words, for every $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule $V$,
every $a\in\mathbb{N}$ and every $b\in\mathbb{N}$, we are going to identify
the $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule $V^{\otimes
a}\otimes V^{\otimes b}$ with the $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule $V^{\otimes\left(  a+b\right)  }$.
\end{condition}

Finally, Proposition \ref{prop.semi.(X)multvsRes} yields:

\begin{proposition}
\label{prop.semi.(X)nvsRes}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$.
Let $n\in\mathbb{N}$.\newline If $V$ is any $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule, then $\operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}\left(  V^{\otimes n}\right)  =\left(  \operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}V\right)  ^{\otimes n}$ as $\mathfrak{h}%
$-modules. This allows us to speak of "the $\mathfrak{h}$-module $V^{\otimes
n}$" without having to worry whether we mean $\operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}\left(  V^{\otimes n}\right)  $ or $\left(
\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V\right)  ^{\otimes n}$
(because it does not matter, since $\operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}\left(  V^{\otimes n}\right)  =\left(  \operatorname{Res}%
_{\mathfrak{h}}^{\mathfrak{g}}V\right)  ^{\otimes n}$).
\end{proposition}

This Proposition \ref{prop.semi.(X)nvsRes} generalizes Proposition
\ref{prop.lie.(X)nvsRes} to $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules.
\end{verlong}

\subsection{The tensor algebra of a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule}

\begin{vershort}
In Definition \ref{defs.lietensor}, we have defined a canonical $\mathfrak{g}%
$-module structure on $\otimes V$ for every $\mathfrak{g}$-module $V$.
Similarly, we can define a canonical $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule structure on $\otimes V$ for every $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule $V$. This will be called the
\textit{tensor }$\left(  \mathfrak{g},\mathfrak{h}\right)  $%
\textit{-semimodule} of the $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule $V$. The obvious analogue of Proposition
\ref{prop.(X)-hom-is-hom} for $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules holds. We will also use the obvious analogue of Definition
\ref{defs.lie.(X).inj}.
\end{vershort}

\begin{verlong}
In Definition \ref{defs.lietensor}, we have defined a canonical $\mathfrak{g}%
$-module structure on $\otimes V$ for every $\mathfrak{g}$-module $V$. Here is
a definition which extends this notion to $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodules:

\begin{definition}
\label{defs.semitensor}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$.\newline Let $V$ be a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule. Since $V^{\otimes i}$ is a $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule for all $i\in\mathbb{N}$ (by Definition
\ref{defs.semi(X)^n}), the direct sum $\bigoplus\limits_{i\in\mathbb{N}%
}V^{\otimes i}$ is also a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule (by Definition \ref{defs.semi(++)}). In other words, the tensor
algebra $\otimes V$ thus becomes a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule (since $\otimes V=\bigoplus\limits_{i\in\mathbb{N}}V^{\otimes i}%
$). This $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule $\otimes V$
is called the \textit{tensor }$\left(  \mathfrak{g},\mathfrak{h}\right)
$\textit{-semimodule} of the $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule $V$.\newline Whenever we will speak of the $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule $\otimes V$, we will be meaning
this $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule (although there
might be many different $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule structures on the $k$-module $\otimes V$).
\end{definition}

Since this Definition \ref{defs.semitensor} is analogous to Definition
\ref{defs.lietensor}, we have:

\begin{proposition}
Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie algebra, and
let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $V$ be any
$\mathfrak{g}$-module. Then, the $\mathfrak{g}$-module $\otimes V$ defined in
Definition \ref{defs.lietensor} is identic with the $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule $\otimes V$ defined in Definition
\ref{defs.semitensor} (which is well-defined since $V$ is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule (which is because Proposition
\ref{prop.semimod.mod} says that every $\mathfrak{g}$-module is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule)). This allows us to speak of
"the $\mathfrak{g}$-module $\otimes V$" without having to worry whether it is
understood according to Definition \ref{defs.lietensor} or according to
Definition \ref{defs.semitensor} (because it doesn't matter, as both
definitions yield the same result).
\end{proposition}

It is very easy to see that:

\begin{proposition}
\label{prop.semi.(X)-hom-is-hom}Let $k$ be a commutative ring. Let
$\mathfrak{g}$ be a $k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of
$\mathfrak{g}$. Let $V$ and $W$ be two $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodules, and let $f:V\rightarrow W$ be a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule homomorphism. Then, $\otimes
f:\otimes V\rightarrow\otimes W$ is a $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule homomorphism.
\end{proposition}

This Proposition \ref{prop.semi.(X)-hom-is-hom} is the analogue of Proposition
\ref{prop.(X)-hom-is-hom} for $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules.

\begin{definition}
\label{defs.semi.(X).inj}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a $k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$.
Let $V$ be a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule. Then,
according to Proposition \ref{prop.semi(++).inj}, we consider $V^{\otimes n}$
as a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-subsemimodule of the direct
sum $\bigoplus\limits_{i\in\mathbb{N}}V^{\otimes i}=\otimes V$ for every
$n\in\mathbb{N}$. In particular, $k=V^{\otimes0}$ and $V=V^{\otimes1}$ become
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-subsemimodules of $\otimes V$
this way.
\end{definition}

This Definition \ref{defs.semi.(X).inj} is completely analogous to Definition
\ref{defs.lie.(X).inj}. Therefore:

\begin{proposition}
Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie algebra, and
let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $V$ be any
$\mathfrak{g}$-module. Let $n\in\mathbb{N}$. Then, we can identify $V^{\otimes
n}$ with a $\mathfrak{g}$-submodule of $\otimes V$ (according to Definition
\ref{defs.lie.(X).inj}), but we can also identify $V^{\otimes n}$ with a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-subsemimodule of $\otimes V$
(according to Definition \ref{defs.semi.(X).inj}, because $V$ is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule (which is because Proposition
\ref{prop.semimod.mod} says that every $\mathfrak{g}$-module is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule)). These two identifications do
\textbf{not} conflict with each other, because they both identify $V^{\otimes
n}$ with \textbf{one and the same} subset of $\otimes V$.
\end{proposition}
\end{verlong}

\subsection{$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semialgebras}

Just as we introduced $\mathfrak{g}$-algebras in Section \ref{sect.intro}, we
can define $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semialgebras:

\begin{definition}
\label{defs.semi.g-alg}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. A
$\left(  \mathfrak{g},\mathfrak{h}\right)  $\textit{-semialgebra} will mean a
$k$-algebra $A$ equipped with a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule structure such that%
\begin{equation}
\left(  a\rightharpoonup\left(  uv\right)  =\left(  a\rightharpoonup u\right)
\cdot v+u\cdot\left(  a\rightharpoonup v\right)  \text{ for every }%
a\in\mathfrak{g}\text{, }u\in A\text{ and }v\in A\right)  .
\label{defs.semi.g-alg.cond}%
\end{equation}

\end{definition}

\begin{remark}
In Definition \ref{defs.semi.g-alg}, when we speak of "a $k$-algebra $A$
equipped with a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule
structure", the words "a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule structure" mean "a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule structure on the underlying $k$-module of the $k$-algebra
$A\ \ \ \ $". This $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule
structure must therefore be $k$-bilinear with respect to the underlying
$k$-module structure of the $k$-algebra $A$.
\end{remark}

\begin{vershort}
The above Definition \ref{defs.semi.g-alg} is a weakening of Definition
\ref{defs.g-alg}, in the sense that every $\mathfrak{g}$-algebra is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semialgebra. Again, many properties of
$\mathfrak{g}$-semialgebras can be extended to $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semialgebras.
\end{vershort}

\begin{verlong}
\begin{remark}
Definition \ref{defs.semi.g-alg} is often rewritten as follows:

\begin{definition}
\label{defs.semi.g-alg.der}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$. A $\left(  \mathfrak{g},\mathfrak{h}\right)  $\textit{-semialgebra} will
mean a $k$-algebra $A$ equipped with a $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule structure such that $\mathfrak{g}$ acts on $A$ by means
of derivations. Here, we say that "$\mathfrak{g}$ acts on $A$ by means of
derivations" if and only if the map $A\rightarrow A,$ $u\mapsto\left(
a\rightharpoonup u\right)  $ is a derivation for every $a\in\mathfrak{g}$.
\end{definition}

This Definition \ref{defs.semi.g-alg.der} is indeed equivalent to Definition
\ref{defs.semi.g-alg} because the condition that $\mathfrak{g}$ acts on $A$ by
means of derivations is equivalent to (\ref{defs.semi.g-alg.cond}) (as can be
easily seen).
\end{remark}

The above Definition \ref{defs.semi.g-alg} is a weakening of Definition
\ref{defs.g-alg}, in the sense that we have:

\begin{proposition}
\label{prop.g-alg-is-semi}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a $k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$.
Then, every $\mathfrak{g}$-algebra is a $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semialgebra.
\end{proposition}

This proposition is completely trivial (since every $\mathfrak{g}$-module is a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule).

It is easy to see that, when $\mathfrak{h}$ is a Lie subalgebra of a $k$-Lie
algebra $\mathfrak{g}$, every $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semialgebra canonically can be made into an $\mathfrak{h}$-algebra:

\begin{definition}
\label{defs.semi.g-as-h-algebra}Let $k$ be a commutative ring. Let
$\mathfrak{g}$ be a $k$-Lie algebra, and let $\mathfrak{h}$ be a Lie
subalgebra of $\mathfrak{g}$. Then, every $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semialgebra $A$ canonically becomes an $\mathfrak{h}$-algebra. (In
fact, the $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule $A$
canonically becomes an $\mathfrak{h}$-module according to Definition
\ref{defs.semi.restrict}, and thus $A$ is a $k$-algebra equipped with an
$\mathfrak{h}$-module structure which satisfies (\ref{defs.g-alg.cond}) with
$\mathfrak{g}$ replaced by $\mathfrak{h}$, so that $A$ thus is an
$\mathfrak{h}$-algebra.). This $\mathfrak{h}$-algebra is called the
\textit{restriction of }$A$ \textit{to }$\mathfrak{h}$, and denoted by
$\operatorname*{Res}\nolimits_{\mathfrak{h}}^{\mathfrak{g}}A$. However, when
there is no possibility of confusion, we will denote this $\mathfrak{h}%
$-algebra by $A$, and we will distinguish it from the original $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semialgebra $A$ by means of referring to
the former one as "the $\mathfrak{h}$-algebra $A$" and referring to the latter
one as "the $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semialgebra $A$".
\end{definition}

This Definition \ref{defs.semi.g-as-h-algebra} is a generalization of
Definition \ref{defs.g-as-h-algebra} to $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semialgebras.

The following proposition generalizes Proposition \ref{prop.g-alg.PQ} to
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semialgebras:

\begin{proposition}
\label{prop.semi.g-alg.PQ}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a $k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$.
Let $A$ be a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semialgebra. Let $P$
and $Q$ be two $\left(  \mathfrak{g},\mathfrak{h}\right)  $-subsemimodules of
$A$. Then, $P\cdot Q$ is a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-subsemimodule of $A$.
\end{proposition}

\begin{remark}
Here, $P\cdot Q$ is to be understood as according to Convention \ref{conv.A*B}
\textbf{(a)}.
\end{remark}

\textit{Proof of Proposition \ref{prop.semi.g-alg.PQ}. }In order to obtain a
proof of Proposition \ref{prop.semi.g-alg.PQ}, it is enough to apply the
following changes to the proof of Proposition \ref{prop.g-alg.PQ}:

\begin{itemize}
\item Replace "Proposition \ref{prop.g-alg.PQ}" by "Proposition
\ref{prop.semi.g-alg.PQ}".

\item Replace the reference to (\ref{defs.g-alg.cond}) by a reference to
(\ref{defs.semi.g-alg.cond}).

\item Replace the words "$\mathfrak{g}$-submodule" by "$\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-subsemimodule".
\end{itemize}

This completes the proof of Proposition \ref{prop.semi.g-alg.PQ}.
\end{verlong}

\subsection{$\otimes V$ is a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semialgebra}

The following proposition is a generalization of Proposition
\ref{prop.lietensor} to $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semialgebras:

\begin{vershort}
\begin{proposition}
\label{prop.semitensor.short}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$.\newline Let $V$ be a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule. If we equip the $k$-algebra $\otimes V$ (this $k$-algebra was
defined in Definition \ref{defs.tensor} \textbf{(a)}) with the $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule structure defined in Definition
\ref{defs.lietensor} (but with "$\mathfrak{g}$-module" replaced by "$\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule"), we obtain a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semialgebra.
\end{proposition}

This allows us to generalize Definition \ref{defs.lietensor.g-alg} to $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodules. The natural generalization of
Proposition \ref{prop.lie.(X)algvsRes} also holds.
\end{vershort}

\begin{verlong}
\begin{proposition}
\label{prop.semitensor}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$.\newline Let $V$ be a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule. If we equip the $k$-algebra $\otimes V$ (this $k$-algebra was
defined in Definition \ref{defs.tensor} \textbf{(a)}) with the $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule structure defined in Definition
\ref{defs.semitensor}, we obtain a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semialgebra.
\end{proposition}

\begin{definition}
\label{defs.semitensor.g-alg}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$.\newline Let $V$ be a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule. The $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semialgebra
$\otimes V$ defined in Proposition \ref{prop.semitensor} is called the
\textit{tensor }$\left(  \mathfrak{g},\mathfrak{h}\right)  $%
\textit{-semialgebra} of the $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule $V$. Whenever we will speak of the $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semialgebra $\otimes V$, we will be meaning this
tensor $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semialgebra $\otimes V$
(unless we explicitly say that we are talking about a different $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semialgebra structure on $\otimes V$).
\end{definition}

This Definition \ref{defs.semitensor.g-alg} is a generalization of Definition
\ref{defs.lietensor.g-alg} to $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules. Therefore:

\begin{proposition}
Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie algebra, and
let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let $V$ be any
$\mathfrak{g}$-module. Then, the $\mathfrak{g}$-algebra $\otimes V$ defined in
Definition \ref{defs.lietensor.g-alg} is identic with the $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semialgebra $\otimes V$ defined in
Definition \ref{defs.semitensor.g-alg} (which is well-defined since $V$ is a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule (which is because
Proposition \ref{prop.semimod.mod} says that every $\mathfrak{g}$-module is a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule)). This allows us to
speak of "the $\mathfrak{g}$-algebra $\otimes V$" without having to worry
whether it is understood according to Definition \ref{defs.lietensor.g-alg} or
according to Definition \ref{defs.semitensor.g-alg} (because it doesn't
matter, as both definitions yield the same result).
\end{proposition}

\textit{Proof of Proposition \ref{prop.semitensor}.} In order to obtain a
proof of Proposition \ref{prop.semitensor}, it is enough to apply the
following changes to the proof of Proposition \ref{prop.lietensor}:

\begin{itemize}
\item Replace "Proposition \ref{prop.lietensor}" by "Proposition
\ref{prop.semitensor}".

\item Replace "Proposition \ref{prop.lie(X)^n.bracketing}" by "Proposition
\ref{prop.semi(X)^n.bracketing}".

\item Replace "Definition \ref{defs.g-alg}" by "Definition
\ref{defs.semi.g-alg}".

\item Replace the words "$\mathfrak{g}$-module" by "$\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule".

\item Replace the words "$\mathfrak{g}$-algebra" by "$\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semialgebra".

\item Replace the references to (\ref{prop.lie(++).form}) by references to
(\ref{prop.semi(++).form}).

\item Replace the references to (\ref{prop.lie(X).form}) by references to
(\ref{prop.semi(X).form}).

\item Replace the references to (\ref{defs.g-alg.cond}) by references to
(\ref{defs.semi.g-alg.cond}).
\end{itemize}

This completes the proof of Proposition \ref{prop.semitensor}.

We notice that the $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semialgebra
$\otimes V$ behaves under restriction as we would want it to:

\begin{proposition}
\label{prop.semi.(X)algvsRes}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$.\newline If $V$ is any $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule, then $\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}\left(
\otimes V\right)  =\otimes\left(  \operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}V\right)  $ as $\mathfrak{h}$-algebras. This allows us to
speak of "the $\mathfrak{h}$-algebra $\otimes V$" without having to worry
whether we mean $\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}\left(
\otimes V\right)  $ or $\otimes\left(  \operatorname{Res}_{\mathfrak{h}%
}^{\mathfrak{g}}V\right)  $ (because it does not matter, since
$\operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}\left(  \otimes V\right)
=\otimes\left(  \operatorname{Res}_{\mathfrak{h}}^{\mathfrak{g}}V\right)  $).
\end{proposition}

This Proposition \ref{prop.semi.(X)algvsRes} is the generalization of
Proposition \ref{prop.lie.(X)algvsRes} to $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodules.
\end{verlong}

\subsection{\label{subsect.semimod.h}Semimodules and $\mathfrak{h}$-module
homomorphisms}

We have devoted a great part of Section \ref{sect.(g,h)} to formulating
properties of $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules which
are analogous to some known properties of $\mathfrak{g}$-modules. Of course,
we have barely scratched the surface - there are many more such properties. In
the present Subsection \ref{subsect.semimod.h}, as well as in Subsection
\ref{subsect.U(g,h)} further below, we are going to present some different
viewpoints on $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules.

The following result is not an analogue of a result from Section
\ref{sect.intro} anymore, but instead rewrites the definition of a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule in terms of the notion of
$\mathfrak{h}$-module homomorphisms:

\begin{proposition}
\label{prop.h-lin-semimod}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a $k$-Lie algebra, and let $\mathfrak{h}$ be a Lie subalgebra of
$\mathfrak{g}$. Let $V$ be an $\mathfrak{h}$-module. Let $\beta:\mathfrak{g}%
\times V\rightarrow V$ denote a $k$-bilinear map such that $\beta
\mid_{\mathfrak{h}\times V}$ is the Lie action of the $\mathfrak{h}$-module
$V$. By the universal property of the tensor product, this $k$-bilinear map
$\beta:\mathfrak{g}\times V\rightarrow V$ gives rise to a $k$-linear map
$\widetilde{\beta}:\mathfrak{g}\otimes V\rightarrow V$ which satisfies%
\[
\left(  \widetilde{\beta}\left(  a\otimes v\right)  =\beta\left(  a,v\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{ and }v\in
V\right)  .
\]
\newline Then, $\left(  V,\beta\right)  $ is a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule if and only if $\widetilde{\beta
}:\mathfrak{g}\otimes V\rightarrow V$ is an $\mathfrak{h}$-module homomorphism.
\end{proposition}

We are neither going to use, nor going to prove this (something the reader
could easily do), but we remark that this can be applied to $\mathfrak{h}%
=\mathfrak{g}$ and results in a property of $\mathfrak{g}$-modules:

\begin{proposition}
\label{prop.h-lin-semimod-cor}Let $k$ be a commutative ring. Let
$\mathfrak{g}$ be a $k$-Lie algebra. Let $V$ be a $\mathfrak{g}$-module. Let
$\beta:\mathfrak{g}\times V\rightarrow V$ denote the Lie action of the
$\mathfrak{g}$-module $V$. By the universal property of the tensor product,
this $k$-bilinear map $\beta:\mathfrak{g}\times V\rightarrow V$ gives rise to
a $k$-linear map $\widetilde{\beta}:\mathfrak{g}\otimes V\rightarrow V$ which
satisfies%
\[
\left(  \widetilde{\beta}\left(  a\otimes v\right)  =\beta\left(  a,v\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{ and }v\in
V\right)  .
\]
\newline Then, $\widetilde{\beta}:\mathfrak{g}\otimes V\rightarrow V$ is a
$\mathfrak{g}$-module homomorphism.
\end{proposition}

\subsection{\label{subsect.U(g,h)}$\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules as modules}

In this Subsection \ref{subsect.U(g,h)}, we are going to relate the notion of
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules to the notion of
$\mathfrak{g}$-modules (but for a different Lie algebra $\mathfrak{g}$) and to
the notion of $A$-modules for an (associative) algebra $A$. This subsection is
not relevant to the rest of the present paper, but it sheds a different light
on the notion of $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules
(more concretely, it provides alternative definitions of the notion of
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules, although I consider
Definition \ref{defs.semimod} to be the simplest and most explanatory one).

First, let us give an analogue of Definition \ref{defs.U} tailored to $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodules:

\begin{definition}
\label{defs.U.semi}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. We
define the algebra $U\left(  \mathfrak{g},\mathfrak{h}\right)  $ to be the
factor algebra $\left(  \otimes\mathfrak{g}\right)  \diagup I_{\mathfrak{g}%
,\mathfrak{h}}$, where $I_{\mathfrak{g},\mathfrak{h}}$ is the two-sided ideal%
\[
\left(  \otimes\mathfrak{g}\right)  \cdot\left\langle v\otimes w-w\otimes
v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}%
\times\mathfrak{h}\right\rangle \cdot\left(  \otimes\mathfrak{g}\right)
\]
of the algebra $\otimes\mathfrak{g}$.
\end{definition}

\begin{remark}
In Definition \ref{defs.U.semi}, the term $\left\langle v\otimes w-w\otimes
v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}%
\times\mathfrak{h}\right\rangle $ is to be understood according to Convention
\ref{conv.<>}, and the multiplication sign $\cdot$ in
\[
\left(  \otimes\mathfrak{g}\right)  \cdot\left\langle v\otimes w-w\otimes
v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}%
\times\mathfrak{h}\right\rangle \cdot\left(  \otimes\mathfrak{g}\right)
\]
is to be understood according to Convention \ref{conv.A*B}. We note that,
although the multiplication in $\otimes\mathfrak{g}$ is related to the tensor
product by (\ref{*=(X)}), the product $\left(  \otimes\mathfrak{g}\right)
\cdot\left\langle v\otimes w-w\otimes v-\left[  v,w\right]  \ \mid\ \left(
v,w\right)  \in\mathfrak{g}\times\mathfrak{h}\right\rangle \cdot\left(
\otimes\mathfrak{g}\right)  $ has nothing to do with the tensor product
$\left(  \otimes\mathfrak{g}\right)  \otimes\left\langle v\otimes w-w\otimes
v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}%
\times\mathfrak{h}\right\rangle \otimes\left(  \otimes\mathfrak{g}\right)  $ !
\end{remark}

\begin{remark}
In Definition \ref{defs.U.semi}, the ideal $I_{\mathfrak{g},\mathfrak{h}}$ is
the same as the ideal $J$ from Theorem \ref{thm.l25}.
\end{remark}

Now, the following generalization of Proposition \ref{prop.U.equiv} allows us
to consider $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules as
$A$-modules for $A=U\left(  \mathfrak{g},\mathfrak{h}\right)  $:

\begin{proposition}
\label{prop.U.equiv.semi}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a $k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$.
Consider the algebra $U\left(  \mathfrak{g},\mathfrak{h}\right)  $ and the
ideal $I_{\mathfrak{g},\mathfrak{h}}$ defined in Definition \ref{defs.U.semi}%
.\newline\textbf{(a)} For every $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule $V$, there is one and only one $U\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-module structure on $V$ satisfying%
\[
\left(  \overline{a}\cdot v=a\rightharpoonup v\ \ \ \ \ \ \ \ \ \ \text{for
every }a\in\mathfrak{g}\text{ and }v\in V\right)
\]
(where $\overline{a}$ denotes the projection of $a\in\mathfrak{g}%
\subseteq\otimes\mathfrak{g}$ on $\left(  \otimes\mathfrak{g}\right)  \diagup
I_{\mathfrak{g},\mathfrak{h}}=U\left(  \mathfrak{g},\mathfrak{h}\right)  $).
This $U\left(  \mathfrak{g},\mathfrak{h}\right)  $-module structure is
canonical. Thus, every $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule
$V$ canonically becomes a $U\left(  \mathfrak{g},\mathfrak{h}\right)
$-module.\newline\textbf{(b)} Conversely, for every $U\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-module $V$, we can define a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule structure on $V$ by%
\[
\left(  a\rightharpoonup v=\overline{a}\cdot v\ \ \ \ \ \ \ \ \ \ \text{for
every }a\in\mathfrak{g}\text{ and }v\in V\right)
\]
(where $\overline{a}$ denotes the projection of $a\in\mathfrak{g}%
\subseteq\otimes\mathfrak{g}$ on $\left(  \otimes\mathfrak{g}\right)  \diagup
I_{\mathfrak{g},\mathfrak{h}}=U\left(  \mathfrak{g},\mathfrak{h}\right)  $).
This $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule structure is
canonical. Thus, every $U\left(  \mathfrak{g},\mathfrak{h}\right)  $-module
$V$ canonically becomes a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule.\newline\textbf{(c)} Let $V$ and $W$ be two $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodules. Then, according to
Proposition \ref{prop.U.equiv.semi} \textbf{(a)}, each of $V$ and $W$
canonically becomes a $U\left(  \mathfrak{g},\mathfrak{h}\right)  $-module.
Let $f:V\rightarrow W$ be a map. Then, $f$ is a homomorphism of $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodules if and only if $f$ is a
homomorphism of $U\left(  \mathfrak{g},\mathfrak{h}\right)  $-modules.\newline%
\textbf{(d)} Let $V$ and $W$ be two $U\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-modules. Then, according to Proposition \ref{prop.U.equiv.semi}
\textbf{(b)}, each of $V$ and $W$ canonically becomes a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule. Let $f:V\rightarrow W$ be a
map. Then, $f$ is a homomorphism of $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules if and only if $f$ is a homomorphism of $U\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-modules.\newline\textbf{(e)} We can define a functor
$U_{1}$ from the category of $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules to the category of $U\left(  \mathfrak{g},\mathfrak{h}\right)
$-modules as follows: For every $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule $V$, let $U_{1}\left(  V\right)  $ be the $U\left(
\mathfrak{g},\mathfrak{h}\right)  $-module $V$ defined in Proposition
\ref{prop.U.equiv.semi} \textbf{(a)}. For every homomorphism $f$ between
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules, let $U_{1}\left(
f\right)  $ be the same homomorphism $f$, but considered as a homomorphism
between $U\left(  \mathfrak{g},\mathfrak{h}\right)  $-modules this time (this
is legitimate due to Proposition \ref{prop.U.equiv.semi} \textbf{(c)}%
).\newline\textbf{(f)} We can define a functor $U_{2}$ from the category of
$U\left(  \mathfrak{g},\mathfrak{h}\right)  $-modules to the category of
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules as follows: For every
$U\left(  \mathfrak{g},\mathfrak{h}\right)  $-module $V$, let $U_{2}\left(
V\right)  $ be the $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule $V$
defined in Proposition \ref{prop.U.equiv.semi} \textbf{(b)}. For every
homomorphism $f$ between $U\left(  \mathfrak{g},\mathfrak{h}\right)
$-modules, let $U_{2}\left(  f\right)  $ be the same homomorphism $f$, but
considered as a homomorphism between $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodules this time (this is legitimate due to Proposition
\ref{prop.U.equiv.semi} \textbf{(d)}).\newline\textbf{(g)} The two functors
$U_{1}$ and $U_{2}$ defined in Proposition \ref{prop.U.equiv.semi}
\textbf{(e)} and \textbf{(f)} are mutually inverse.\newline\textbf{(h)} Both
functors $U_{1}$ and $U_{2}$ are additive, exact and preserve kernels,
cokernels and direct sums.
\end{proposition}

This proposition provides for an easy way to obtain results about $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodules from results about $A$-modules
(over every associative algebra $A$), at least when the latter results are
basic enough to remain valid under an invertible, additive and exact functor
which preserves kernels, cokernels and direct sums. We can also add tensor
products to this list by making $U\left(  \mathfrak{g},\mathfrak{h}\right)  $
into a Hopf algebra (in analogy to Proposition \ref{prop.tensorhopf}) and
proving an analogue of Proposition \ref{prop.Uhopf}. We will not delve into
details here as these analogues are straightforward to obtain and automatic to
prove (by taking the proofs of Propositions \ref{prop.tensorhopf} and
\ref{prop.Uhopf} and applying the obvious changes).

Thus we have found a way to see $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules as $A$-modules for an associative algebra $A$. One could wonder
whether we can also see them as $\mathfrak{g}$-modules for some \textit{other}
Lie algebra $\mathfrak{g}$. The answer is, again, positive:

\begin{proposition}
\label{prop.h^(1)}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$.\newline Let $\operatorname*{FreeLie}\mathfrak{g}$ denote the free Lie
algebra on the $k$-module $\mathfrak{g}$, and let $\iota:\mathfrak{g}%
\rightarrow\operatorname*{FreeLie}\mathfrak{g}$ be the corresponding
homomorphism. (A definition of a "free Lie algebra" along with the
corresponding homomorphism is given in various sources, e. g., in
\cite[\S 1.11.2]{sternberg}. For us it is only important that it satisfies the
following universal property: For every $k$-Lie algebra $\mathfrak{u}$ and
every $k$-module homomorphism $p:\mathfrak{g}\rightarrow\mathfrak{u}$, there
exists one and only one $k$-Lie algebra homomorphism
$P:\operatorname*{FreeLie}\mathfrak{g}\rightarrow\mathfrak{u}$ satisfying
$P\circ\iota=p$.) Let $\mathfrak{i}$ denote the Lie ideal of
$\operatorname*{FreeLie}\mathfrak{g}$ generated by the $k$-submodule
$\left\langle \left[  \iota\left(  v\right)  ,\iota\left(  w\right)  \right]
-\iota\left(  \left[  v,w\right]  \right)  \ \mid\ \left(  v,w\right)
\in\mathfrak{h}\times\mathfrak{g}\right\rangle $ of $\operatorname*{FreeLie}%
\mathfrak{g}$. Let $\mathfrak{h}^{(1)}$ denote the $k$-Lie algebra $\left(
\operatorname*{FreeLie}\mathfrak{g}\right)  \diagup\mathfrak{i}$%
.\newline\textbf{(a)} For every $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule $V$, there is one and only one $\mathfrak{h}^{(1)}$-module
structure on $V$ satisfying%
\begin{equation}
\left(  \overline{\iota\left(  a\right)  }\rightharpoonup v=a\rightharpoonup
v\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{ and }v\in
V\right)  \label{prop.h^(1).a}%
\end{equation}
(where $\overline{\iota\left(  a\right)  }$ denotes the projection of
$\iota\left(  a\right)  \in\operatorname*{FreeLie}\mathfrak{g}$ on $\left(
\operatorname*{FreeLie}\mathfrak{g}\right)  \diagup\mathfrak{i}=\mathfrak{h}%
^{(1)}$). This $\mathfrak{h}^{(1)}$-module structure is canonical. Thus, every
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule $V$ canonically
becomes an $\mathfrak{h}^{(1)}$-module.\newline\textbf{(b)} Conversely, for
every $\mathfrak{h}^{(1)}$-module $V$, we can define a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule structure on $V$ by%
\[
\left(  a\rightharpoonup v=\overline{\iota\left(  a\right)  }\rightharpoonup
v\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{ and }v\in
V\right)
\]
(where $\overline{\iota\left(  a\right)  }$ denotes the projection of
$\iota\left(  a\right)  \in\operatorname*{FreeLie}\mathfrak{g}$ on $\left(
\operatorname*{FreeLie}\mathfrak{g}\right)  \diagup\mathfrak{i}=\mathfrak{h}%
^{(1)}$). This $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule
structure is canonical. Thus, every $\mathfrak{h}^{(1)}$-module $V$
canonically becomes a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule.\newline\textbf{(c)} Let $V$ and $W$ be two $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodules. Then, according to
Proposition \ref{prop.h^(1)} \textbf{(a)}, each of $V$ and $W$ canonically
becomes an $\mathfrak{h}^{(1)}$-module. Let $f:V\rightarrow W$ be a map. Then,
$f$ is a homomorphism of $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodules if and only if $f$ is a homomorphism of $\mathfrak{h}^{(1)}%
$-modules.\newline\textbf{(d)} Let $V$ and $W$ be two $\mathfrak{h}^{(1)}%
$-modules. Then, according to Proposition \ref{prop.h^(1)} \textbf{(b)}, each
of $V$ and $W$ canonically becomes a $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule. Let $f:V\rightarrow W$ be a map. Then, $f$ is a
homomorphism of $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules if
and only if $f$ is a homomorphism of $\mathfrak{h}^{(1)}$-modules.\newline%
\textbf{(e)} We can define a functor $U_{1}$ from the category of $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodules to the category of
$\mathfrak{h}^{(1)}$-modules as follows: For every $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule $V$, let $U_{1}\left(  V\right)  $ be the
$\mathfrak{h}^{(1)}$-module $V$ defined in Proposition \ref{prop.h^(1)}
\textbf{(a)}. For every homomorphism $f$ between $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules, let $U_{1}\left(  f\right)  $ be the same
homomorphism $f$, but considered as a homomorphism between $\mathfrak{h}%
^{(1)}$-modules this time (this is legitimate due to Proposition
\ref{prop.h^(1)} \textbf{(c)}).\newline\textbf{(f)} We can define a functor
$U_{2}$ from the category of $\mathfrak{h}^{(1)}$-modules to the category of
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules as follows: For every
$\mathfrak{h}^{(1)}$-module $V$, let $U_{2}\left(  V\right)  $ be the $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule $V$ defined in Proposition
\ref{prop.h^(1)} \textbf{(b)}. For every homomorphism $f$ between
$\mathfrak{h}^{(1)}$-modules, let $U_{2}\left(  f\right)  $ be the same
homomorphism $f$, but considered as a homomorphism between $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodules this time (this is legitimate
due to Proposition \ref{prop.h^(1)} \textbf{(d)}).\newline\textbf{(g)} The two
functors $U_{1}$ and $U_{2}$ defined in Proposition \ref{prop.h^(1)}
\textbf{(e)} and \textbf{(f)} are mutually inverse.\newline\textbf{(h)} Both
functors $U_{1}$ and $U_{2}$ are additive, exact and preserve kernels,
cokernels and direct sums.
\end{proposition}

Additionally:

\begin{proposition}
\label{prop.h^(1).restr}Consider the situation of Proposition \ref{prop.h^(1)}%
.\newline\textbf{(a)} The composition%
\[%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{8pc}
%\xymatrix{
%\mathfrak{h} \ar[r]^{\text{canonical inclusion}} & \mathfrak{g} \ar
%[r]^-{\iota} & \operatorname*{FreeLie}\mathfrak{g} \ar[r]^-{\text
%{canonical projection}} & \mathfrak{h}^{(1)}
%}}}%
%BeginExpansion
\xymatrixcolsep{8pc}
\xymatrix{
\mathfrak{h} \ar[r]^{\text{canonical inclusion}} & \mathfrak{g} \ar
[r]^-{\iota} & \operatorname*{FreeLie}\mathfrak{g} \ar[r]^-{\text
{canonical projection}} & \mathfrak{h}^{(1)}
}%
%EndExpansion
\]
is an injective $k$-Lie algebra homomorphism $\mathfrak{h}\rightarrow
\mathfrak{h}^{(1)}$.\newline\textbf{(b)} Let $V$ be a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule. Then, according to Proposition
\ref{prop.h^(1)} \textbf{(a)}, this $V$ canonically becomes an $\mathfrak{h}%
^{(1)}$-module. The restriction of this $\mathfrak{h}^{(1)}$-module to
$\mathfrak{h}$ (by means of the injective $k$-Lie algebra homomorphism
$\mathfrak{h}\rightarrow\mathfrak{h}^{(1)}$ constructed in part \textbf{(a)})
is identic with the restriction of the $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule $V$ to $\mathfrak{h}$.
\end{proposition}

As a consequence, an $\mathfrak{h}$-module is the restriction of some $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule to $\mathfrak{h}$ if and only
if it is the restriction of some $\mathfrak{h}^{(1)}$-module to $\mathfrak{h}$.

The paper \cite{calaque1} mostly views $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodules through the lens of Proposition \ref{prop.h^(1)}, thus
considering them as $\mathfrak{h}^{(1)}$-modules. The notion of a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule does not even occur in this paper.

We can view $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules as
$U\left(  \mathfrak{g},\mathfrak{h}\right)  $-modules (by Proposition
\ref{prop.U.equiv.semi}), but on the other hand we can also view them as
$\mathfrak{h}^{(1)}$-modules (by Proposition \ref{prop.h^(1)}) and therefore
as $U\left(  \mathfrak{h}^{(1)}\right)  $-modules (by Proposition
\ref{prop.U.equiv}, applied to $\mathfrak{h}^{(1)}$ instead of $\mathfrak{g}%
$). This leaves us wondering whether there is a relation between $U\left(
\mathfrak{g},\mathfrak{h}\right)  $ and $U\left(  \mathfrak{h}^{(1)}\right)
$. The answer is as simple as it could be: there exists an isomorphism:

\begin{proposition}
\label{prop.U(g,h)=U(h^(1))}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}%
$. Consider the $k$-algebra $U\left(  \mathfrak{g},\mathfrak{h}\right)  $
defined in Definition \ref{defs.U.semi}, and consider the $k$-Lie algebra
$\mathfrak{h}^{(1)}$ defined in Proposition \ref{prop.h^(1)}. Then, there
exists a unique $k$-algebra homomorphism $P:U\left(  \mathfrak{g}%
,\mathfrak{h}\right)  \rightarrow U\left(  \mathfrak{h}^{(1)}\right)  $ which
satisfies $\left(  P\left(  \overline{a}\right)  =\overline{\iota\left(
a\right)  }\text{ for every }a\in\mathfrak{g}\right)  $. This homomorphism $P$
is a $k$-algebra isomorphism.
\end{proposition}

Before we continue, let us contour the proof of Proposition \ref{prop.h^(1)}.
It needs the following fact about Lie algebra modules:

\begin{proposition}
\label{prop.liemod.curry}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a $k$-Lie algebra.\newline\textbf{(a)} Let $V$ be a $\mathfrak{g}$-module. For
every $a\in\mathfrak{g}$, let $\beta_{a}:V\rightarrow V$ be the map which
sends every $v\in V$ to $a\rightharpoonup v\in V$. Then, $\beta_{a}%
\in\operatorname{End}V$. Now, define a map $\operatorname{cur}_{V}%
:\mathfrak{g}\rightarrow\operatorname{End}V$ by%
\[
\left(  \operatorname{cur}_{V}\left(  a\right)  =\beta_{a}%
\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\right)  .
\]
Then, $\operatorname{cur}_{V}$ is a Lie algebra homomorphism. (Here,
$\operatorname*{End}V$ is a $k$-Lie algebra with the commutator of
endomorphisms as Lie bracket.)\newline\textbf{(b)} Let $V$ be a $k$-module,
and let $P:\mathfrak{g}\rightarrow\operatorname{End}V$ be a Lie algebra
homomorphism. Then, we can make $V$ into a $\mathfrak{g}$-module by setting%
\[
\left(  a\rightharpoonup v=\left(  P\left(  a\right)  \right)  \left(
v\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{ and
}v\in V\right)  .
\]
\newline\textbf{(c)} Let $V$ be a $k$-module. Part \textbf{(a)} of this
proposition assigns a Lie algebra homomorphism $\operatorname{cur}%
_{V}:\mathfrak{g}\rightarrow\operatorname{End}V$ to any $\mathfrak{g}$-module
structure on $V$, whereas part \textbf{(b)} of this proposition assigns a
$\mathfrak{g}$-module structure on $V$ to any Lie algebra homomorphism
$P:\mathfrak{g}\rightarrow\operatorname{End}V$. These two assignments are
mutually inverse. This means that if we start with a $\mathfrak{g}$-module
structure on $V$, then assign a Lie algebra homomorphism $\operatorname{cur}%
_{V}:\mathfrak{g}\rightarrow\operatorname{End}V$ to it according to part
\textbf{(a)}, and then assign a $\mathfrak{g}$-module structure to it
according to part \textbf{(b)} (applied to $P=\operatorname{cur}_{V}$), then
we get our original $\mathfrak{g}$-module structure on $V$ back, and the same
holds the other way round.\newline This yields that if $V$ is a $k$-module,
then Lie algebra homomorphisms $\mathfrak{g}\rightarrow\operatorname{End}V$
stand in 1-to-1 correspondence with $\mathfrak{g}$-module structures on $V$.
\end{proposition}

This is a known fact, and has an even better known counterpart about
associative algebras (which is proven similarly), so we see no need to prove
it here. Moreover, we need the following lemma:

\begin{lemma}
\label{lem.h^(1).equality}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a $k$-Lie algebra. Let $S$ be a subset of $\mathfrak{g}$ such that the Lie
algebra $\mathfrak{g}$ is generated (as a Lie algebra) by the subset
$S$.\newline Let $V$ be a $k$-module, and let $\mu_{1}:\mathfrak{g}\times
V\rightarrow V$ and $\mu_{2}:\mathfrak{g}\times V\rightarrow V$ be two maps
such that $\left(  V,\mu_{1}\right)  $ and $\left(  V,\mu_{2}\right)  $ are
two $\mathfrak{g}$-modules. Assume that every $s\in S$ and every $v\in V$
satisfy $\mu_{1}\left(  s,v\right)  =\mu_{2}\left(  s,v\right)  $. Then,
$\mu_{1}=\mu_{2}$.
\end{lemma}

\begin{vershort}
\textit{Proof of Lemma \ref{lem.h^(1).equality}.} Let $\mathfrak{h}$ be the
subset%
\[
\left\{  a\in\mathfrak{g}\ \mid\ \mu_{1}\left(  a,v\right)  =\mu_{2}\left(
a,v\right)  \text{ for every }v\in V\right\}
\]
of $\mathfrak{g}$. We can then easily prove that $\mathfrak{h}$ is a
$k$-submodule of $\mathfrak{g}$ (since $\mu_{1}$ and $\mu_{2}$ are
$k$-bilinear, which is because $\left(  V,\mu_{1}\right)  $ and $\left(
V,\mu_{2}\right)  $ are two $\mathfrak{g}$-modules). Besides, any
$x\in\mathfrak{h}$ and $y\in\mathfrak{h}$ satisfy $\left[  x,y\right]
\in\mathfrak{h}$ (since%
\begin{align*}
\mu_{1}\left(  \left[  x,y\right]  ,v\right)   &  =\underbrace{\mu_{1}\left(
x,\mu_{1}\left(  y,v\right)  \right)  }_{\substack{=\mu_{2}\left(  x,\mu
_{1}\left(  y,v\right)  \right)  \\\text{(since }x\in\mathfrak{h}\text{)}%
}}-\underbrace{\mu_{1}\left(  y,\mu_{1}\left(  x,v\right)  \right)
}_{\substack{=\mu_{2}\left(  y,\mu_{1}\left(  x,v\right)  \right)
\\\text{(since }y\in\mathfrak{h}\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{due to (\ref{L-mod}) (applied to }\mu_{1}\text{, }x\text{ and }y\text{
instead of }\mu\text{, }a\text{ and }b\text{),}\\
\text{since }\left(  V,\mu_{1}\right)  \text{ is a }\mathfrak{g}\text{-module}%
\end{array}
\right) \\
&  =\mu_{2}\left(  x,\underbrace{\mu_{1}\left(  y,v\right)  }_{\substack{=\mu
_{2}\left(  y,v\right)  \\\text{(since }y\in\mathfrak{h}\text{)}}}\right)
-\mu_{2}\left(  y,\underbrace{\mu_{1}\left(  x,v\right)  }_{\substack{=\mu
_{2}\left(  x,v\right)  \\\text{(since }x\in\mathfrak{h}\text{)}}}\right) \\
&  =\mu_{2}\left(  x,\mu_{2}\left(  y,v\right)  \right)  -\mu_{2}\left(
y,\mu_{2}\left(  x,v\right)  \right)  =\mu_{2}\left(  \left[  x,y\right]
,v\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{due to (\ref{L-mod}) (applied to }\mu_{2}\text{, }x\text{ and }y\text{
instead of }\mu\text{, }a\text{ and }b\text{),}\\
\text{since }\left(  V,\mu_{2}\right)  \text{ is a }\mathfrak{g}\text{-module}%
\end{array}
\right)
\end{align*}
for every $v\in V$), which shows that $\mathfrak{h}$ is actually a Lie
subalgebra of $\mathfrak{g}$. On the other hand, $S\subseteq\mathfrak{h}$
(since every $s\in S$ satisfies $s\in\mathfrak{h}$, as the definition of
$\mathfrak{h}$ and the condition of Lemma \ref{lem.h^(1).equality} show).

But we know that the Lie algebra $\mathfrak{g}$ is generated (as a Lie
algebra) by the subset $S$. This means that $\mathfrak{g}$ is identical with
the Lie subalgebra of $\mathfrak{g}$ generated by $S$. So we have%
\begin{align*}
\mathfrak{g}  &  =\left(  \text{the Lie subalgebra of }\mathfrak{g}\text{
generated by }S\right) \\
&  =\left(  \text{the smallest Lie subalgebra of }\mathfrak{g}\text{ which
contains }S\text{ as a subset}\right) \\
&  \subseteq\mathfrak{h}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\mathfrak{h}\text{ is a Lie subalgebra of }\mathfrak{g}\text{ which contains
}S\text{ as a subset}\right)  .
\end{align*}
Thus, every $x\in\mathfrak{g}$ satisfies $\mu_{1}\left(  x,v\right)  =\mu
_{2}\left(  x,v\right)  $ for every $v\in V$. In other words, $\mu_{1}=\mu
_{2}$. Lemma \ref{lem.h^(1).equality} is now proven.
\end{vershort}

\begin{verlong}
\textit{Proof of Lemma \ref{lem.h^(1).equality}.} First, we notice that
$\mu_{1}$ and $\mu_{2}$ are two $k$-bilinear maps (since $\left(  V,\mu
_{1}\right)  $ and $\left(  V,\mu_{2}\right)  $ are two $\mathfrak{g}$-modules).

Let $\mathfrak{h}$ be the subset%
\[
\left\{  a\in\mathfrak{g}\ \mid\ \mu_{1}\left(  a,v\right)  =\mu_{2}\left(
a,v\right)  \text{ for every }v\in V\right\}
\]
of $\mathfrak{g}$.

We note that every $s\in S$ satisfies $\mu_{1}\left(  s,v\right)  =\mu
_{2}\left(  s,v\right)  $ for every $v\in V$. In other words, every $s\in S$
satisfies $s\in\left\{  a\in\mathfrak{g}\ \mid\ \mu_{1}\left(  a,v\right)
=\mu_{2}\left(  a,v\right)  \text{ for every }v\in V\right\}  =\mathfrak{h}$.
This yields $S\subseteq\mathfrak{h}$.

Then,%
\begin{equation}
\left(  \text{every }x\in\mathfrak{h}\text{ and every }\lambda\in k\text{
satisfy }\lambda x\in\mathfrak{h}\right)  . \label{pf.h^(1).equality.1}%
\end{equation}


\textit{Proof of (\ref{pf.h^(1).equality.1}).} Let $x\in\mathfrak{h}$ and
$\lambda\in k$ be arbitrary. Then,
\[
x\in\mathfrak{h}=\left\{  a\in\mathfrak{g}\ \mid\ \mu_{1}\left(  a,v\right)
=\mu_{2}\left(  a,v\right)  \text{ for every }v\in V\right\}  .
\]
Thus, $\mu_{1}\left(  x,v\right)  =\mu_{2}\left(  x,v\right)  $ for every
$v\in V$. Thus,%
\begin{align*}
\mu_{1}\left(  \lambda x,v\right)   &  =\lambda\mu_{1}\left(  x,v\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu_{1}\text{ is }k\text{-bilinear}%
\right) \\
&  =\lambda\mu_{2}\left(  x,v\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\mu_{1}\left(  x,v\right)  =\mu_{2}\left(  x,v\right)  \right) \\
&  =\mu_{2}\left(  \lambda x,v\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\mu_{2}\text{ is }k\text{-bilinear}\right)
\end{align*}
for every $v\in V$. This yields that $\lambda x\in\left\{  a\in\mathfrak{g}%
\ \mid\ \mu_{1}\left(  a,v\right)  =\mu_{2}\left(  a,v\right)  \text{ for
every }v\in V\right\}  =\mathfrak{h}$. This proves (\ref{pf.h^(1).equality.1}).

Besides,%
\begin{equation}
\left(  \text{every }x\in\mathfrak{h}\text{ and every }y\in\mathfrak{h}\text{
satisfy }x+y\in\mathfrak{h}\right)  . \label{pf.h^(1).equality.2}%
\end{equation}


\textit{Proof of (\ref{pf.h^(1).equality.2}).} Let $x\in\mathfrak{h}$ and
$y\in\mathfrak{h}$ be arbitrary. Then, $\mu_{1}\left(  x,v\right)  =\mu
_{2}\left(  x,v\right)  $ for every $v\in V$ (because $x\in\mathfrak{h}%
=\left\{  a\in\mathfrak{g}\ \mid\ \mu_{1}\left(  a,v\right)  =\mu_{2}\left(
a,v\right)  \text{ for every }v\in V\right\}  $). On the other hand, $\mu
_{1}\left(  y,v\right)  =\mu_{2}\left(  y,v\right)  $ for every $v\in V$
(because \newline$y\in\mathfrak{h}=\left\{  a\in\mathfrak{g}\ \mid\ \mu
_{1}\left(  a,v\right)  =\mu_{2}\left(  a,v\right)  \text{ for every }v\in
V\right\}  $). Thus,%
\begin{align*}
\mu_{1}\left(  x+y,v\right)   &  =\underbrace{\mu_{1}\left(  x,v\right)
}_{=\mu_{2}\left(  x,v\right)  }+\underbrace{\mu_{1}\left(  y,v\right)
}_{=\mu_{2}\left(  y,v\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\mu_{1}\text{ is }k\text{-bilinear}\right) \\
&  =\mu_{2}\left(  x,v\right)  +\mu_{2}\left(  y,v\right)  =\mu_{2}\left(
x+y,v\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mu_{2}\text{ is
}k\text{-bilinear}\right)
\end{align*}
for every $v\in V$. This yields that $x+y\in\left\{  a\in\mathfrak{g}%
\ \mid\ \mu_{1}\left(  a,v\right)  =\mu_{2}\left(  a,v\right)  \text{ for
every }v\in V\right\}  =\mathfrak{h}$. This proves (\ref{pf.h^(1).equality.2}).

Finally,%
\begin{equation}
\left(  \text{every }x\in\mathfrak{h}\text{ and every }y\in\mathfrak{h}\text{
satisfy }\left[  x,y\right]  \in\mathfrak{h}\right)  .
\label{pf.h^(1).equality.3}%
\end{equation}


\textit{Proof of (\ref{pf.h^(1).equality.3}).} Let $x\in\mathfrak{h}$ and
$y\in\mathfrak{h}$ be arbitrary. Then,
\begin{equation}
\mu_{1}\left(  x,v\right)  =\mu_{2}\left(  x,v\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }v\in V \label{pf.h^(1).equality.3a}%
\end{equation}
(because $x\in\mathfrak{h}=\left\{  a\in\mathfrak{g}\ \mid\ \mu_{1}\left(
a,v\right)  =\mu_{2}\left(  a,v\right)  \text{ for every }v\in V\right\}  $).
On the other hand,
\begin{equation}
\mu_{1}\left(  y,v\right)  =\mu_{2}\left(  y,v\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }v\in V \label{pf.h^(1).equality.3b}%
\end{equation}
(because $y\in\mathfrak{h}=\left\{  a\in\mathfrak{g}\ \mid\ \mu_{1}\left(
a,v\right)  =\mu_{2}\left(  a,v\right)  \text{ for every }v\in V\right\}  $).
Thus, every $v\in V$ satisfies%
\begin{align*}
\mu_{1}\left(  \left[  x,y\right]  ,v\right)   &  =\underbrace{\mu_{1}\left(
x,\mu_{1}\left(  y,v\right)  \right)  }_{\substack{=\mu_{2}\left(  x,\mu
_{1}\left(  y,v\right)  \right)  \\\text{(due to (\ref{pf.h^(1).equality.3a}%
),}\\\text{applied to }\mu_{1}\left(  y,v\right)  \text{ instead of }%
v\text{)}}}-\underbrace{\mu_{1}\left(  y,\mu_{1}\left(  x,v\right)  \right)
}_{\substack{=\mu_{2}\left(  y,\mu_{1}\left(  x,v\right)  \right)
\\\text{(due to (\ref{pf.h^(1).equality.3b}),}\\\text{applied to }\mu
_{1}\left(  x,v\right)  \text{ instead of }v\text{)}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{due to (\ref{L-mod}) (applied to }\mu_{1}\text{, }x\text{ and }y\text{
instead of }\mu\text{, }a\text{ and }b\text{),}\\
\text{since }\left(  V,\mu_{1}\right)  \text{ is a }\mathfrak{g}\text{-module}%
\end{array}
\right) \\
&  =\mu_{2}\left(  x,\underbrace{\mu_{1}\left(  y,v\right)  }_{\substack{=\mu
_{2}\left(  y,v\right)  \\\text{(due to (\ref{pf.h^(1).equality.3b}))}%
}}\right)  -\mu_{2}\left(  y,\underbrace{\mu_{1}\left(  x,v\right)
}_{\substack{=\mu_{2}\left(  x,v\right)  \\\text{(due to
(\ref{pf.h^(1).equality.3a}))}}}\right) \\
&  =\mu_{2}\left(  x,\mu_{2}\left(  y,v\right)  \right)  -\mu_{2}\left(
y,\mu_{2}\left(  x,v\right)  \right)  .
\end{align*}
Comparing this with%
\begin{align*}
\mu_{2}\left(  \left[  x,y\right]  ,v\right)   &  =\mu_{2}\left(  x,\mu
_{2}\left(  y,v\right)  \right)  -\mu_{2}\left(  y,\mu_{2}\left(  x,v\right)
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{due to (\ref{L-mod}) (applied to }\mu_{2}\text{, }x\text{ and }y\text{
instead of }\mu\text{, }a\text{ and }b\text{),}\\
\text{since }\left(  V,\mu_{2}\right)  \text{ is a }\mathfrak{g}\text{-module}%
\end{array}
\right)  ,
\end{align*}
we conclude that $\mu_{1}\left(  \left[  x,y\right]  ,v\right)  =\mu
_{2}\left(  \left[  x,y\right]  ,v\right)  $ for every $v\in V$. In other
words, $\left[  x,y\right]  \in\left\{  a\in\mathfrak{g}\ \mid\ \mu_{1}\left(
a,v\right)  =\mu_{2}\left(  a,v\right)  \text{ for every }v\in V\right\}
=\mathfrak{h}$. This proves (\ref{pf.h^(1).equality.3}).

From (\ref{pf.h^(1).equality.1}) and (\ref{pf.h^(1).equality.2}), we conclude
that $\mathfrak{h}$ is a $k$-submodule of $\mathfrak{g}$.

Since we now know that $\mathfrak{h}$ is a $k$-submodule of $\mathfrak{g}$
which satisfies (\ref{pf.h^(1).equality.3}), we conclude that $\mathfrak{h}$
is a Lie subalgebra of $\mathfrak{g}$. Since $S\subseteq\mathfrak{h}$, this
shows that $\mathfrak{h}$ is a Lie subalgebra of $\mathfrak{g}$ which contains
$S$ as a subset.

But we know that the Lie algebra $\mathfrak{g}$ is generated (as a Lie
algebra) by the subset $S$. This means that $\mathfrak{g}$ is identical with
the Lie subalgebra of $\mathfrak{g}$ generated by $S$. So we have%
\begin{align*}
\mathfrak{g}  &  =\left(  \text{the Lie subalgebra of }\mathfrak{g}\text{
generated by }S\right) \\
&  =\left(  \text{the smallest Lie subalgebra of }\mathfrak{g}\text{ which
contains }S\text{ as a subset}\right) \\
&  \subseteq\mathfrak{h}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\mathfrak{h}\text{ is a Lie subalgebra of }\mathfrak{g}\text{ which contains
}S\text{ as a subset}\right)  .
\end{align*}
Thus, every $x\in\mathfrak{g}$ satisfies $\mu_{1}\left(  x,v\right)  =\mu
_{2}\left(  x,v\right)  $ for every $v\in V$ (because $x\in\mathfrak{g}%
\subseteq\mathfrak{h}=\left\{  a\in\mathfrak{g}\ \mid\ \mu_{1}\left(
a,v\right)  =\mu_{2}\left(  a,v\right)  \text{ for every }v\in V\right\}  $).
In other words, every $\left(  x,v\right)  \in\mathfrak{g}\times V$ satisfies
$\mu_{1}\left(  x,v\right)  =\mu_{2}\left(  x,v\right)  $. This yields that
$\mu_{1}=\mu_{2}$. Lemma \ref{lem.h^(1).equality} is now proven.
\end{verlong}

Our next lemma is proven in the same spirit:

\begin{lemma}
\label{lem.h^(1).linearity}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie algebra. Let $S$ be a subset of $\mathfrak{g}$ such that the Lie
algebra $\mathfrak{g}$ is generated (as a Lie algebra) by the subset
$S$.\newline Let $V$ and $W$ be two $\mathfrak{g}$-modules, and let
$f:V\rightarrow W$ be a $k$-module homomorphism.\newline Assume that every
$s\in S$ and every $v\in V$ satisfy $f\left(  s\rightharpoonup v\right)
=s\rightharpoonup\left(  f\left(  v\right)  \right)  $.\newline Then, $f$ is a
$\mathfrak{g}$-module homomorphism.
\end{lemma}

\begin{vershort}
\textit{Proof of Lemma \ref{lem.h^(1).linearity}.} The proof follows the same
track as that of Lemma \ref{lem.h^(1).equality}, except that $\mathfrak{h}$ is
defined as the subset%
\[
\left\{  a\in\mathfrak{g}\ \mid\ f\left(  a\rightharpoonup v\right)
=a\rightharpoonup\left(  f\left(  v\right)  \right)  \text{ for every }v\in
V\right\}  \subseteq\mathfrak{g}%
\]
this time.
\end{vershort}

\begin{verlong}
\textit{Proof of Lemma \ref{lem.h^(1).linearity}.} Let $\mathfrak{h}$ be the
subset%
\[
\left\{  a\in\mathfrak{g}\ \mid\ f\left(  a\rightharpoonup v\right)
=a\rightharpoonup\left(  f\left(  v\right)  \right)  \text{ for every }v\in
V\right\}
\]
of $\mathfrak{g}$.

We note that every $s\in S$ satisfies $f\left(  s\rightharpoonup v\right)
=s\rightharpoonup\left(  f\left(  v\right)  \right)  $ for every $v\in V$. In
other words, every $s\in S$ satisfies $s\in\left\{  a\in\mathfrak{g}%
\ \mid\ f\left(  a\rightharpoonup v\right)  =a\rightharpoonup\left(  f\left(
v\right)  \right)  \text{ for every }v\in V\right\}  =\mathfrak{h}$. This
yields $S\subseteq\mathfrak{h}$.

Then,%
\begin{equation}
\left(  \text{every }x\in\mathfrak{h}\text{ and every }\lambda\in k\text{
satisfy }\lambda x\in\mathfrak{h}\right)  . \label{pf.h^(1).linearity.1}%
\end{equation}


\textit{Proof of (\ref{pf.h^(1).linearity.1}).} Let $x\in\mathfrak{h}$ and
$\lambda\in k$ be arbitrary. Then,%
\[
x\in\mathfrak{h}=\left\{  a\in\mathfrak{g}\ \mid\ f\left(  a\rightharpoonup
v\right)  =a\rightharpoonup\left(  f\left(  v\right)  \right)  \text{ for
every }v\in V\right\}  .
\]
Thus, $f\left(  x\rightharpoonup v\right)  =x\rightharpoonup\left(  f\left(
v\right)  \right)  $ for every $v\in V$. Thus,%
\begin{align*}
f\left(  \underbrace{\left(  \lambda x\right)  \rightharpoonup v}%
_{\substack{=\lambda\left(  x\rightharpoonup v\right)  \\\text{(since the Lie
action}\\\text{on }V\text{ is }k\text{-bilinear)}}}\right)   &  =f\left(
\lambda\left(  x\rightharpoonup v\right)  \right)  =\lambda
\underbrace{f\left(  x\rightharpoonup v\right)  }_{=x\rightharpoonup\left(
f\left(  v\right)  \right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{
is }k\text{-linear}\right) \\
&  =\lambda\left(  x\rightharpoonup\left(  f\left(  v\right)  \right)
\right)  =\left(  \lambda x\right)  \rightharpoonup\left(  f\left(  v\right)
\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Lie action on }W\text{ is
}k\text{-bilinear}\right)
\end{align*}
for every $v\in V$. This yields that $\lambda x\in\left\{  a\in\mathfrak{g}%
\ \mid\ f\left(  a\rightharpoonup v\right)  =a\rightharpoonup\left(  f\left(
v\right)  \right)  \text{ for every }v\in V\right\}  =\mathfrak{h}$. This
proves (\ref{pf.h^(1).linearity.1}).

Besides,%
\begin{equation}
\left(  \text{every }x\in\mathfrak{h}\text{ and every }y\in\mathfrak{h}\text{
satisfy }x+y\in\mathfrak{h}\right)  . \label{pf.h^(1).linearity.2}%
\end{equation}


\textit{Proof of (\ref{pf.h^(1).linearity.2}).} Let $x\in\mathfrak{h}$ and
$y\in\mathfrak{h}$ be arbitrary. Then, $f\left(  x\rightharpoonup v\right)
=x\rightharpoonup\left(  f\left(  v\right)  \right)  $ for every $v\in V$
(because $x\in\mathfrak{h}=\left\{  a\in\mathfrak{g}\ \mid\ f\left(
a\rightharpoonup v\right)  =a\rightharpoonup\left(  f\left(  v\right)
\right)  \text{ for every }v\in V\right\}  $). On the other hand, $f\left(
y\rightharpoonup v\right)  =y\rightharpoonup\left(  f\left(  v\right)
\right)  $ for every $v\in V$ (because \newline$y\in\mathfrak{h}=\left\{
a\in\mathfrak{g}\ \mid\ f\left(  a\rightharpoonup v\right)  =a\rightharpoonup
\left(  f\left(  v\right)  \right)  \text{ for every }v\in V\right\}  $).
Thus,%
\begin{align*}
f\left(  \underbrace{\left(  x+y\right)  \rightharpoonup v}%
_{\substack{=x\rightharpoonup v+y\rightharpoonup v\\\text{(since the Lie
action}\\\text{on }V\text{ is }k\text{-bilinear)}}}\right)   &  =f\left(
x\rightharpoonup v+y\rightharpoonup v\right)  =\underbrace{f\left(
x\rightharpoonup v\right)  }_{=x\rightharpoonup\left(  f\left(  v\right)
\right)  }+\underbrace{f\left(  y\rightharpoonup v\right)  }%
_{=y\rightharpoonup\left(  f\left(  v\right)  \right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is }k\text{-linear}\right) \\
&  =x\rightharpoonup\left(  f\left(  v\right)  \right)  +y\rightharpoonup
\left(  f\left(  v\right)  \right)  =\left(  x+y\right)  \rightharpoonup
\left(  f\left(  v\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Lie action on }V\text{ is
}k\text{-bilinear}\right)
\end{align*}
for every $v\in V$. This yields that $x+y\in\left\{  a\in\mathfrak{g}%
\ \mid\ f\left(  a\rightharpoonup v\right)  =a\rightharpoonup\left(  f\left(
v\right)  \right)  \text{ for every }v\in V\right\}  =\mathfrak{h}$. This
proves (\ref{pf.h^(1).linearity.2}).

Finally,%
\begin{equation}
\left(  \text{every }x\in\mathfrak{h}\text{ and every }y\in\mathfrak{h}\text{
satisfy }\left[  x,y\right]  \in\mathfrak{h}\right)  .
\label{pf.h^(1).linearity.3}%
\end{equation}


\textit{Proof of (\ref{pf.h^(1).linearity.3}).} Let $x\in\mathfrak{h}$ and
$y\in\mathfrak{h}$ be arbitrary. Then,
\begin{equation}
f\left(  x\rightharpoonup v\right)  =x\rightharpoonup\left(  f\left(
v\right)  \right)  \ \ \ \ \ \ \ \ \ \ \text{for every }v\in V
\label{pf.h^(1).linearity.3a}%
\end{equation}
(because $x\in\mathfrak{h}=\left\{  a\in\mathfrak{g}\ \mid\ f\left(
a\rightharpoonup v\right)  =a\rightharpoonup\left(  f\left(  v\right)
\right)  \text{ for every }v\in V\right\}  $). On the other hand,
\begin{equation}
f\left(  y\rightharpoonup v\right)  =y\rightharpoonup\left(  f\left(
v\right)  \right)  \ \ \ \ \ \ \ \ \ \ \text{for every }v\in V
\label{pf.h^(1).linearity.3b}%
\end{equation}
(because $y\in\mathfrak{h}=\left\{  a\in\mathfrak{g}\ \mid\ f\left(
a\rightharpoonup v\right)  =a\rightharpoonup\left(  f\left(  v\right)
\right)  \text{ for every }v\in V\right\}  $). Thus, every $v\in V$ satisfies%
\begin{align*}
f\left(  \left[  x,y\right]  \rightharpoonup v\right)   &  =f\left(
x\rightharpoonup\left(  y\rightharpoonup v\right)  -y\rightharpoonup\left(
x\rightharpoonup v\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since (\ref{[L-mod]}) (applied to }x\text{ and }y\text{ instead of
}a\text{ and }b\text{)}\\
\text{yields }\left[  x,y\right]  \rightharpoonup v=x\rightharpoonup\left(
y\rightharpoonup v\right)  -y\rightharpoonup\left(  x\rightharpoonup v\right)
\\
\text{(since }V\text{ is a }\mathfrak{g}\text{-module)}%
\end{array}
\right) \\
&  =\underbrace{f\left(  x\rightharpoonup\left(  y\rightharpoonup v\right)
\right)  }_{\substack{=x\rightharpoonup\left(  f\left(  y\rightharpoonup
v\right)  \right)  \\\text{(due to (\ref{pf.h^(1).linearity.3a}),}%
\\\text{applied to }y\rightharpoonup v\text{ instead of }v\text{)}%
}}-\underbrace{f\left(  y\rightharpoonup\left(  x\rightharpoonup v\right)
\right)  }_{\substack{=y\rightharpoonup\left(  f\left(  x\rightharpoonup
v\right)  \right)  \\\text{(due to (\ref{pf.h^(1).linearity.3b}),}%
\\\text{applied to }x\rightharpoonup v\text{ instead of }v\text{)}%
}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is }k\text{-linear }\right)
\\
&  =x\rightharpoonup\underbrace{\left(  f\left(  y\rightharpoonup v\right)
\right)  }_{\substack{=y\rightharpoonup\left(  f\left(  v\right)  \right)
\\\text{(due to (\ref{pf.h^(1).linearity.3a}))}}}-\underbrace{y\rightharpoonup
\left(  f\left(  x\rightharpoonup v\right)  \right)  }%
_{\substack{=x\rightharpoonup\left(  f\left(  v\right)  \right)  \\\text{(due
to (\ref{pf.h^(1).linearity.3b}))}}}\\
&  =x\rightharpoonup\left(  y\rightharpoonup\left(  f\left(  v\right)
\right)  \right)  -y\rightharpoonup\left(  x\rightharpoonup\left(  f\left(
v\right)  \right)  \right)  .
\end{align*}
Comparing this with%
\begin{align*}
\left[  x,y\right]  \rightharpoonup\left(  f\left(  v\right)  \right)   &
=x\rightharpoonup\left(  y\rightharpoonup\left(  f\left(  v\right)  \right)
\right)  -y\rightharpoonup\left(  x\rightharpoonup\left(  f\left(  v\right)
\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{due to (\ref{[L-mod]}) (applied to }W\text{, }f\left(  v\right)  \text{,
}x\text{ and }y\text{ instead of }V\text{, }v\text{, }a\text{ and }%
b\text{),}\\
\text{since }W\text{ is a }\mathfrak{g}\text{-module}%
\end{array}
\right)  ,
\end{align*}
we conclude that $f\left(  \left[  x,y\right]  \rightharpoonup v\right)
=\left[  x,y\right]  \rightharpoonup\left(  f\left(  v\right)  \right)  $ for
every $v\in V$. In other words, $\left[  x,y\right]  \in\left\{
a\in\mathfrak{g}\ \mid\ f\left(  a\rightharpoonup v\right)  =a\rightharpoonup
\left(  f\left(  v\right)  \right)  \text{ for every }v\in V\right\}
=\mathfrak{h}$. This proves (\ref{pf.h^(1).linearity.3}).

From (\ref{pf.h^(1).linearity.1}) and (\ref{pf.h^(1).linearity.2}), we
conclude that $\mathfrak{h}$ is a $k$-submodule of $\mathfrak{g}$.

Since we now know that $\mathfrak{h}$ is a $k$-submodule of $\mathfrak{g}$
which satisfies (\ref{pf.h^(1).linearity.3}), we conclude that $\mathfrak{h}$
is a Lie subalgebra of $\mathfrak{g}$. Since $S\subseteq\mathfrak{h}$, this
shows that $\mathfrak{h}$ is a Lie subalgebra of $\mathfrak{g}$ which contains
$S$ as a subset.

But we know that the Lie algebra $\mathfrak{g}$ is generated (as a Lie
algebra) by the subset $S$. This means that $\mathfrak{g}$ is identical with
the Lie subalgebra of $\mathfrak{g}$ generated by $S$. So we have%
\begin{align*}
\mathfrak{g}  &  =\left(  \text{the Lie subalgebra of }\mathfrak{g}\text{
generated by }S\right) \\
&  =\left(  \text{the smallest Lie subalgebra of }\mathfrak{g}\text{ which
contains }S\text{ as a subset}\right) \\
&  \subseteq\mathfrak{h}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\mathfrak{h}\text{ is a Lie subalgebra of }\mathfrak{g}\text{ which contains
}S\text{ as a subset}\right)  .
\end{align*}
Thus, every $x\in\mathfrak{g}$ satisfies $f\left(  x\rightharpoonup v\right)
=x\rightharpoonup\left(  f\left(  v\right)  \right)  $ for every $v\in V$
(because $x\in\mathfrak{g}\subseteq\mathfrak{h}=\left\{  a\in\mathfrak{g}%
\ \mid\ f\left(  a\rightharpoonup v\right)  =a\rightharpoonup\left(  f\left(
v\right)  \right)  \text{ for every }v\in V\right\}  $). In other words, every
$\left(  x,v\right)  \in\mathfrak{g}\times V$ satisfies $f\left(
x\rightharpoonup v\right)  =x\rightharpoonup\left(  f\left(  v\right)
\right)  $. This yields that $f$ is a $\mathfrak{g}$-module homomorphism.
Lemma \ref{lem.h^(1).linearity} is now proven.
\end{verlong}

Finally:

\begin{lemma}
\label{lem.h^(1).lieideal}Let $k$ be a commutative ring. Let $\mathfrak{g}$
and $\mathfrak{h}$ be two $k$-Lie algebras. Let $f:\mathfrak{g}\rightarrow
\mathfrak{h}$ be a $k$-Lie algebra homomorphism.\newline Then,
$\operatorname{Ker}f$ is a Lie ideal of $\mathfrak{g}$.
\end{lemma}

\begin{vershort}
The proof is omitted for triviality.
\end{vershort}

\begin{verlong}
\textit{Proof of Lemma \ref{lem.h^(1).lieideal}.} Let $v\in\mathfrak{g}$ and
$x\in\operatorname{Ker}f$ be arbitrary. Then, $f\left(  x\right)  =0$ (since
$x\in\operatorname{Ker}f$) and%
\begin{align*}
f\left(  \left[  v,x\right]  \right)   &  =\left[  f\left(  v\right)
,\underbrace{f\left(  x\right)  }_{=0}\right]  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }f\text{ is a }k\text{-Lie algebra homomorphism}\right) \\
&  =\left[  f\left(  v\right)  ,0\right]  =0\ \ \ \ \ \ \ \ \ \ \left(
\text{since the Lie bracket on }\mathfrak{h}\text{ is }k\text{-bilinear}%
\right)  .
\end{align*}
Thus, $\left[  v,x\right]  \in\operatorname{Ker}f$.

We have thus proven that $\left[  v,x\right]  \in\operatorname{Ker}f$ for
every $v\in\mathfrak{g}$ and $x\in\operatorname{Ker}f$. Since we know that
$\operatorname{Ker}f$ is a $k$-submodule of $\mathfrak{g}$ (because $f$ is
$k$-linear), this yields that $\operatorname{Ker}f$ is a Lie ideal of
$\mathfrak{g}$.
\end{verlong}

\textit{Proof of Proposition \ref{prop.h^(1)} (sketched).} We record the
universal property of the free Lie algebra $\operatorname*{FreeLie}%
\mathfrak{g}$ on $\mathfrak{g}$: It claims that%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{for every }k\text{-Lie algebra }\mathfrak{u}\text{ and every
}k\text{-module homomorphism }p:\mathfrak{g}\rightarrow\mathfrak{u}\text{,}\\
\text{there exists one and only one }k\text{-Lie algebra homomorphism}\\
P:\operatorname*{FreeLie}\mathfrak{g}\rightarrow\mathfrak{u}\text{ satisfying
}P\circ\iota=p
\end{array}
\right)  . \label{pf.h^(1).free}%
\end{equation}


According to the construction of the free Lie algebra $\operatorname{FreeLie}%
\mathfrak{g}$, every element of $\operatorname{FreeLie}\mathfrak{g}$ can be
obtained by repeated addition, scalar multiplication (i. e., multiplication
with elements of $k$) and forming the Lie bracket from elements of the form
$\iota\left(  a\right)  $ with $a\in\mathfrak{g}$. Since $\mathfrak{h}^{(1)}$
is a quotient of this Lie algebra $\operatorname{FreeLie}\mathfrak{g}$, we
thus conclude that every element of $\mathfrak{h}^{(1)}$ can be obtained by
repeated addition, scalar multiplication (i. e., multiplication with elements
of $k$) and forming the Lie bracket from elements of the form $\overline
{\iota\left(  a\right)  }$ with $a\in\mathfrak{g}$. In other words, the
$k$-Lie algebra $\mathfrak{h}^{(1)}$ is generated (as a Lie algebra) by the
elements $\overline{\iota\left(  a\right)  }$ with $a\in\mathfrak{g}$. In
other words, the $k$-Lie algebra $\mathfrak{h}^{(1)}$ is generated (as a Lie
algebra) by the subset $S$, where $S=\left\{  \overline{\iota\left(  a\right)
}\ \mid\ a\in\mathfrak{g}\right\}  $.

\begin{noncompile}
This assertion can be understood in two equivalent ways: First,%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{the smallest Lie subalgebra of }\mathfrak{h}^{(1)}\text{ which contains
}S\\
\text{(as a subset) is }\mathfrak{h}^{(1)}\text{ itself}%
\end{array}
\right)  . \label{pf.h^(1).gen1}%
\end{equation}
Second,%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{every element of }\mathfrak{h}^{(1)}\text{ can be obtained by repeated
addition,}\\
\text{scalar multiplication (i. e., multiplication with elements of }%
k\text{)}\\
\text{and forming of the Lie bracket from elements of the form}\\
\overline{\iota\left(  a\right)  }\text{ with }a\in\mathfrak{g}%
\end{array}
\right)  . \label{pf.h^(1).gen2}%
\end{equation}

\end{noncompile}

\textbf{(a)} Let $V$ be a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule. We are now going to prove the following two assertions:

\textit{Assertion }$\mathcal{X}$\textit{:} There exists a canonical
$\mathfrak{h}^{(1)}$-module structure on $V$ satisfying (\ref{prop.h^(1).a}).

\textit{Assertion }$\mathcal{Y}$\textit{:} There exists at most one
$\mathfrak{h}^{(1)}$-module structure on $V$ satisfying (\ref{prop.h^(1).a}).

\textit{Proof of Assertion }$\mathcal{X}$\textit{:} For every $a\in
\mathfrak{g}$, let $\beta_{a}:V\rightarrow V$ be the map which sends every
$v\in V$ to $a\rightharpoonup v\in V$. Then, $\beta_{a}\in\operatorname{End}%
V$. Now, define a map $\beth:\mathfrak{g}\rightarrow\operatorname{End}V$ by%
\[
\left(  \beth\left(  a\right)  =\beta_{a}\ \ \ \ \ \ \ \ \ \ \text{for every
}a\in\mathfrak{g}\right)  .
\]
We consider $\operatorname{End}V$ as a $k$-Lie algebra, with the commutator of
endomorphisms as its Lie bracket. Every $\left(  a,b\right)  \in
\mathfrak{h}\times\mathfrak{g}$ and every $v\in V$ satisfy%
\begin{align*}
&  \left(  \underbrace{\left[  \beth\left(  a\right)  ,\beth\left(  b\right)
\right]  }_{=\left(  \beth\left(  a\right)  \right)  \circ\left(  \beth\left(
b\right)  \right)  -\left(  \beth\left(  b\right)  \right)  \circ\left(
\beth\left(  a\right)  \right)  }-\beth\left(  \left[  a,b\right]  \right)
\right)  \left(  v\right) \\
&  =\left(  \left(  \beth\left(  a\right)  \right)  \circ\left(  \beth\left(
b\right)  \right)  -\left(  \beth\left(  b\right)  \right)  \circ\left(
\beth\left(  a\right)  \right)  -\beth\left(  \left[  a,b\right]  \right)
\right)  \left(  v\right) \\
&  =\underbrace{\left(  \beth\left(  a\right)  \right)  }_{=\beta_{a}}\left(
\underbrace{\left(  \beth\left(  b\right)  \right)  }_{=\beta_{b}}\left(
v\right)  \right)  -\underbrace{\left(  \beth\left(  b\right)  \right)
}_{=\beta_{b}}\left(  \underbrace{\left(  \beth\left(  a\right)  \right)
}_{=\beta_{a}}\left(  v\right)  \right)  -\underbrace{\left(  \beth\left(
\left[  a,b\right]  \right)  \right)  }_{=\beta_{\left[  a,b\right]  }}\left(
v\right) \\
&  =\beta_{a}\left(  \underbrace{\beta_{b}\left(  v\right)  }%
_{=b\rightharpoonup v}\right)  -\beta_{b}\left(  \underbrace{\beta_{a}\left(
v\right)  }_{=a\rightharpoonup v}\right)  -\underbrace{\beta_{\left[
a,b\right]  }\left(  v\right)  }_{=\left[  a,b\right]  \rightharpoonup
v}=\underbrace{\beta_{a}\left(  b\rightharpoonup v\right)  }%
_{=a\rightharpoonup\left(  b\rightharpoonup v\right)  }-\underbrace{\beta
_{b}\left(  a\rightharpoonup v\right)  }_{=b\rightharpoonup\left(
a\rightharpoonup v\right)  }-\underbrace{\left[  a,b\right]  \rightharpoonup
v}_{\substack{=a\rightharpoonup\left(  b\rightharpoonup v\right)
-b\rightharpoonup\left(  a\rightharpoonup v\right)  \\\text{(by
(\ref{defs.semimod.->}))}}}\\
&  =a\rightharpoonup\left(  b\rightharpoonup v\right)  -b\rightharpoonup
\left(  a\rightharpoonup v\right)  -\left(  a\rightharpoonup\left(
b\rightharpoonup v\right)  -b\rightharpoonup\left(  a\rightharpoonup v\right)
\right)  =0.
\end{align*}
This means that%
\begin{equation}
\text{every }\left(  a,b\right)  \in\mathfrak{h}\times\mathfrak{g}\text{
satisfies }\left[  \beth\left(  a\right)  ,\beth\left(  b\right)  \right]
-\beth\left(  \left[  a,b\right]  \right)  =0. \label{pf.h^(1).beth}%
\end{equation}


Applying (\ref{pf.h^(1).free}) to $\mathfrak{u}=\operatorname{End}V$ and
$p=\beth$, we see that there exists one and only one $k$-Lie algebra
homomorphism $P:\operatorname*{FreeLie}\mathfrak{g}\rightarrow
\operatorname{End}V$ satisfying $P\circ\iota=\beth$. Let $P_{1}$ be this
homomorphism. Then, $P_{1}\circ\iota=\beth$.

We now are going to show that $P_{1}\left(  \mathfrak{i}\right)  =0$.

Since $P_{1}$ is a $k$-Lie algebra homomorphism, Lemma
\ref{lem.h^(1).lieideal} (applied to $\operatorname{FreeLie}\mathfrak{g}$,
$\operatorname{End}V$ and $P_{1}$ instead of $\mathfrak{g}$, $\mathfrak{h}$
and $f$) yields that $\operatorname{Ker}P_{1}$ is a Lie ideal of
$\operatorname*{FreeLie}\mathfrak{g}$.

Now, let $T$ be the $k$-submodule $\left\langle \left[  \iota\left(  v\right)
,\iota\left(  w\right)  \right]  -\iota\left(  \left[  v,w\right]  \right)
\ \mid\ \left(  v,w\right)  \in\mathfrak{h}\times\mathfrak{g}\right\rangle $
of $\operatorname*{FreeLie}\mathfrak{g}$. Then, $\mathfrak{i}$ is the Lie
ideal of $\operatorname*{FreeLie}\mathfrak{g}$ generated by the subset $T$
(since $\mathfrak{i}$ was defined as the Lie ideal of $\operatorname*{FreeLie}%
\mathfrak{g}$ generated by the $k$-submodule $\left\langle \left[
\iota\left(  v\right)  ,\iota\left(  w\right)  \right]  -\iota\left(  \left[
v,w\right]  \right)  \ \mid\ \left(  v,w\right)  \in\mathfrak{h}%
\times\mathfrak{g}\right\rangle $ of $\operatorname*{FreeLie}\mathfrak{g}$).
In other words, $\mathfrak{i}$ is the smallest Lie ideal of
$\operatorname{FreeLie}\mathfrak{g}$ which contains $T$ as a subset.

Now,%
\begin{align*}
&  P_{1}\left(  \left\{  \left[  \iota\left(  v\right)  ,\iota\left(
w\right)  \right]  -\iota\left(  \left[  v,w\right]  \right)  \ \mid\ \left(
v,w\right)  \in\mathfrak{h}\times\mathfrak{g}\right\}  \right) \\
&  =P_{1}\left(  \left\{  \left[  \iota\left(  a\right)  ,\iota\left(
b\right)  \right]  -\iota\left(  \left[  a,b\right]  \right)  \ \mid\ \left(
a,b\right)  \in\mathfrak{h}\times\mathfrak{g}\right\}  \right)  =0,
\end{align*}
because every $\left(  a,b\right)  \in\mathfrak{h}\times\mathfrak{g}$
satisfies%
\begin{align*}
P_{1}\left(  \left[  \iota\left(  a\right)  ,\iota\left(  b\right)  \right]
-\iota\left(  \left[  a,b\right]  \right)  \right)   &  =\underbrace{P_{1}%
\left(  \left[  \iota\left(  a\right)  ,\iota\left(  b\right)  \right]
\right)  }_{\substack{=\left[  P_{1}\left(  \iota\left(  a\right)  \right)
,P_{1}\left(  \iota\left(  b\right)  \right)  \right]  \\\text{(since }%
P_{1}\text{ is a }k\text{-Lie algebra}\\\text{homomorphism)}}%
}-\underbrace{P_{1}\left(  \iota\left(  \left[  a,b\right]  \right)  \right)
}_{=\left(  P_{1}\circ\iota\right)  \left(  \left[  a,b\right]  \right)  }\\
&  =\left[  \underbrace{P_{1}\left(  \iota\left(  a\right)  \right)
}_{=\left(  P_{1}\circ\iota\right)  \left(  a\right)  },\underbrace{P_{1}%
\left(  \iota\left(  b\right)  \right)  }_{=\left(  P_{1}\circ\iota\right)
\left(  b\right)  }\right]  -\left(  P_{1}\circ\iota\right)  \left(  \left[
a,b\right]  \right) \\
&  =\left[  \underbrace{\left(  P_{1}\circ\iota\right)  }_{=\beth}\left(
a\right)  ,\underbrace{\left(  P_{1}\circ\iota\right)  }_{=\beth}\left(
b\right)  \right]  -\underbrace{\left(  P_{1}\circ\iota\right)  }_{=\beth
}\left(  \left[  a,b\right]  \right) \\
&  =\left[  \beth\left(  a\right)  ,\beth\left(  b\right)  \right]
-\beth\left(  \left[  a,b\right]  \right)  =0\ \ \ \ \ \ \ \ \ \ \left(
\text{by (\ref{pf.h^(1).beth})}\right)  .
\end{align*}


Now,
\begin{align*}
&  P_{1}\left(  T\right)  =P_{1}\left(  \left\langle \left[  \iota\left(
v\right)  ,\iota\left(  w\right)  \right]  -\iota\left(  \left[  v,w\right]
\right)  \ \mid\ \left(  v,w\right)  \in\mathfrak{h}\times\mathfrak{g}%
\right\rangle \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }T=\left\langle \left[
\iota\left(  v\right)  ,\iota\left(  w\right)  \right]  -\iota\left(  \left[
v,w\right]  \right)  \ \mid\ \left(  v,w\right)  \in\mathfrak{h}%
\times\mathfrak{g}\right\rangle \right) \\
&  =P_{1}\left(  \left\langle \left\{  \left[  \iota\left(  v\right)
,\iota\left(  w\right)  \right]  -\iota\left(  \left[  v,w\right]  \right)
\ \mid\ \left(  v,w\right)  \in\mathfrak{h}\times\mathfrak{g}\right\}
\right\rangle \right) \\
&  =\left\langle \underbrace{P_{1}\left(  \left\{  \left[  \iota\left(
v\right)  ,\iota\left(  w\right)  \right]  -\iota\left(  \left[  v,w\right]
\right)  \ \mid\ \left(  v,w\right)  \in\mathfrak{h}\times\mathfrak{g}%
\right\}  \right)  }_{=0}\right\rangle \ \ \ \ \ \ \ \ \ \ \left(  \text{since
}P_{1}\text{ is }k\text{-linear}\right) \\
&  =\left\langle 0\right\rangle =0,
\end{align*}
so that $T\subseteq\operatorname{Ker}P_{1}$. Since $\operatorname{Ker}P_{1}$
is a Lie ideal of $\operatorname*{FreeLie}\mathfrak{g}$, we thus conclude that
$\operatorname{Ker}P_{1}$ is a Lie ideal of $\operatorname*{FreeLie}%
\mathfrak{g}$ which contains $T$ as a subset. Thus,%
\begin{align*}
\operatorname{Ker}P_{1}  &  \supseteq\left(  \text{the smallest Lie ideal of
}\operatorname{FreeLie}\mathfrak{g}\text{ which contains }T\text{ as a
subset}\right) \\
&  =\mathfrak{i}.
\end{align*}
In other words, $P_{1}\left(  \mathfrak{i}\right)  =0$.

We thus have constructed a Lie algebra homomorphism $P_{1}%
:\operatorname*{FreeLie}\mathfrak{g}\rightarrow\operatorname{End}V$ satisfying
$P_{1}\circ\iota=\beth$, and we have shown that $P_{1}\left(  \mathfrak{i}%
\right)  =0$. Hence, by the homomorphism theorem (for Lie algebras), the
homomorphism $P_{1}$ factors through the Lie algebra $\left(
\operatorname*{FreeLie}\mathfrak{g}\right)  \diagup\mathfrak{i}=\mathfrak{h}%
^{(1)}$. That is, there exists a Lie algebra homomorphism $P_{2}%
:\mathfrak{h}^{(1)}\rightarrow\operatorname{End}V$ satisfying $P_{2}\circ
z=P_{1}$, where $z$ denotes the canonical projection $\operatorname*{FreeLie}%
\mathfrak{g}\rightarrow\mathfrak{h}^{(1)}$. Consider this $P_{2}$. Combining
$P_{2}\circ z=P_{1}$ with $P_{1}\circ\iota=\beth$, we obtain
$\underbrace{P_{2}\circ z}_{=P_{1}}\circ\iota=P_{1}\circ\iota=\beth$.

Now, Proposition \ref{prop.liemod.curry} \textbf{(c)} (applied to
$\mathfrak{h}^{(1)}$ instead of $\mathfrak{g}$) shows that $k$-Lie algebra
homomorphisms $\mathfrak{h}^{(1)}\rightarrow\operatorname{End}V$ stand in
1-to-1 correspondence with $\mathfrak{h}^{(1)}$-module structures on $V$.
Therefore, the Lie algebra homomorphism $P_{2}:\mathfrak{h}^{(1)}%
\rightarrow\operatorname{End}V$ gives rise to an $\mathfrak{h}^{(1)}$-module
structure on $V$. This structure satisfies%
\[
x\rightharpoonup v=\left(  P_{2}\left(  x\right)  \right)  \left(  v\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }x\in\mathfrak{h}^{(1)}\text{ and }v\in
V.
\]
Therefore, this structure satisfies%
\begin{align*}
\overline{\iota\left(  a\right)  }\rightharpoonup v  &  =\left(  P_{2}\left(
\overline{\iota\left(  a\right)  }\right)  \right)  \left(  v\right)  =\left(
P_{2}\left(  z\left(  \iota\left(  a\right)  \right)  \right)  \right)
\left(  v\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }z\text{ is the canonical
projection }\operatorname*{FreeLie}\mathfrak{g}\rightarrow\mathfrak{h}%
^{(1)}\text{, and thus }\overline{\iota\left(  a\right)  }=z\left(
\iota\left(  a\right)  \right)  \right) \\
&  =\left(  \underbrace{\left(  P_{2}\circ z\circ\iota\right)  }_{=\beth
}\left(  a\right)  \right)  \left(  v\right)  =\underbrace{\left(
\beth\left(  a\right)  \right)  }_{=\beta_{a}}\left(  v\right)  =\beta
_{a}\left(  v\right)  =a\rightharpoonup v
\end{align*}
for every $a\in\mathfrak{g}$ and $v\in V$.

We have thus proven that there exists a canonical $\mathfrak{h}^{(1)}$-module
structure on $V$ satisfying (\ref{prop.h^(1).a}). This proves Assertion
$\mathcal{X}$.

\textit{Proof of Assertion $\mathcal{Y}$:} Let $\mu_{1}$ and $\mu_{2}$ be two
$\mathfrak{h}^{(1)}$-module structures on $V$ which both satisfy
(\ref{prop.h^(1).a}). We are now going to show that $\mu_{1}=\mu_{2}$.

In fact, we have assumed that the $\mathfrak{h}^{(1)}$-module structure
$\mu_{1}$ satisfies (\ref{prop.h^(1).a}). In other words, we have%
\[
\left(  \overline{\iota\left(  a\right)  }\rightharpoonup v=a\rightharpoonup
v\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{ and }v\in
V\right)
\]
if we understand $\overline{\iota\left(  a\right)  }\rightharpoonup v$ to mean
$\mu_{1}\left(  \overline{\iota\left(  a\right)  },v\right)  $. In other
words, we have%
\[
\left(  \mu_{1}\left(  \overline{\iota\left(  a\right)  },v\right)
=a\rightharpoonup v\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{
and }v\in V\right)  .
\]
The same argument, applied to $\mu_{2}$ instead of $\mu_{1}$, shows that%
\[
\left(  \mu_{2}\left(  \overline{\iota\left(  a\right)  },v\right)
=a\rightharpoonup v\ \ \ \ \ \ \ \ \ \ \text{for every }a\in\mathfrak{g}\text{
and }v\in V\right)  .
\]
Thus, every $s\in S$ satisfies $\mu_{1}\left(  s,v\right)  =\mu_{2}\left(
s,v\right)  $ for every $v\in V$.\ \ \ \ {\footnotemark}%
\footnotetext{\textit{Proof.} Let $s\in S$ be arbitrary. Then, $s\in
S=\left\{  \overline{\iota\left(  a\right)  }\ \mid\ a\in\mathfrak{g}\right\}
$. Hence, there exists some $a\in\mathfrak{g}$ such that $s=\overline
{\iota\left(  a\right)  }$. Consider this $a$. Then, $\mu_{1}\left(
\underbrace{s}_{=\overline{\iota\left(  a\right)  }},v\right)  =\mu_{1}\left(
\overline{\iota\left(  a\right)  },v\right)  =a\rightharpoonup v=\mu
_{2}\left(  \underbrace{\overline{\iota\left(  a\right)  }}_{=s},v\right)
=\mu_{2}\left(  s,v\right)  $ for every $v\in V$, qed.}

But $\left(  V,\mu_{1}\right)  $ and $\left(  V,\mu_{2}\right)  $ are two
$\mathfrak{h}^{(1)}$-modules (since $\mu_{1}$ and $\mu_{2}$ are two
$\mathfrak{h}^{(1)}$-module structures on $V$). Thus, Lemma
\ref{lem.h^(1).equality} (applied to $\mathfrak{h}^{(1)}$ instead of
$\mathfrak{g}$) yields that $\mu_{1}=\mu_{2}$.

We have thus shown that whenever $\mu_{1}$ and $\mu_{2}$ are two
$\mathfrak{h}^{(1)}$-module structures on $V$ which both satisfy
(\ref{prop.h^(1).a}), we must necessarily have $\mu_{1}=\mu_{2}$. In other
words, we have shown that any two $\mathfrak{h}^{(1)}$-module structures on
$V$ which both satisfy (\ref{prop.h^(1).a}) must be equal to each other. In
other words, we have proven Assertion $\mathcal{Y}$.

(There is also an alternative proof of Assertion $\mathcal{Y}$, which proceeds
by tracking down the universal properties used in the above proof of Assertion
$\mathcal{X}$, and applying the uniqueness assertions of these properties.)

Combining Assertions $\mathcal{X}$ and $\mathcal{Y}$, we see that there is one
and only one $\mathfrak{h}^{(1)}$-module structure on $V$ satisfying
(\ref{prop.h^(1).a}), and that this structure is canonical. This proves
Proposition \ref{prop.h^(1)} \textbf{(a)}.

\textbf{(b)} The only thing we must prove to show the validity of Proposition
\ref{prop.h^(1)} \textbf{(b)} is that%
\[
\left[  a,b\right]  \rightharpoonup v=a\rightharpoonup\left(  b\rightharpoonup
v\right)  -b\rightharpoonup\left(  a\rightharpoonup v\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }\left(  a,b\right)  \in\mathfrak{h}%
\times\mathfrak{g}.
\]
But this can be seen by simple computation: In fact, every $\left(
a,b\right)  \in\mathfrak{h}\times\mathfrak{g}$ satisfies $\left[  \iota\left(
a\right)  ,\iota\left(  b\right)  \right]  -\iota\left(  \left[  a,b\right]
\right)  \in\mathfrak{i}$ (since $\mathfrak{i}$ is the Lie ideal of
$\operatorname{FreeLie}\mathfrak{g}$ generated by $\left\langle \left[
\iota\left(  v\right)  ,\iota\left(  w\right)  \right]  -\iota\left(  \left[
v,w\right]  \right)  \ \mid\ \left(  v,w\right)  \in\mathfrak{h}%
\times\mathfrak{g}\right\rangle $, and therefore contains $\left[
\iota\left(  v\right)  ,\iota\left(  w\right)  \right]  -\iota\left(  \left[
v,w\right]  \right)  $ for every $\left(  v,w\right)  \in\mathfrak{h}%
\times\mathfrak{g}$). Thus, every $\left(  a,b\right)  \in\mathfrak{h}%
\times\mathfrak{g}$ satisfies $\overline{\left[  \iota\left(  a\right)
,\iota\left(  b\right)  \right]  }=\overline{\iota\left(  \left[  a,b\right]
\right)  }$ in $\left(  \operatorname{FreeLie}\mathfrak{g}\right)
\diagup\mathfrak{i}$. Consequently, every $\left(  a,b\right)  \in
\mathfrak{h}\times\mathfrak{g}$ satisfies
\begin{align*}
&  a\rightharpoonup\left(  b\rightharpoonup v\right)  -b\rightharpoonup\left(
a\rightharpoonup v\right) \\
&  =\overline{\iota\left(  a\right)  }\rightharpoonup\left(  \overline
{\iota\left(  b\right)  }\rightharpoonup v\right)  -\overline{\iota\left(
b\right)  }\rightharpoonup\left(  \overline{\iota\left(  a\right)
}\rightharpoonup v\right)  =\underbrace{\left[  \overline{\iota\left(
a\right)  },\overline{\iota\left(  b\right)  }\right]  }_{=\overline{\left[
\iota\left(  a\right)  ,\iota\left(  b\right)  \right]  }=\overline
{\iota\left(  \left[  a,b\right]  \right)  }}\rightharpoonup v\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{due to (\ref{[L-mod]}) (applied to }\overline{\iota\left(  a\right)
}\text{, }\overline{\iota\left(  b\right)  }\text{ and }\mathfrak{h}%
^{(1)}\text{ instead of }a\text{, }b\text{ and}\mathfrak{\ g}\text{),}\\
\text{since }V\text{ is an }\mathfrak{h}^{(1)}\text{-module}%
\end{array}
\right) \\
&  =\overline{\iota\left(  \left[  a,b\right]  \right)  }\rightharpoonup
v=\left[  a,b\right]  \rightharpoonup v,
\end{align*}
qed.

\textbf{(c)} It is trivial that if $f$ is a homomorphism of $\mathfrak{h}%
^{(1)}$-modules, then $f$ is a homomorphism of $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules. What is less trivial is the converse
direction: that if $f$ is a homomorphism of $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodules, then $f$ is a homomorphism of
$\mathfrak{h}^{(1)}$-modules. Here is a fast way to see this: Assume that $f$
is a homomorphism of $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules.
Then, every $a\in\mathfrak{g}$ and every $v\in V$ satisfy%
\begin{align*}
f\left(  \overline{\iota\left(  a\right)  }\rightharpoonup v\right)   &
=f\left(  a\rightharpoonup v\right) \\
&  =a\rightharpoonup\left(  f\left(  v\right)  \right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }f\text{ is a homomorphism of }\left(
\mathfrak{g},\mathfrak{h}\right)  \text{-semimodules}\right) \\
&  =\overline{\iota\left(  a\right)  }\rightharpoonup\left(  f\left(
v\right)  \right)  .
\end{align*}
In other words, every $s\in S$ and every $v\in V$ satisfy $f\left(
s\rightharpoonup v\right)  =s\rightharpoonup\left(  f\left(  v\right)
\right)  $ (because $s\in S=\left\{  \overline{\iota\left(  a\right)  }%
\ \mid\ a\in\mathfrak{g}\right\}  $, so that for every $s\in S$ there exists
some $a\in\mathfrak{g}$ such that $s=\overline{\iota\left(  a\right)  }$).
According to Lemma \ref{lem.h^(1).linearity} (applied to $\mathfrak{h}^{(1)}$
instead of $\mathfrak{g}$), this yields that $f$ is a homomorphism of
$\mathfrak{h}^{(1)}$-modules, qed.

\textbf{(d)} This is proven exactly in the same way as \textbf{(c)} (and is
actually equivalent to \textbf{(c)} in light of part \textbf{(g)}, which we
prove below).

\textbf{(e)} and \textbf{(f)} are trivial.

\textbf{(g)} It is clear that $U_{2}\circ U_{1}=\operatorname{id}$. That
$U_{1}\circ U_{2}=\operatorname{id}$ can be shown easily:

\begin{itemize}
\item For every $\mathfrak{h}^{(1)}$-module $V$, we have $V=\left(  U_{1}\circ
U_{2}\right)  \left(  V\right)  $ as $k$-modules (trivially, because neither
$U_{1}$ nor $U_{2}$ change the underlying $k$-module), and the $\mathfrak{h}%
^{(1)}$-module structures of these $\mathfrak{h}^{(1)}$-modules $V$ and
$\left(  U_{1}\circ U_{2}\right)  \left(  V\right)  $ are identical (because
both of them are $\mathfrak{h}^{(1)}$-module structures on $U_{2}\left(
V\right)  $ satisfying (\ref{prop.h^(1).a}), but the Assertion $\mathcal{Y}$
that we showed above (in the proof of \textbf{(a)}) shows that there exists at
most one $\mathfrak{h}^{(1)}$-module structure on $U_{2}\left(  V\right)  $
satisfying (\ref{prop.h^(1).a})). Thus, for every $\mathfrak{h}^{(1)}$-module
$V$, we have $V=\left(  U_{1}\circ U_{2}\right)  \left(  V\right)  $ as
$\mathfrak{h}^{(1)}$-modules. In other words, $U_{1}\circ U_{2}%
=\operatorname{id}$ on objects.

\item We also have $U_{1}\circ U_{2}=\operatorname{id}$ on morphisms (because
both functors $U_{1}$ and $U_{2}$ leave morphisms unchanged).
\end{itemize}

Therefore, $U_{1}\circ U_{2}=\operatorname{id}$ is shown, qed.

\textbf{(h)} This is left to the reader.

\section{\label{sect.split}The splitting of the filtration of $\left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  $}

\subsection{Statement of the theorem}

In this section we are going to show a certain strengthening of Theorem
\ref{thm.l25} under an additional condition:

\begin{theorem}
\label{thm.l37ba}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra, and let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$.
Assume that the inclusion $\mathfrak{h}\hookrightarrow\mathfrak{g}$ splits
\textit{as a }$k$\textit{-module inclusion} (but not necessarily as an
$\mathfrak{h}$-module inclusion).\newline Let $J$ be the two-sided ideal%
\[
\left(  \otimes\mathfrak{g}\right)  \cdot\left\langle v\otimes w-w\otimes
v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}%
\times\mathfrak{h}\right\rangle \cdot\left(  \otimes\mathfrak{g}\right)
\]
of the $k$-algebra $\otimes\mathfrak{g}$.\newline As we know, $\mathfrak{g}$
is a $\mathfrak{g}$-module, and thus also an $\mathfrak{h}$-module (by
Definition \ref{defs.g-as-h-module}). Let $\mathfrak{n}=\mathfrak{g}%
\diagup\mathfrak{h}$. This $\mathfrak{n}$ is an $\mathfrak{h}$-module (because
both $\mathfrak{g}$ and $\mathfrak{h}$ are $\mathfrak{h}$-modules).
\textbf{Assume that this }$\mathfrak{h}$\textbf{-module }$\mathfrak{n}$
\textbf{is actually the restriction of some }$\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $\textbf{-semimodule to }$\mathfrak{h}$\textbf{.}%
\newline Let $\pi:\mathfrak{g}\rightarrow\mathfrak{n}$ be the canonical
projection with kernel $\mathfrak{h}$. Obviously, $\pi$ is an $\mathfrak{h}%
$-module homomorphism. Thus, $\otimes\pi:\otimes\mathfrak{g}\rightarrow
\otimes\mathfrak{n}$ is also an $\mathfrak{h}$-module homomorphism (according
to Proposition \ref{prop.(X)-hom-is-hom}).\newline We consider $\mathfrak{h}$
as an $\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$ by means of the
embedding $\mathfrak{h}\hookrightarrow\mathfrak{g}\hookrightarrow
\otimes\mathfrak{g}$.\newline\textbf{(a)} Both $J$ and $\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}$ are $\mathfrak{h}$-submodules of
$\otimes\mathfrak{g}$. Thus, $\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
$ is an $\mathfrak{h}$-module. Let $\zeta:\otimes\mathfrak{g}\rightarrow
\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ be the canonical projection.
Then, $\zeta$ is an $\mathfrak{h}$-module homomorphism.\newline\textbf{(b)}
For every $n\in\mathbb{N}$, let $F_{n}$ be the $\mathfrak{h}$-submodule
$\zeta\left(  \mathfrak{g}^{\otimes\leq n}\right)  $ of $\left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  $. (That $F_{n}$ indeed is an
$\mathfrak{h}$-submodule was proven in Theorem \ref{thm.l25} already.) Then,
$\left(  F_{n}\right)  _{n\geq0}$ is an $\mathfrak{h}$-module filtration of
$\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ and satisfies $F_{n}%
\cong\mathfrak{n}^{\otimes\leq n}$ as $\mathfrak{h}$-modules for every
$n\in\mathbb{N}$.\newline\textbf{(c)} There exists an $\mathfrak{h}$-module
isomorphism $\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \rightarrow
\otimes\mathfrak{n}$ such that for every $n\in\mathbb{N}$, the image of
$F_{n}$ under this isomorphism is $\mathfrak{n}^{\otimes\leq n}$%
.\newline\textbf{(d)} The filtration $\left(  F_{n}\right)  _{n\geq0}$ of
$\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ is $\mathfrak{h}$-split.
\end{theorem}

This theorem provides the main ingredient of Lemma 3.9 in \cite{calaque1}.

\subsection{Preparations for the proof}

Before we embark upon the proof of this fact (which will be shorter than that
of Theorem \ref{thm.l25}, since we have already paved some of the way), let us
fix a convention that we are going to use throughout the rest of Section
\ref{sect.split}:

\begin{vershort}
\begin{condition}
\label{conv.split.short}\textbf{(a)} Throughout the whole Section
\ref{sect.split}, we are going to work under the conditions of Theorem
\ref{thm.l37ba}.\newline\textbf{(b)} According to the conditions of Theorem
\ref{thm.l37ba}, the $\mathfrak{h}$-module $\mathfrak{n}$ is actually the
restriction of some $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule to
$\mathfrak{h}$. Let us fix one such $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule for the rest of Section \ref{sect.split}. We are going to denote
this $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule by $\mathfrak{n}%
$. (This is allowed since this $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule is equal to $\mathfrak{n}$ as an $\mathfrak{h}$-module, because
its restriction to $\mathfrak{h}$ is $\mathfrak{n}$).\newline Since
$\mathfrak{n}$ is a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule,
Proposition \ref{prop.semitensor.short} (applied to $V=\mathfrak{n}$) yields
that $\otimes\mathfrak{n}$ is a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semialgebra. In particular, $\otimes\mathfrak{n}$ is thus a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule.\newline We are going to use
Convention \ref{conv.semimod.->} as liberally as we can. This means that when
$a$ is an element of $\mathfrak{g}$ whereas $v$ is an element of a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule $V$ (for example, $V$ can be
one of the $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules
$\mathfrak{g}$, $\mathfrak{n}$ or $\otimes\mathfrak{n}$), we will denote by
$a\rightharpoonup v$ the Lie action of the $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule $V$ applied to $\left(  a,v\right)  $.
\end{condition}
\end{vershort}

\begin{verlong}
\begin{condition}
\label{conv.split}\textbf{(a)} Throughout the whole Section \ref{sect.split},
we are going to work under the conditions of Theorem \ref{thm.l37ba}%
.\newline\textbf{(b)} According to the conditions of Theorem \ref{thm.l37ba},
the $\mathfrak{h}$-module $\mathfrak{n}$ is actually the restriction of some
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule to $\mathfrak{h}$. Let
us fix one such $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule for
the rest of Section \ref{sect.split}. We are going to denote this $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule by $\mathfrak{n}$. (This is
allowed since this $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule is
equal to $\mathfrak{n}$ as an $\mathfrak{h}$-module, because its restriction
to $\mathfrak{h}$ is $\mathfrak{n}$).\newline Since $\mathfrak{n}$ is a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule, Proposition
\ref{prop.semitensor} (applied to $V=\mathfrak{n}$) yields that $\otimes
\mathfrak{n}$ is a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semialgebra.
In particular, $\otimes\mathfrak{n}$ is thus a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule.\newline We are going to use Convention
\ref{conv.semimod.->} as liberally as we can. This means that when $a$ is an
element of $\mathfrak{g}$ whereas $v$ is an element of a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule $V$ (for example, $V$ can be
one of the $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodules
$\mathfrak{g}$, $\mathfrak{n}$ or $\otimes\mathfrak{n}$), we will denote by
$a\rightharpoonup v$ the Lie action of the $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule $V$ applied to $\left(  a,v\right)  $.
\end{condition}
\end{verlong}

\subsection{Definitions and basic properties of $\gamma$}

We will now construct a homomorphism from $\otimes\mathfrak{g}$ to
$\otimes\mathfrak{n}$ which will later give rise to the required
$\mathfrak{h}$-module isomorphism from $F_{n}$ to $\mathfrak{n}^{\otimes\leq
n}$:

\begin{definition}
\label{defs.l37.core}Consider the situation of Theorem \ref{thm.l37ba}%
.\newline\textbf{(a)} Let us define a $k$-linear map $\gamma_{p}%
:\mathfrak{g}^{\otimes p}\rightarrow\otimes\mathfrak{n}$ for every
$p\in\mathbb{N}$. We are going to define this map $\gamma_{p}$ by induction
over $p$:\newline\textit{Induction base:} For $p=0$, define the map
$\gamma_{p}:\mathfrak{g}^{\otimes p}\rightarrow\otimes\mathfrak{n}$ by%
\begin{equation}
\left(  \gamma_{0}\left(  \lambda\right)  =\lambda
\ \ \ \ \ \ \ \ \ \ \text{for every }\lambda\in\mathfrak{g}^{\otimes0}\right)
\label{defs.gamma.0}%
\end{equation}
(this definition makes sense since $\mathfrak{g}^{\otimes0}=k\subseteq
\otimes\mathfrak{n}$).\newline\textit{Induction step:} For any $p>0$, we
assume that the map $\gamma_{p-1}:\mathfrak{g}^{\otimes\left(  p-1\right)
}\rightarrow\otimes\mathfrak{n}$ is already defined, and now we define a map
$\gamma_{p}:\mathfrak{g}^{\otimes p}\rightarrow\otimes\mathfrak{n}$ as
follows: The map%
\[
\mathfrak{g}\times\mathfrak{g}^{\otimes\left(  p-1\right)  }\rightarrow
\otimes\mathfrak{n},\ \ \ \ \ \ \ \ \ \ \left(  u,U\right)  \mapsto\pi\left(
u\right)  \cdot\gamma_{p-1}\left(  U\right)  +u\rightharpoonup\left(
\gamma_{p-1}\left(  U\right)  \right)
\]
{\footnotemark} is $k$-bilinear (because the maps $\gamma_{p-1}$ and $\pi$ are
$k$-linear and the Lie action of $\otimes\mathfrak{n}$ is $k$-bilinear). Thus,
by the universal property of the tensor product, this map gives rise to a
$k$-linear map $\mathfrak{g}\otimes\mathfrak{g}^{\otimes\left(  p-1\right)
}\rightarrow\otimes\mathfrak{n}$ which sends $u\otimes U$ to $\pi\left(
u\right)  \cdot\gamma_{p-1}\left(  U\right)  +u\rightharpoonup\left(
\gamma_{p-1}\left(  U\right)  \right)  $ for every $\left(  u,U\right)
\in\mathfrak{g}\times\mathfrak{g}^{\otimes\left(  p-1\right)  }$. This
$k$-linear map is going to be denoted by $\gamma_{p}$. It is a map from
$\mathfrak{g}^{\otimes p}$ to $\otimes\mathfrak{n}$ because $\mathfrak{g}%
\otimes\mathfrak{g}^{\otimes\left(  p-1\right)  }=\mathfrak{g}^{\otimes p}%
$.\newline This completes the inductive definition of $\gamma_{p}$ for every
$p\in\mathbb{N}$.\newline\textbf{(b)} Now, we define a $k$-linear map
$\gamma:\otimes\mathfrak{g}\rightarrow\otimes\mathfrak{n}$ as follows: The sum
$\sum\limits_{i\in\mathbb{N}}\gamma_{i}$ of the maps $\gamma_{i}%
:\mathfrak{g}^{\otimes i}\rightarrow\otimes\mathfrak{n}$ is a map from
$\bigoplus\limits_{i\in\mathbb{N}}\mathfrak{g}^{\otimes i}$ to $\otimes
\mathfrak{n}$. Since $\bigoplus\limits_{i\in\mathbb{N}}\mathfrak{g}^{\otimes
i}=\otimes\mathfrak{g}$, the sum $\sum\limits_{i\in\mathbb{N}}\gamma_{i}$ of
the maps $\gamma_{i}:\mathfrak{g}^{\otimes i}\rightarrow\otimes\mathfrak{n}$
is thus a map from $\otimes\mathfrak{g}$ to $\otimes\mathfrak{n}$. Denote this
map by $\gamma$.
\end{definition}

\begin{vershort}
\footnotetext{Here, of course, the term $u\rightharpoonup\left(  \gamma
_{p-1}\left(  U\right)  \right)  $ denotes the Lie action \textit{of the
}$\left(  \mathfrak{g},\mathfrak{h}\right)  $\textit{-semimodule }%
$\otimes\mathfrak{n}$, applied to $\left(  u,\gamma_{p-1}\left(  U\right)
\right)  $.}
\end{vershort}

\begin{verlong}
\footnotetext{Here, according to Convention \ref{conv.split} \textbf{(b)}, the
term $u\rightharpoonup\left(  \gamma_{p-1}\left(  U\right)  \right)  $ denotes
the Lie action \textit{of the }$\left(  \mathfrak{g},\mathfrak{h}\right)
$\textit{-semimodule }$\otimes\mathfrak{n}$, applied to $\left(
u,\gamma_{p-1}\left(  U\right)  \right)  $.}
\end{verlong}

\begin{condition}
\label{conv.l37.core}Throughout the rest of Section \ref{sect.split}, we are
going to work in the situation of Definition \ref{defs.l37.core}. So, for
example, when we refer to $\mathfrak{h}$, we mean the Lie subalgebra
$\mathfrak{h}$ of Theorem \ref{thm.l37ba}, and when we refer to $\gamma$, we
mean the map $\gamma$ of Definition \ref{defs.l37.core}.
\end{condition}

\begin{remark}
\label{rmk.l37.alpha}\textbf{(a)} As a consequence of the inductive step in
the definition of $\gamma_{p}$ (in Definition \ref{defs.l37.core}
\textbf{(a)}), we know that for every $p>0$, the map $\gamma_{p}$ is the
$k$-linear map $\mathfrak{g}\otimes\mathfrak{g}^{\otimes\left(  p-1\right)
}\rightarrow\otimes\mathfrak{n}$ which sends $u\otimes U$ to $\pi\left(
u\right)  \cdot\gamma_{p-1}\left(  U\right)  +u\rightharpoonup\left(
\gamma_{p-1}\left(  U\right)  \right)  $ for every $\left(  u,U\right)
\in\mathfrak{g}\times\mathfrak{g}^{\otimes\left(  p-1\right)  }$. In other
words,%
\begin{equation}
\gamma_{p}\left(  u\otimes U\right)  =\pi\left(  u\right)  \cdot\gamma
_{p-1}\left(  U\right)  +u\rightharpoonup\left(  \gamma_{p-1}\left(  U\right)
\right)  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(  u,U\right)
\in\mathfrak{g}\times\mathfrak{g}^{\otimes\left(  p-1\right)  }.
\label{defs.gamma.p}%
\end{equation}
\newline\textbf{(b)} According to Definition \ref{defs.l37.core} \textbf{(b)},
the map $\gamma:\otimes\mathfrak{g}\rightarrow\otimes\mathfrak{n}$ is the sum
$\sum\limits_{i\in\mathbb{N}}\gamma_{i}$ of the maps $\gamma_{i}%
:\mathfrak{g}^{\otimes i}\rightarrow\otimes\mathfrak{n}$. Hence,
\begin{equation}
\gamma\left(  T\right)  =\left(  \sum\limits_{i\in\mathbb{N}}\gamma
_{i}\right)  \left(  T\right)  =\gamma_{p}\left(  T\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }p\in\mathbb{N}\text{ and every }%
T\in\mathfrak{g}^{\otimes p}. \label{rmk.gamma-p}%
\end{equation}
\textbf{(c)} Every $\lambda\in k$ satisfies%
\begin{align}
\gamma\left(  \lambda\right)   &  =\gamma_{0}\left(  \lambda\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{rmk.gamma-p}) (applied to
}p=0\text{ and }T=\lambda\text{), because }\lambda\in k=\mathfrak{g}%
^{\otimes0}\right) \nonumber\\
&  =\lambda\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{defs.gamma.0})}\right)
. \label{rmk.gamma-0}%
\end{align}
This yields, in particular, that $\gamma\left(  k\right)  =k$.\newline%
\textbf{(d)} Every $u\in\mathfrak{g}$ satisfies%
\begin{align}
\gamma\left(  u\right)   &  =\gamma_{1}\left(  u\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{rmk.gamma-p}) (applied to
}p=1\text{ and }T=u\text{), because }u\in\mathfrak{g}=\mathfrak{g}^{\otimes
1}\right) \nonumber\\
&  =\gamma_{1}\left(  u\otimes1\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{since }u=u\otimes1\text{ under the identification }\mathfrak{g}%
\cong\mathfrak{g}\otimes k\right) \nonumber\\
&  =\pi\left(  u\right)  \cdot\underbrace{\gamma_{1-1}}_{=\gamma_{0}}\left(
1\right)  +u\rightharpoonup\left(  \underbrace{\gamma_{1-1}}_{=\gamma_{0}%
}\left(  1\right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{defs.gamma.p}), applied to
}p=1\text{ and }U=1\right) \nonumber\\
&  =\pi\left(  u\right)  \cdot\underbrace{\gamma_{0}\left(  1\right)
}_{\substack{=1\text{ (by (\ref{defs.gamma.0}),}\\\text{applied to }%
\lambda=1\text{)}}}+u\rightharpoonup\left(  \underbrace{\gamma_{0}\left(
1\right)  }_{\substack{=1\text{ (by (\ref{defs.gamma.0}),}\\\text{applied to
}\lambda=1\text{)}}}\right) \nonumber\\
&  =\pi\left(  u\right)  \cdot1+\underbrace{u\rightharpoonup1}%
_{\substack{=0\text{ (since the Lie}\\\text{action of }k\text{ is }0\text{)}%
}}=\pi\left(  u\right)  \cdot1+0=\pi\left(  u\right)  . \label{rmk.gamma-1}%
\end{align}
This yields $\gamma\left(  \mathfrak{g}\right)  =\pi\left(  \mathfrak{g}%
\right)  =\mathfrak{n}$ (since $\pi$ is a projection of $\mathfrak{g}$ on
$\mathfrak{n}$).\newline\textbf{(e)} We now know that $\gamma\left(  k\right)
=k$ and $\gamma\left(  \mathfrak{g}\right)  \subseteq\mathfrak{n}$. But in
general, we cannot generalize this to $\gamma\left(  \mathfrak{g}^{\otimes
p}\right)  \subseteq\mathfrak{n}^{\otimes p}$ for all $p\in\mathbb{N}$.
However, Proposition \ref{prop.l37.gamma.filt} will give us a weaker result
that is actually true.
\end{remark}

The next proposition (a kind of analogue of Proposition
\ref{prop.l25.alpha.rec} for $\gamma$ instead of $\varphi$) generalizes
(\ref{defs.gamma.p}) to arbitrary $U$:

\begin{proposition}
\label{prop.l37.gamma.rec}Every $u\in\mathfrak{g}$ and $U\in\otimes
\mathfrak{g}$ satisfy%
\begin{equation}
\gamma\left(  u\cdot U\right)  =\pi\left(  u\right)  \cdot\gamma\left(
U\right)  +u\rightharpoonup\left(  \gamma\left(  U\right)  \right)  .
\label{prop.gamma.rec}%
\end{equation}

\end{proposition}

\begin{vershort}
\textit{Proof of Proposition \ref{prop.l37.gamma.rec}.} This is analogous to
the proof we gave for Proposition \ref{prop.l25.alpha.rec}, so we will not elaborate.
\end{vershort}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.l37.gamma.rec}.} Let $u\in\mathfrak{g}$
and $U\in\otimes\mathfrak{g}$ be arbitrary. Since $U\in\otimes\mathfrak{g}%
=\bigoplus\limits_{i\in\mathbb{N}}\mathfrak{g}^{\otimes i}$, we can write $U$
as a family $U=\left(  U_{i}\right)  _{i\in\mathbb{N}}$, where $\left(
U_{i}\in\mathfrak{g}^{\otimes i}\text{ for every }i\in\mathbb{N}\right)  $.
Taking into account that we consider $\mathfrak{g}^{\otimes p}$ as a
$k$-submodule of $\bigoplus\limits_{i\in\mathbb{N}}\mathfrak{g}^{\otimes i}$
for every $p\in\mathbb{N}$, we have $\left(  U_{i}\right)  _{i\in\mathbb{N}%
}=\sum\limits_{i\in\mathbb{N}}U_{i}$, and thus $U=\left(  U_{i}\right)
_{i\in\mathbb{N}}=\sum\limits_{i\in\mathbb{N}}U_{i}$.

On the other hand, every $i\in\mathbb{N}$ satisfies $\underbrace{u}%
_{\in\mathfrak{g}}\otimes\underbrace{U_{i}}_{\in\mathfrak{g}^{\otimes i}}%
\in\mathfrak{g}\otimes\mathfrak{g}^{\otimes i}=\mathfrak{g}^{\otimes\left(
i+1\right)  }$. Also, since $\gamma=\sum\limits_{i\in\mathbb{N}}\gamma
_{i}=\sum\limits_{p\in\mathbb{N}}\gamma_{p}$ is the sum of the maps
$\gamma_{p}:\mathfrak{g}^{\otimes p}\rightarrow\otimes\mathfrak{n}$ for all
$p\in\mathbb{N}$, we have
\begin{equation}
\gamma\left(  v\right)  =\gamma_{i}\left(  v\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }v\in\mathfrak{g}^{\otimes i}.
\label{pf.l37.gamma.rec.1}%
\end{equation}
Similarly,%
\begin{equation}
\gamma\left(  v\right)  =\gamma_{i+1}\left(  v\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }v\in\mathfrak{g}^{\otimes\left(
i+1\right)  }. \label{pf.l37.gamma.rec.2}%
\end{equation}
Every $i\in\mathbb{N}$ satisfies $\gamma\left(  u\otimes U_{i}\right)
=\gamma_{i+1}\left(  u\otimes U_{i}\right)  $ according to
(\ref{pf.l37.gamma.rec.2}) (applied to $u\otimes U_{i}$ instead of $v$),
because $u\otimes U_{i}\in\mathfrak{g}^{\otimes\left(  i+1\right)  }$. Every
$i\in\mathbb{N}$ satisfies $\gamma\left(  U_{i}\right)  =\gamma_{i}\left(
U_{i}\right)  $ according to (\ref{pf.l37.gamma.rec.1}) (applied to $U_{i}$
instead of $v$), since $U_{i}\in\mathfrak{g}^{\otimes i}$. Now, $U=\sum
\limits_{i\in\mathbb{N}}U_{i}$ leads to
\begin{align*}
&  u\cdot U\\
&  =u\cdot\sum\limits_{i\in\mathbb{N}}U_{i}=\sum\limits_{i\in\mathbb{N}}u\cdot
U_{i}=\sum\limits_{i\in\mathbb{N}}u\otimes U_{i}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since every }i\in\mathbb{N}\text{ satisfies }u\in\mathfrak{g}%
=\mathfrak{g}^{\otimes1}\text{ and }U_{i}\in\mathfrak{g}^{\otimes i}\text{ and
therefore}\\
u\cdot U_{i}=u\otimes U_{i}\text{ (due to (\ref{*=(X)}), applied to }u\text{,
}U_{i}\text{, }1\text{ and }i\text{ instead of }a\text{, }b\text{, }n\text{
and }m\text{)}%
\end{array}
\right)  ,
\end{align*}
and therefore%
\begin{align*}
\gamma\left(  u\cdot U\right)   &  =\gamma\left(  \sum\limits_{i\in\mathbb{N}%
}u\otimes U_{i}\right)  =\sum\limits_{i\in\mathbb{N}}\underbrace{\gamma\left(
u\otimes U_{i}\right)  }_{=\gamma_{i+1}\left(  u\otimes U_{i}\right)
}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\gamma\text{ is }k\text{-linear}%
\right) \\
&  =\sum_{i\in\mathbb{N}}\gamma_{i+1}\left(  u\otimes U_{i}\right)
=\sum_{i\in\mathbb{N}}\left(  \pi\left(  u\right)  \cdot\underbrace{\gamma
_{\left(  i+1\right)  -1}}_{=\gamma_{i}}\left(  U_{i}\right)
+u\rightharpoonup\left(  \underbrace{\gamma_{\left(  i+1\right)  -1}}%
_{=\gamma_{i}}\left(  U_{i}\right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since every }i\in\mathbb{N}\text{ satisfies}\\
\gamma_{i+1}\left(  u\otimes U_{i}\right)  =\pi\left(  u\right)  \cdot
\gamma_{\left(  i+1\right)  -1}\left(  U_{i}\right)  +u\rightharpoonup\left(
\gamma_{\left(  i+1\right)  -1}\left(  U_{i}\right)  \right) \\
\text{according to (\ref{defs.gamma.p}) (applied to }i+1\text{ and }%
U_{i}\text{ instead of }p\text{ and }U\text{)}%
\end{array}
\right) \\
&  =\sum_{i\in\mathbb{N}}\left(  \pi\left(  u\right)  \cdot\gamma_{i}\left(
U_{i}\right)  +u\rightharpoonup\left(  \gamma_{i}\left(  U_{i}\right)
\right)  \right)  =\sum_{i\in\mathbb{N}}\pi\left(  u\right)  \cdot
\underbrace{\gamma_{i}\left(  U_{i}\right)  }_{=\gamma\left(  U_{i}\right)
}+\sum_{i\in\mathbb{N}}u\rightharpoonup\left(  \underbrace{\gamma_{i}\left(
U_{i}\right)  }_{=\gamma\left(  U_{i}\right)  }\right) \\
&  =\sum_{i\in\mathbb{N}}\pi\left(  u\right)  \cdot\gamma\left(  U_{i}\right)
+\sum_{i\in\mathbb{N}}u\rightharpoonup\left(  \gamma\left(  U_{i}\right)
\right)  .
\end{align*}
Since%
\[
\sum_{i\in\mathbb{N}}\pi\left(  u\right)  \cdot\gamma\left(  U_{i}\right)
=\pi\left(  u\right)  \cdot\underbrace{\sum_{i\in\mathbb{N}}\gamma\left(
U_{i}\right)  }_{\substack{=\gamma\left(  \sum\limits_{i\in\mathbb{N}}%
U_{i}\right)  \\\text{(since }\gamma\text{ is }k\text{-linear)}}}=\pi\left(
u\right)  \cdot\gamma\left(  \underbrace{\sum\limits_{i\in\mathbb{N}}U_{i}%
}_{=U}\right)  =\pi\left(  u\right)  \cdot\gamma\left(  U\right)
\]
and%
\begin{align*}
\sum_{i\in\mathbb{N}}u\rightharpoonup\left(  \gamma\left(  U_{i}\right)
\right)   &  =u\rightharpoonup\underbrace{\left(  \sum_{i\in\mathbb{N}}%
\gamma\left(  U_{i}\right)  \right)  }_{\substack{=\gamma\left(
\sum\limits_{i\in\mathbb{N}}U_{i}\right)  \\\text{(since }\gamma\text{ is
}k\text{-linear)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since the Lie action of
}\otimes\mathfrak{n}\text{ is linear}\right) \\
&  =u\rightharpoonup\left(  \gamma\left(  \underbrace{\sum\limits_{i\in
\mathbb{N}}U_{i}}_{=U}\right)  \right)  =u\rightharpoonup\left(  \gamma\left(
U\right)  \right)  ,
\end{align*}
this becomes%
\[
\gamma\left(  u\cdot U\right)  =\underbrace{\sum_{i\in\mathbb{N}}\pi\left(
u\right)  \cdot\gamma\left(  U_{i}\right)  }_{=\pi\left(  u\right)
\cdot\gamma\left(  U\right)  }+\underbrace{\sum_{i\in\mathbb{N}}%
u\rightharpoonup\left(  \gamma\left(  U_{i}\right)  \right)  }%
_{=u\rightharpoonup\left(  \gamma\left(  U\right)  \right)  }=\pi\left(
u\right)  \cdot\gamma\left(  U\right)  +u\rightharpoonup\left(  \gamma\left(
U\right)  \right)  .
\]
This proves Proposition \ref{prop.l37.gamma.rec}.
\end{verlong}

Now we come to a corollary of Proposition \ref{prop.l37.gamma.rec} which is
similar to Proposition \ref{cor.l25.alpha.rec.h} (note that an analogue to
Corollary \ref{cor.l25.alpha.rec.N} does not seem to exist):

\begin{corollary}
\label{cor.l37.gamma.rec.h}Every $u\in\mathfrak{h}$ and $U\in\otimes
\mathfrak{g}$ satisfy%
\begin{equation}
\gamma\left(  u\cdot U\right)  =u\rightharpoonup\left(  \gamma\left(
U\right)  \right)  . \label{prop.gamma.rec.h}%
\end{equation}

\end{corollary}

\begin{vershort}
\textit{Proof of Corollary \ref{cor.l37.gamma.rec.h}.} This corollary
trivially follows from (\ref{prop.gamma.rec}) (just as Corollary
\ref{cor.l25.alpha.rec.h} followed from (\ref{prop.alpha.rec})).
\end{vershort}

\begin{verlong}
\textit{Proof of Corollary \ref{cor.l37.gamma.rec.h}.} Since $u\in
\mathfrak{h}$, we have $\pi\left(  u\right)  =0$ (since $\pi$ is a projection
with kernel $\mathfrak{h}$). Thus, (\ref{prop.gamma.rec}) yields%
\[
\gamma\left(  u\cdot U\right)  =\underbrace{\pi\left(  u\right)  }_{=0}%
\cdot\gamma\left(  U\right)  +u\rightharpoonup\left(  \gamma\left(  U\right)
\right)  =\underbrace{0\cdot\gamma\left(  U\right)  }_{=0}+u\rightharpoonup
\left(  \gamma\left(  U\right)  \right)  =u\rightharpoonup\left(
\gamma\left(  U\right)  \right)  .
\]
This proves Corollary \ref{cor.l37.gamma.rec.h}.
\end{verlong}

Here is an analogue of Proposition \ref{prop.l25.alpha.filt}:

\begin{proposition}
\label{prop.l37.gamma.filt}The map $\gamma:\otimes\mathfrak{g}\rightarrow
\otimes\mathfrak{n}$ respects the filtration. Here, the filtration on
$\otimes\mathfrak{g}$ is the degree filtration $\left(  \mathfrak{g}%
^{\otimes\leq n}\right)  _{n\geq0}$, and the filtration on $\otimes
\mathfrak{n}$ is the degree filtration $\left(  \mathfrak{n}^{\otimes\leq
n}\right)  _{n\geq0}$.
\end{proposition}

\begin{vershort}
\textit{Proof of Proposition \ref{prop.l37.gamma.filt}.} We skip this proof,
for it is an obvious variation on that of Proposition
\ref{prop.l25.alpha.filt}.
\end{vershort}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.l37.gamma.filt}.} Let us use Convention
\ref{conv.left-ind2}.

We are going to show that%
\begin{equation}
\gamma\left(  \mathfrak{g}^{\otimes p}\right)  \subseteq\mathfrak{n}%
^{\otimes\leq p}\text{ for every }p\in\mathbb{N}. \label{pf.l37.gamma.filt.1}%
\end{equation}


\textit{Proof of (\ref{pf.l37.gamma.filt.1}).} We will prove
(\ref{pf.l37.gamma.filt.1}) by induction over $p$:

\textit{Induction base:} We know that $\mathfrak{g}^{\otimes0}=k$ and
$\mathfrak{n}^{\otimes\leq0}=\bigoplus\limits_{i=0}^{0}\mathfrak{n}^{\otimes
i}=\mathfrak{n}^{\otimes0}=k$, and we know (from Remark \ref{rmk.l37.alpha}
\textbf{(c)}) that $\gamma\left(  k\right)  =k$. Thus, $\mathfrak{g}%
^{\otimes0}=k$ yields $\gamma\left(  \mathfrak{g}^{\otimes0}\right)
=\gamma\left(  k\right)  =k=\mathfrak{n}^{\otimes\leq0}$. In other words,
(\ref{pf.l37.gamma.filt.1}) holds for $p=0$. This completes the induction base.

\textit{Induction step:} Let $q\in\mathbb{N}_{+}$. Assume that
(\ref{pf.l37.gamma.filt.1}) holds for $p=q-1$. Now we must prove that
(\ref{pf.l37.gamma.filt.1}) holds for $p=q$.

We have assumed that (\ref{pf.l37.gamma.filt.1}) holds for $p=q-1$. In other
words, $\gamma\left(  \mathfrak{g}^{\otimes\left(  q-1\right)  }\right)
\subseteq\mathfrak{n}^{\otimes\leq\left(  q-1\right)  }$.

Now we are going to prove that every $U\in\gamma\left(  \mathfrak{g}%
_{\operatorname*{lind}}^{\otimes q}\right)  $ satisfies $U\in\mathfrak{n}%
^{\otimes\leq q}$.

In fact, let $U\in\gamma\left(  \mathfrak{g}_{\operatorname*{lind}}^{\otimes
q}\right)  $ be arbitrary. Then, there exists a $U^{\prime}\in\mathfrak{g}%
_{\operatorname*{lind}}^{\otimes q}$ such that $U=\gamma\left(  U^{\prime
}\right)  $ (since $U\in\gamma\left(  \mathfrak{g}_{\operatorname*{lind}%
}^{\otimes q}\right)  $). Consider this $U^{\prime}$. This $U^{\prime}$ is a
left-induced tensor in $\mathfrak{g}^{\otimes q}$ (since $U^{\prime}%
\in\mathfrak{g}_{\operatorname*{lind}}^{\otimes q}$); this means that there
exist some $v\in\mathfrak{g}$ and $T\in\mathfrak{g}^{\otimes\left(
q-1\right)  }$ such that $U^{\prime}=v\otimes T$. Consider these $v$ and $T$.
Since $v\in\mathfrak{g}=\mathfrak{g}^{\otimes1}$ and $T\in\mathfrak{g}%
^{\otimes\left(  q-1\right)  }$, we have $v\cdot T=v\otimes T$ (due to
(\ref{*=(X)}), applied to $v$, $T$, $1$ and $q-1$ instead of $a$, $b$, $n$ and
$m$).

We notice that $\gamma\left(  \underbrace{T}_{\in\mathfrak{g}^{\otimes\left(
q-1\right)  }}\right)  \in\gamma\left(  \mathfrak{g}^{\otimes\left(
q-1\right)  }\right)  \subseteq\mathfrak{n}^{\otimes\leq\left(  q-1\right)  }%
$, and thus $v\rightharpoonup\left(  \gamma\left(  T\right)  \right)
\in\mathfrak{n}^{\otimes\leq\left(  q-1\right)  }$ (since $\mathfrak{n}%
^{\otimes\leq\left(  q-1\right)  }$ is a $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule, which is because $\mathfrak{n}$ is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule).

Now,%
\begin{align*}
U  &  =\gamma\left(  \underbrace{U^{\prime}}_{=v\otimes T=v\cdot T}\right)
=\gamma\left(  v\cdot T\right)  =\underbrace{\pi\left(  v\right)  }%
_{\in\mathfrak{n}=\mathfrak{n}^{\otimes1}\subseteq\mathfrak{n}^{\otimes\leq1}%
}\cdot\gamma\left(  \underbrace{T}_{\in\mathfrak{g}^{\otimes\left(
q-1\right)  }}\right)  +\underbrace{v\rightharpoonup\left(  \gamma\left(
T\right)  \right)  }_{\in\mathfrak{n}^{\otimes\leq\left(  q-1\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{prop.gamma.rec}), applied to
}v\text{ and }T\text{ instead of }u\text{ and }U\right) \\
&  \in\mathfrak{n}^{\otimes\leq1}\cdot\underbrace{\gamma\left(  \mathfrak{g}%
^{\otimes\left(  q-1\right)  }\right)  }_{\subseteq\mathfrak{n}^{\otimes
\leq\left(  q-1\right)  }}+\underbrace{\mathfrak{n}^{\otimes\leq\left(
q-1\right)  }}_{\substack{\subseteq\mathfrak{n}^{\otimes\leq q}\\\text{(since
}\left(  \mathfrak{n}^{\otimes\leq n}\right)  _{n\geq0}\text{ is a
filtration)}}}\subseteq\mathfrak{n}^{\otimes\leq1}\cdot\mathfrak{n}%
^{\otimes\leq\left(  q-1\right)  }+\mathfrak{n}^{\otimes\leq q}.
\end{align*}
Since Proposition \ref{prop.tensor.filt-is-alg-filt} \textbf{(b)} (applied to
$\mathfrak{n}$, $1$ and $q-1$ instead of $V$, $n$ and $m$) yields
$\mathfrak{n}^{\otimes\leq1}\cdot\mathfrak{n}^{\otimes\leq\left(  q-1\right)
}\subseteq\mathfrak{n}^{\otimes\leq\left(  1+\left(  q-1\right)  \right)
}=\mathfrak{n}^{\otimes\leq q}$, this becomes%
\[
U\in\underbrace{\mathfrak{n}^{\otimes\leq1}\cdot\mathfrak{n}^{\otimes
\leq\left(  q-1\right)  }}_{\subseteq\mathfrak{n}^{\otimes\leq q}%
}+\mathfrak{n}^{\otimes\leq q}\subseteq\mathfrak{n}^{\otimes\leq
q}+\mathfrak{n}^{\otimes\leq q}\subseteq\mathfrak{n}^{\otimes\leq
q}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathfrak{n}^{\otimes\leq q}\text{
is a }k\text{-module}\right)  .
\]


We have thus proven that every $U\in\gamma\left(  \mathfrak{g}%
_{\operatorname*{lind}}^{\otimes q}\right)  $ satisfies $U\in\mathfrak{n}%
^{\otimes\leq q}$. In other words, $\gamma\left(  \mathfrak{g}%
_{\operatorname*{lind}}^{\otimes q}\right)  \subseteq\mathfrak{n}^{\otimes\leq
q}$.

Now, Proposition \ref{prop.left-ind2} (applied to $q$ and $\mathfrak{g}$
instead of $p$ and $V$) says $\mathfrak{g}^{\otimes q}=\left\langle
\mathfrak{g}_{\operatorname*{lind}}^{\otimes q}\right\rangle $, and thus%
\begin{align*}
\gamma\left(  \mathfrak{g}^{\otimes q}\right)   &  =\gamma\left(  \left\langle
\mathfrak{g}_{\operatorname*{lind}}^{\otimes q}\right\rangle \right)
=\left\langle \gamma\left(  \mathfrak{g}_{\operatorname*{lind}}^{\otimes
q}\right)  \right\rangle \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Proposition \ref{prop.<>} \textbf{(b)} (applied to}\\
\mathfrak{g}^{\otimes q}\text{, }\mathfrak{g}_{\operatorname*{lind}}^{\otimes
q}\text{, }\otimes\mathfrak{n}\text{ and }\gamma\text{ instead of }M\text{,
}S\text{, }R\text{ and }f\text{)}%
\end{array}
\right) \\
&  \subseteq\mathfrak{n}^{\otimes\leq q}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by Proposition \ref{prop.<>} \textbf{(a)} (applied to }\otimes
\mathfrak{n}\text{, }\gamma\left(  \mathfrak{g}_{\operatorname*{lind}%
}^{\otimes q}\right)  \text{ and }\mathfrak{n}^{\otimes\leq q}\\
\text{instead of }M\text{, }S\text{ and }Q\text{), since }\gamma\left(
\mathfrak{g}_{\operatorname*{lind}}^{\otimes q}\right)  \subseteq
\mathfrak{n}^{\otimes\leq q}%
\end{array}
\right)  .
\end{align*}
In other words, (\ref{pf.l37.gamma.filt.1}) holds for $p=q$. This completes
the induction step, and thus the induction proof of (\ref{pf.l37.gamma.filt.1}%
) is done.

Now, let $m\in\mathbb{N}$. The definition of $\mathfrak{g}^{\otimes\leq m}$
says $\mathfrak{g}^{\otimes\leq m}=\bigoplus\limits_{i=0}^{m}\mathfrak{g}%
^{\otimes i}$, so that $\mathfrak{g}^{\otimes\leq m}=\bigoplus\limits_{i=0}%
^{m}\mathfrak{g}^{\otimes i}=\sum\limits_{i=0}^{m}\mathfrak{g}^{\otimes i}$
(since direct sums are sums) and thus%
\begin{align*}
\gamma\left(  \mathfrak{g}^{\otimes\leq m}\right)   &  =\gamma\left(
\sum\limits_{i=0}^{m}\mathfrak{g}^{\otimes i}\right)  =\sum\limits_{i=0}%
^{m}\underbrace{\gamma\left(  \mathfrak{g}^{\otimes i}\right)  }%
_{\substack{\subseteq\mathfrak{n}^{\otimes\leq i}\\\text{(by
(\ref{pf.l37.gamma.filt.1}), applied}\\\text{to }i\text{ instead of }%
p\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\gamma\text{ is
}k\text{-linear}\right) \\
&  \subseteq\sum\limits_{i=0}^{m}\underbrace{\mathfrak{n}^{\otimes\leq i}%
}_{\substack{\subseteq\mathfrak{n}^{\otimes\leq m}\text{ (since }i\leq
m\\\text{and since }\left(  \mathfrak{n}^{\otimes\leq n}\right)  _{n\geq
0}\text{ is}\\\text{a filtration)}}}\subseteq\sum\limits_{i=0}^{m}%
\mathfrak{n}^{\otimes\leq m}\subseteq\mathfrak{n}^{\otimes\leq m}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathfrak{n}^{\otimes\leq m}\text{ is
a }k\text{-module}\right)  .
\end{align*}
Since this holds for every $m\in\mathbb{N}$, this yields that $\gamma$
respects the filtration. This proves Proposition \ref{prop.l37.gamma.filt}.
\end{verlong}

\subsection{$\gamma$ is an $\mathfrak{h}$-module map}

Next we will show a property of $\gamma$ which distinguishes it from the map
$\varphi$ of Section \ref{sect.L25}:

\begin{proposition}
\label{prop.l37.gamma.h}The map $\gamma:\otimes\mathfrak{g}\rightarrow
\otimes\mathfrak{n}$ is an $\mathfrak{h}$-module homomorphism.
\end{proposition}

\textit{Proof of Proposition \ref{prop.l37.gamma.h}.} We know that
$\otimes\mathfrak{n}$ is a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule (because $\mathfrak{n}$ is a $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule). Thus, (according to Definition \ref{defs.semimod}) the
relation (\ref{defs.semimod.->}) with $V$ replaced by $\otimes\mathfrak{n}$ holds.

Let $a\in\mathfrak{h}$. Then, the map $\otimes\mathfrak{g}\rightarrow
\otimes\mathfrak{g},$ $T\mapsto a\rightharpoonup T$ is $k$-linear (because the
Lie action of $\otimes\mathfrak{g}$ is $k$-bilinear), and the map
$\otimes\mathfrak{n}\rightarrow\otimes\mathfrak{n},$ $T\mapsto
a\rightharpoonup T$ is $k$-linear (because the Lie action of $\otimes
\mathfrak{n}$ is $k$-bilinear).

Let $Z:\otimes\mathfrak{g}\rightarrow\otimes\mathfrak{n}$ be the map defined
by%
\begin{equation}
\left(  Z\left(  T\right)  =a\rightharpoonup\left(  \gamma\left(  T\right)
\right)  -\gamma\left(  a\rightharpoonup T\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }T\in\otimes\mathfrak{g}\right)  .
\label{pf.l37.gamma.h.1}%
\end{equation}
This map $Z$ is $k$-linear (since the map $\otimes\mathfrak{g}\rightarrow
\otimes\mathfrak{g},$ $T\mapsto a\rightharpoonup T$, the map $\otimes
\mathfrak{n}\rightarrow\otimes\mathfrak{n},$ $T\mapsto a\rightharpoonup T$ and
the map $\gamma$ are all $k$-linear). Thus, $\operatorname{Ker}Z$ is a
$k$-submodule of $\otimes\mathfrak{g}$.

Now we are going to prove that%
\begin{equation}
\mathfrak{g}^{\otimes n}\subseteq\operatorname{Ker}%
Z\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}\text{.}
\label{pf.l37.gamma.h.2}%
\end{equation}


\textit{Proof of (\ref{pf.l37.gamma.h.2}).} We are going to prove
(\ref{pf.l37.gamma.h.2}) by induction over $n$:

\begin{vershort}
\textit{Induction base:} For the induction base, we must prove that
$\mathfrak{g}^{\otimes0}\subseteq\operatorname{Ker}Z$. This is rather trivial
using $\mathfrak{g}^{\otimes0}=k$ and (\ref{rmk.gamma-0}).
\end{vershort}

\begin{verlong}
\textit{Induction base:} Every $\lambda\in\mathfrak{g}^{\otimes\leq0}$
satisfies $\lambda\in k$ (since $\mathfrak{g}^{\otimes\leq0}=\bigoplus
\limits_{i=0}^{0}\mathfrak{g}^{\otimes i}=\mathfrak{g}^{\otimes0}=k$).
Therefore, every $\lambda\in\mathfrak{g}^{\otimes\leq0}$ satisfies
$\gamma\left(  \lambda\right)  =\lambda$ (according to (\ref{rmk.gamma-0}),
since $\lambda\in k$) and thus%
\begin{align*}
Z\left(  \lambda\right)   &  =a\rightharpoonup\left(  \underbrace{\gamma
\left(  \lambda\right)  }_{=\lambda}\right)  -\gamma\left(  a\rightharpoonup
\lambda\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.l37.gamma.h.1}),
applied to }T=\lambda\right) \\
&  =\underbrace{a\rightharpoonup\lambda}_{\substack{=0\text{ (since the
Lie}\\\text{action of }k\text{ is }0\text{)}}}-\gamma\left(
\underbrace{a\rightharpoonup\lambda}_{\substack{=0\text{ (since the
Lie}\\\text{action of }k\text{ is }0\text{)}}}\right)  =0-\underbrace{\gamma
\left(  0\right)  }_{=0\text{ (since }\gamma\text{ is }k\text{-linear)}%
}=0-0=0.
\end{align*}
We have thus shown that every $\lambda\in\mathfrak{g}^{\otimes\leq0}$
satisfies $Z\left(  \lambda\right)  =0$. In other words, every $\lambda
\in\mathfrak{g}^{\otimes\leq0}$ satisfies $\lambda\in\operatorname{Ker}Z$.
Hence, $\mathfrak{g}^{\otimes\leq0}\subseteq\operatorname*{Ker}Z$. In other
words, (\ref{pf.l37.gamma.h.2}) holds for $n=0$. This completes the induction base.
\end{verlong}

\textit{Induction step:} Let $p\in\mathbb{N}$. Assume that
(\ref{pf.l37.gamma.h.2}) holds for $n=p$. We now must show that
(\ref{pf.l37.gamma.h.2}) also holds for $n=p+1$.

Since (\ref{pf.l37.gamma.h.2}) holds for $n=p$, we have $\mathfrak{g}^{\otimes
p}\subseteq\operatorname*{Ker}Z$.

We are now going to show that $\mathfrak{g}^{\otimes\left(  p+1\right)
}\subseteq\operatorname*{Ker}Z$.

\begin{vershort}
Since $\operatorname*{Ker}Z$ is a $k$-submodule of $\otimes\mathfrak{g}$,
while $\mathfrak{g}^{\otimes\left(  p+1\right)  }=\mathfrak{g}\otimes
\mathfrak{g}^{\otimes p}$ is generated by left-induced tensors, we will be
done once we have shown that every left-induced tensor in $\mathfrak{g}%
^{\otimes\left(  p+1\right)  }$ lies in $\operatorname*{Ker}Z$. So let
$V\in\mathfrak{g}^{\otimes\left(  p+1\right)  }$ be some left-induced tensor.
Thus, there exist $u\in\mathfrak{g}$ and $U\in\mathfrak{g}^{\otimes p}$ such
that $V=u\otimes U$. Consider these $u$ and $U$. Since $u\in\mathfrak{g}%
=\mathfrak{g}^{\otimes1}$ and $U\in\mathfrak{g}^{\otimes p}$, we have $u\cdot
U=u\otimes U$ (due to (\ref{*=(X)})).
\end{vershort}

\begin{verlong}
Proposition \ref{prop.left-ind2} (applied to $p+1$ and $\mathfrak{g}$ instead
of $p$ and $V$) yields $\mathfrak{g}^{\otimes\left(  p+1\right)
}=\left\langle \mathfrak{g}_{\operatorname*{lind}}^{\otimes\left(  p+1\right)
}\right\rangle $ (where we are using Convention \ref{conv.left-ind2}).

Now we are going to prove that $\mathfrak{g}_{\operatorname*{lind}}%
^{\otimes\left(  p+1\right)  }\subseteq\operatorname{Ker}Z$. Indeed, let
$V\in\mathfrak{g}_{\operatorname*{lind}}^{\otimes\left(  p+1\right)  }$ be
arbitrary. Then, $V$ is a left-induced tensor in $\mathfrak{g}^{\otimes\left(
p+1\right)  }$ (since $V\in\mathfrak{g}_{\operatorname*{lind}}^{\otimes\left(
p+1\right)  }$), and thus there exist $u\in\mathfrak{g}$ and $U\in
\mathfrak{g}^{\otimes p}$ such that $V=u\otimes U$. Consider these $u$ and
$U$. Since $u\in\mathfrak{g}=\mathfrak{g}^{\otimes1}$ and $U\in\mathfrak{g}%
^{\otimes p}$, we have $u\cdot U=u\otimes U$ (due to (\ref{*=(X)}), applied to
$u$, $U$, $1$ and $p$ instead of $a$, $b$, $n$ and $m$).
\end{verlong}

Since $U\in\mathfrak{g}^{\otimes p}\subseteq\operatorname*{Ker}Z$, we have
$Z\left(  U\right)  =0$. But applying (\ref{pf.l37.gamma.h.1}) to $T=U$, we
obtain $Z\left(  U\right)  =a\rightharpoonup\left(  \gamma\left(  U\right)
\right)  -\gamma\left(  a\rightharpoonup U\right)  $. Thus, $a\rightharpoonup
\left(  \gamma\left(  U\right)  \right)  -\gamma\left(  a\rightharpoonup
U\right)  =Z\left(  U\right)  =0$, so that $a\rightharpoonup\left(
\gamma\left(  U\right)  \right)  =\gamma\left(  a\rightharpoonup U\right)  $.

We know that the relation (\ref{defs.semimod.->}) with $V$ replaced by
$\otimes\mathfrak{n}$ holds. Thus, we can apply this relation
(\ref{defs.semimod.->}) to $u$, $\gamma\left(  U\right)  $ and $\otimes
\mathfrak{n}$ instead of $b$, $v$ and $V$, and thus we obtain%
\[
\left[  a,u\right]  \rightharpoonup\left(  \gamma\left(  U\right)  \right)
=a\rightharpoonup\left(  u\rightharpoonup\left(  \gamma\left(  U\right)
\right)  \right)  -u\rightharpoonup\left(  a\rightharpoonup\left(
\gamma\left(  U\right)  \right)  \right)  .
\]


Now, $V=u\otimes U=u\cdot U$, so that%
\begin{align*}
a\rightharpoonup V  &  =a\rightharpoonup\left(  u\cdot U\right)  =\left(
a\rightharpoonup u\right)  \cdot U+u\cdot\left(  a\rightharpoonup U\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{by (\ref{defs.g-alg.cond}) (applied to }\otimes\mathfrak{g}\text{ and
}U\text{ instead of }A\text{ and }v\text{), }\\
\text{because }\otimes\mathfrak{g}\text{ is a }\mathfrak{g}\text{-algebra}%
\end{array}
\right)  .
\end{align*}
Hence,%
\begin{align}
&  \gamma\left(  a\rightharpoonup V\right) \nonumber\\
&  =\gamma\left(  \left(  a\rightharpoonup u\right)  \cdot U+u\cdot\left(
a\rightharpoonup U\right)  \right) \nonumber\\
&  =\underbrace{\gamma\left(  \left(  a\rightharpoonup u\right)  \cdot
U\right)  }_{\substack{=\pi\left(  a\rightharpoonup u\right)  \cdot
\gamma\left(  U\right)  +\left(  a\rightharpoonup u\right)  \rightharpoonup
\left(  \gamma\left(  U\right)  \right)  \\\text{(by (\ref{prop.gamma.rec}),
applied to }a\rightharpoonup u\text{ instead of }u\text{)}}%
}+\underbrace{\gamma\left(  u\cdot\left(  a\rightharpoonup U\right)  \right)
}_{\substack{=\pi\left(  u\right)  \cdot\gamma\left(  a\rightharpoonup
U\right)  +u\rightharpoonup\left(  \gamma\left(  a\rightharpoonup U\right)
\right)  \\\text{(by (\ref{prop.gamma.rec}), applied to }a\rightharpoonup
U\text{ instead of }U\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }%
\gamma\text{ is linear}\right) \nonumber\\
&  =\underbrace{\pi\left(  a\rightharpoonup u\right)  }%
_{\substack{=a\rightharpoonup\left(  \pi\left(  u\right)  \right)
\\\text{(since }\pi\text{ is an }\mathfrak{h}\text{-module}%
\\\text{homomorphism, while }a\in\mathfrak{h}\text{)}}}\cdot\gamma\left(
U\right)  +\underbrace{\left(  a\rightharpoonup u\right)  }%
_{\substack{=\left[  a,u\right]  \\\text{(due to
(\ref{prop.g-is-g-module.form}), applied to}\\a\text{ and }u\text{ instead of
}v\text{ and }w\text{)}}}\rightharpoonup\left(  \gamma\left(  U\right)
\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ +\pi\left(  u\right)  \cdot\gamma\left(
a\rightharpoonup U\right)  +u\rightharpoonup\underbrace{\left(  \gamma\left(
a\rightharpoonup U\right)  \right)  }_{=a\rightharpoonup\left(  \gamma\left(
U\right)  \right)  }\nonumber\\
&  =\left(  a\rightharpoonup\left(  \pi\left(  u\right)  \right)  \right)
\cdot\gamma\left(  U\right)  +\left[  a,u\right]  \rightharpoonup\left(
\gamma\left(  U\right)  \right)  +\pi\left(  u\right)  \cdot\gamma\left(
a\rightharpoonup U\right)  +u\rightharpoonup\left(  a\rightharpoonup\left(
\gamma\left(  U\right)  \right)  \right)  . \label{pf.l37.gamma.h.3}%
\end{align}
On the other hand, $V=u\cdot U$ leads to%
\begin{align*}
a\rightharpoonup\left(  \gamma\left(  V\right)  \right)   &  =a\rightharpoonup
\left(  \gamma\left(  u\cdot U\right)  \right)  =a\rightharpoonup\left(
\pi\left(  u\right)  \cdot\gamma\left(  U\right)  +u\rightharpoonup\left(
\gamma\left(  U\right)  \right)  \right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{prop.gamma.rec})}\right) \\
&  =\underbrace{a\rightharpoonup\left(  \pi\left(  u\right)  \cdot
\gamma\left(  U\right)  \right)  }_{\substack{=\left(  a\rightharpoonup\left(
\pi\left(  u\right)  \right)  \right)  \cdot\gamma\left(  U\right)
+\pi\left(  u\right)  \cdot\left(  a\rightharpoonup\left(  \gamma\left(
U\right)  \right)  \right)  \\\text{(by (\ref{prop.gamma.rec}), applied to
}\pi\left(  u\right)  \text{ and }\gamma\left(  U\right)  \text{ instead of
}u\text{ and }v\text{)}}}+a\rightharpoonup\left(  u\rightharpoonup\left(
\gamma\left(  U\right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Lie action of }\otimes
\mathfrak{n}\text{ is }k\text{-bilinear}\right) \\
&  =\left(  a\rightharpoonup\left(  \pi\left(  u\right)  \right)  \right)
\cdot\gamma\left(  U\right)  +\pi\left(  u\right)  \cdot\underbrace{\left(
a\rightharpoonup\left(  \gamma\left(  U\right)  \right)  \right)  }%
_{=\gamma\left(  a\rightharpoonup U\right)  }+a\rightharpoonup\left(
u\rightharpoonup\left(  \gamma\left(  U\right)  \right)  \right) \\
&  =\left(  a\rightharpoonup\left(  \pi\left(  u\right)  \right)  \right)
\cdot\gamma\left(  U\right)  +\pi\left(  u\right)  \cdot\gamma\left(
a\rightharpoonup U\right)  +a\rightharpoonup\left(  u\rightharpoonup\left(
\gamma\left(  U\right)  \right)  \right)  .
\end{align*}
Subtracting (\ref{pf.l37.gamma.h.3}) from this equation, we obtain%
\begin{align*}
&  a\rightharpoonup\left(  \gamma\left(  V\right)  \right)  -\gamma\left(
a\rightharpoonup V\right) \\
&  =\left(  \left(  a\rightharpoonup\left(  \pi\left(  u\right)  \right)
\right)  \cdot\gamma\left(  U\right)  +\pi\left(  u\right)  \cdot\gamma\left(
a\rightharpoonup U\right)  +a\rightharpoonup\left(  u\rightharpoonup\left(
\gamma\left(  U\right)  \right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  \left(  a\rightharpoonup\left(  \pi\left(
u\right)  \right)  \right)  \cdot\gamma\left(  U\right)  +\left[  a,u\right]
\rightharpoonup\left(  \gamma\left(  U\right)  \right)  \right. \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.  +\pi\left(  u\right)
\cdot\gamma\left(  a\rightharpoonup U\right)  +u\rightharpoonup\left(
a\rightharpoonup\left(  \gamma\left(  U\right)  \right)  \right)  \right) \\
&  =a\rightharpoonup\left(  u\rightharpoonup\left(  \gamma\left(  U\right)
\right)  \right)  -u\rightharpoonup\left(  a\rightharpoonup\left(
\gamma\left(  U\right)  \right)  \right)  -\underbrace{\left[  a,u\right]
\rightharpoonup\left(  \gamma\left(  U\right)  \right)  }%
_{\substack{=a\rightharpoonup\left(  u\rightharpoonup\left(  \gamma\left(
U\right)  \right)  \right)  -u\rightharpoonup\left(  a\rightharpoonup\left(
\gamma\left(  U\right)  \right)  \right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{after some cancellation of terms}\right)
\\
&  =\left(  a\rightharpoonup\left(  u\rightharpoonup\left(  \gamma\left(
U\right)  \right)  \right)  -u\rightharpoonup\left(  a\rightharpoonup\left(
\gamma\left(  U\right)  \right)  \right)  \right)  -\left(  a\rightharpoonup
\left(  u\rightharpoonup\left(  \gamma\left(  U\right)  \right)  \right)
-u\rightharpoonup\left(  a\rightharpoonup\left(  \gamma\left(  U\right)
\right)  \right)  \right) \\
&  =0.
\end{align*}
But (\ref{pf.l37.gamma.h.1}) (applied to $T=V$) yields%
\[
Z\left(  V\right)  =a\rightharpoonup\left(  \gamma\left(  V\right)  \right)
-\gamma\left(  a\rightharpoonup V\right)  =0.
\]
Thus, $V\in\operatorname{Ker}Z$.

\begin{vershort}
We have thus shown that every left-induced tensor $V$ in $\mathfrak{g}%
^{\otimes\left(  p+1\right)  }$ lies in $\operatorname*{Ker}Z$. As we have
said before, this shows that $\mathfrak{g}^{\otimes\left(  p+1\right)
}\subseteq\operatorname{Ker}Z$. In other words, (\ref{pf.l37.gamma.h.2}) holds
for $n=p+1$. This completes the induction step, and thus the induction proof
of (\ref{pf.l37.gamma.h.2}) is done.
\end{vershort}

\begin{verlong}
We have thus shown that every $V\in\mathfrak{g}_{\operatorname*{lind}%
}^{\otimes\left(  p+1\right)  }$ satisfies $V\in\operatorname{Ker}Z$. In other
words, $\mathfrak{g}_{\operatorname*{lind}}^{\otimes\left(  p+1\right)
}\subseteq\operatorname{Ker}Z$.

Consequently, Proposition \ref{prop.<>} \textbf{(a)} (applied to
$\otimes\mathfrak{g}$, $\mathfrak{g}_{\operatorname*{lind}}^{\otimes\left(
p+1\right)  }$ and $\operatorname{Ker}Z$ instead of $M$, $S$ and $Q$) yields
$\left\langle \mathfrak{g}_{\operatorname*{lind}}^{\otimes\left(  p+1\right)
}\right\rangle \subseteq\operatorname{Ker}Z$. Since $\mathfrak{g}%
^{\otimes\left(  p+1\right)  }=\left\langle \mathfrak{g}_{\operatorname*{lind}%
}^{\otimes\left(  p+1\right)  }\right\rangle $, this becomes $\mathfrak{g}%
^{\otimes\left(  p+1\right)  }\subseteq\operatorname{Ker}Z$. In other words,
(\ref{pf.l37.gamma.h.2}) holds for $n=p+1$. This completes the induction step,
and thus the induction proof of (\ref{pf.l37.gamma.h.2}) is done.
\end{verlong}

\begin{vershort}
Now that (\ref{pf.l37.gamma.h.2}) is proven, we conclude that $\otimes
\mathfrak{g}\subseteq\operatorname*{Ker}Z$ (since $\otimes\mathfrak{g}%
=\bigoplus\limits_{n\in\mathbb{N}}\mathfrak{g}^{\otimes n}$). Thus, $Z=0$. By
the definition of $Z$, this means that $\gamma\left(  a\rightharpoonup
T\right)  =a\rightharpoonup\left(  \gamma\left(  T\right)  \right)  $ for
every $T\in\otimes\mathfrak{g}$. Since this holds for every $a\in\mathfrak{h}%
$, this shows that $\gamma$ is an $\mathfrak{h}$-module homomorphism. This
proves Proposition \ref{prop.l37.gamma.h}.
\end{vershort}

\begin{verlong}
Now that (\ref{pf.l37.gamma.h.2}) is proven, we notice that%
\begin{align*}
\otimes\mathfrak{g}  &  =\bigoplus_{n\in\mathbb{N}}\mathfrak{g}^{\otimes
n}=\sum_{n\in\mathbb{N}}\underbrace{\mathfrak{g}^{\otimes n}}%
_{\substack{\subseteq\operatorname{Ker}Z\\\text{(by (\ref{pf.l37.gamma.h.2}%
))}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since direct sums are sums}\right) \\
&  \subseteq\sum_{n\in\mathbb{N}}\operatorname{Ker}Z\subseteq
\operatorname{Ker}Z\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\operatorname{Ker}%
Z\text{ is a }k\text{-module}\right)  .
\end{align*}
In other words, $Z=0$. Thus, $Z\left(  T\right)  =0$ for every $T\in
\otimes\mathfrak{g}$. Using (\ref{pf.l37.gamma.h.1}), this rewrites as%
\[
a\rightharpoonup\left(  \gamma\left(  T\right)  \right)  -\gamma\left(
a\rightharpoonup T\right)  =0\ \ \ \ \ \ \ \ \ \ \text{for every }T\in
\otimes\mathfrak{g}.
\]
In other words, $\gamma\left(  a\rightharpoonup T\right)  =a\rightharpoonup
\left(  \gamma\left(  T\right)  \right)  $ for every $T\in\otimes\mathfrak{g}%
$. Since this holds for every $a\in\mathfrak{h}$, this shows that $\gamma$ is
an $\mathfrak{h}$-module homomorphism. This proves Proposition
\ref{prop.l37.gamma.h}.
\end{verlong}

\subsection{A lemma on $\gamma$ and $k$-submodules of $\otimes\mathfrak{g}$}

We now set forth for a proof of $\gamma\left(  J\right)  =0$ and
$\gamma\left(  \left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
=0$. The first step will be a "little brother" of Lemma \ref{lem.l25.C} (but
notice that it has a weaker condition that Lemma \ref{lem.l25.C}):

\begin{lemma}
\label{lem.l37.C}Let $C$ be a $k$-submodule of $\otimes\mathfrak{g}$
satisfying $\gamma\left(  C\right)  =0$. Then, $\gamma\left(  \left(
\otimes\mathfrak{g}\right)  \cdot C\right)  =0$.
\end{lemma}

\begin{vershort}
The proof of this lemma is a calque of our proof of Lemma \ref{lem.l25.C},
except that it is even simpler. We are leaving it to the reader.
\end{vershort}

\begin{verlong}
\textit{Proof of Lemma \ref{lem.l37.C}.} We are going to prove that%
\begin{equation}
\gamma\left(  \mathfrak{g}^{\otimes p}\cdot C\right)
=0\ \ \ \ \ \ \ \ \ \ \text{for every }p\in\mathbb{N}\text{.}
\label{pf.l37.C.1}%
\end{equation}
\textit{Proof of (\ref{pf.l37.C.1}).} We will prove (\ref{pf.l37.C.1}) by
induction over $p$:

\textit{Induction base:} We have $\mathfrak{g}^{\otimes0}=k$ and thus
$\mathfrak{g}^{\otimes0}\cdot C=k\cdot C=C$ (since $C$ is a $k$-submodule), so
that $\gamma\left(  \mathfrak{g}^{\otimes0}\cdot C\right)  =\gamma\left(
C\right)  =0$. Thus, (\ref{pf.l37.C.1}) holds for $p=0$. This completes the
induction base.

\textit{Induction step:} Let $q\in\mathbb{N}_{+}$. Assume that
(\ref{pf.l37.C.1}) holds for $p=q-1$. Now we must prove that (\ref{pf.l37.C.1}%
) holds for $p=q$.

We have assumed that (\ref{pf.l37.C.1}) holds for $p=q-1$. In other words,
$\gamma\left(  \mathfrak{g}^{\otimes\left(  q-1\right)  }\cdot C\right)  =0$.

Let $D=\mathfrak{g}^{\otimes\left(  q-1\right)  }\cdot C$. Then,
$D=\mathfrak{g}^{\otimes\left(  q-1\right)  }\cdot C$ yields $\gamma\left(
D\right)  =\gamma\left(  \mathfrak{g}^{\otimes\left(  q-1\right)  }\cdot
C\right)  =0$.

On the other hand, Proposition \ref{prop.tensor.filt-is-alg-filt} \textbf{(a)}
(applied to $V=\mathfrak{g}$, $i=1$ and $j=q-1$) yields $\mathfrak{g}%
^{\otimes1}\cdot\mathfrak{g}^{\otimes\left(  q-1\right)  }=\mathfrak{g}%
^{\otimes\left(  1+\left(  q-1\right)  \right)  }=\mathfrak{g}^{\otimes q}$.
Thus,%
\[
\underbrace{\mathfrak{g}^{\otimes q}}_{=\mathfrak{g}^{\otimes1}\cdot
\mathfrak{g}^{\otimes\left(  q-1\right)  }}\cdot C=\underbrace{\mathfrak{g}%
^{\otimes1}}_{=\mathfrak{g}}\cdot\underbrace{\mathfrak{g}^{\otimes\left(
q-1\right)  }\cdot C}_{=D}=\mathfrak{g}\cdot D.
\]


We are now in the following situation: We know that $D$ is a $k$-submodule of
$\otimes\mathfrak{g}$ satisfying $\gamma\left(  D\right)  =0$. We want to
prove that $\gamma\left(  \mathfrak{g}\cdot D\right)  =0$.

Let $E$ be the subset $\left\{  u\cdot U\ \mid\ \left(  u,U\right)
\in\mathfrak{g}\times D\right\}  $ of $\otimes\mathfrak{g}$. Then,%
\[
\mathfrak{g}\cdot D=\left\langle u\cdot U\ \mid\ \left(  u,U\right)
\in\mathfrak{g}\times D\right\rangle =\left\langle \underbrace{\left\{  u\cdot
U\ \mid\ \left(  u,U\right)  \in\mathfrak{g}\times D\right\}  }_{=E}%
\right\rangle =\left\langle E\right\rangle .
\]
Now we are going to show that $E\subseteq\operatorname*{Ker}\gamma$.

In fact, let $T\in E$ be arbitrary. Then, $T\in E=\left\{  u\cdot
U\ \mid\ \left(  u,U\right)  \in\mathfrak{g}\times D\right\}  $, so that there
exists some $\left(  u,U\right)  \in\mathfrak{g}\times D$ such that $T=u\cdot
U$. Consider this $\left(  u,U\right)  $. Clearly, $u\in\mathfrak{g}$ and
$U\in D$. Obviously $U\in D$ leads to $\gamma\left(  U\right)  =0$ (since
$\gamma\left(  D\right)  =0$). Now, (\ref{prop.gamma.rec}) yields%
\[
\gamma\left(  u\cdot U\right)  =\pi\left(  u\right)  \cdot\underbrace{\gamma
\left(  U\right)  }_{=0}+u\rightharpoonup\underbrace{\left(  \gamma\left(
U\right)  \right)  }_{=0}=\underbrace{\pi\left(  u\right)  \cdot0}%
_{=0}+\underbrace{u\rightharpoonup0}_{\substack{=0\text{ (since the Lie
action} \\\text{of }\otimes\mathfrak{n}\text{ is }k\text{-bilinear)}}}=0+0=0.
\]
Since $u\cdot U=T$, this rewrites as $\gamma\left(  T\right)  =0$. Thus,
$T\in\operatorname*{Ker}\gamma$.

We have therefore shown that every $T\in E$ satisfies $T\in\operatorname*{Ker}%
\gamma$. Thus, $E\subseteq\operatorname*{Ker}\gamma$.

Consequently, Proposition \ref{prop.<>} \textbf{(a)} (applied to
$\otimes\mathfrak{g}$, $E$ and $\operatorname*{Ker}\gamma$ instead of $M$, $S$
and $Q$) yields $\left\langle E\right\rangle \subseteq\operatorname*{Ker}%
\gamma$ (since $\operatorname*{Ker}\gamma$ is a $k$-submodule of
$\otimes\mathfrak{g}$, since $\gamma$ is $k$-linear). Since $\left\langle
E\right\rangle =\mathfrak{g}\cdot D$, this rewrites as $\mathfrak{g}\cdot
D\subseteq\operatorname*{Ker}\gamma$, and thus $\gamma\left(  \mathfrak{g}%
\cdot D\right)  =0$. Since $\mathfrak{g}^{\otimes q}\cdot C=\mathfrak{g}\cdot
D$, this becomes $\gamma\left(  \mathfrak{g}^{\otimes q}\cdot C\right)  =0$.
In other words, (\ref{pf.l37.C.1}) holds for $p=q$. This completes the
induction step, and thus the induction proof of (\ref{pf.l37.C.1}) is done.

Now, $\otimes\mathfrak{g}=\bigoplus\limits_{p\in\mathbb{N}}\mathfrak{g}%
^{\otimes p}=\sum\limits_{p\in\mathbb{N}}\mathfrak{g}^{\otimes p}$ (since
direct sums are sums) yields $\left(  \otimes\mathfrak{g}\right)  \cdot
C=\left(  \sum\limits_{p\in\mathbb{N}}\mathfrak{g}^{\otimes p}\right)  \cdot
C=\sum\limits_{p\in\mathbb{N}}\left(  \mathfrak{g}^{\otimes p}\cdot C\right)
$ and thus
\begin{align*}
\gamma\left(  \left(  \otimes\mathfrak{g}\right)  \cdot C\right)   &
=\gamma\left(  \sum\limits_{p\in\mathbb{N}}\left(  \mathfrak{g}^{\otimes
p}\cdot C\right)  \right)  =\sum\limits_{p\in\mathbb{N}}\underbrace{\gamma
\left(  \mathfrak{g}^{\otimes p}\cdot C\right)  }_{=0\text{ (due to
(\ref{pf.l37.C.1}))}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\gamma\text{ is
}k\text{-linear}\right) \\
&  =\sum\limits_{p\in\mathbb{N}}0=0.
\end{align*}
This proves Lemma \ref{lem.l37.C}.
\end{verlong}

\subsection{$\gamma\left(  J\right)  =0$ and $\gamma\left(  \left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  =0$}

We are now ready to prove the following facts (which are similar to
Propositions \ref{prop.l25.alpha(J)} and \ref{prop.l25.alpha(xgh)}, respectively):

\begin{proposition}
\label{prop.l37.gamma(J)}We have $\gamma\left(  J\right)  =0$.
\end{proposition}

\begin{proposition}
\label{prop.l37.gamma(xgh)}We have $\gamma\left(  \left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  =0$.
\end{proposition}

\textit{Proof of Proposition \ref{prop.l37.gamma(J)}.} Consider the
$k$-submodule $J_{0}$ defined in Proposition \ref{prop.l25a} \textbf{(b)}.

We are going to prove that $\gamma\left(  J_{0}\cdot\left(  \otimes
\mathfrak{g}\right)  \right)  =0$ now (this will quickly yield $\gamma\left(
J\right)  =0$ then, due to $J=\left(  \otimes\mathfrak{g}\right)  \cdot
J_{0}\cdot\left(  \otimes\mathfrak{g}\right)  $ and Lemma \ref{lem.l37.C}).

\begin{vershort}
Let $S_{0}$ denote the subset $\left\{  v\otimes w-w\otimes v-\left[
v,w\right]  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}%
\right\}  $ of $\otimes\mathfrak{g}$. Then, $J_{0}=\left\langle S_{0}%
\right\rangle $ (this is how $J_{0}$ was defined).

Firstly, it is clear that in order to prove $\gamma\left(  J_{0}\cdot\left(
\otimes\mathfrak{g}\right)  \right)  =0$, it is enough to show that
$\gamma\left(  s\cdot T\right)  =0$ for every $s\in J_{0}$ and $T\in
\otimes\mathfrak{g}$ (because $\gamma$ is $k$-linear). Secondly, in order to
show that $\gamma\left(  s\cdot T\right)  =0$ for every $s\in J_{0}$ and
$T\in\otimes\mathfrak{g}$, it is enough to prove that $\gamma\left(  s\cdot
T\right)  =0$ for every $s\in S_{0}$ and $T\in\otimes\mathfrak{g}$ (because
$J_{0}=\left\langle S_{0}\right\rangle $, and because multiplication in
$\otimes\mathfrak{g}$ and the map $\gamma$ are $k$-linear). So let us prove
the latter assertion. Fix some $s\in S_{0}$ and $T\in\otimes\mathfrak{g}$.

By the very definition of $S_{0}$, the relation $s\in S_{0}$ means that there
exists some $\left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}$ such that
$s=v\otimes w-w\otimes v-\left[  v,w\right]  $. Since (\ref{*=(X)}) yields
$v\otimes w=v\cdot w$ and $w\otimes v=w\cdot v$, this rewrites as $s=v\cdot
w-w\cdot v-\left[  v,w\right]  $.

We are now going to show $\gamma\left(  s\cdot T\right)  =0$ through
straightforward computation:

We have $\gamma\left(  v\cdot T\right)  =\pi\left(  v\right)  \cdot
\gamma\left(  T\right)  +v\rightharpoonup\left(  \gamma\left(  T\right)
\right)  $ (according to (\ref{prop.gamma.rec}), applied to $u=v$ and $U=T$)
and $\gamma\left(  w\cdot T\right)  =w\rightharpoonup\left(  \gamma\left(
T\right)  \right)  $ (according to (\ref{prop.gamma.rec.h}), applied to $u=w$
and $U=T$).

Now, (\ref{prop.gamma.rec}) (applied to $u=v$ and $U=w\cdot T$) yields%
\begin{align*}
\gamma\left(  v\cdot w\cdot T\right)   &  =\pi\left(  v\right)  \cdot
\underbrace{\gamma\left(  w\cdot T\right)  }_{=w\rightharpoonup\left(
\gamma\left(  T\right)  \right)  }+v\rightharpoonup\left(  \underbrace{\gamma
\left(  w\cdot T\right)  }_{=w\rightharpoonup\left(  \gamma\left(  T\right)
\right)  }\right) \\
&  =\pi\left(  v\right)  \cdot\left(  w\rightharpoonup\left(  \gamma\left(
T\right)  \right)  \right)  +v\rightharpoonup\left(  w\rightharpoonup\left(
\gamma\left(  T\right)  \right)  \right)  .
\end{align*}
On the other hand, (\ref{prop.gamma.rec.h}) (applied to $u=w$ and $U=v\cdot
T$) yields%
\begin{align*}
\gamma\left(  w\cdot v\cdot T\right)   &  =w\rightharpoonup\left(
\underbrace{\gamma\left(  v\cdot T\right)  }_{=\pi\left(  v\right)
\cdot\gamma\left(  T\right)  +v\rightharpoonup\left(  \gamma\left(  T\right)
\right)  }\right)  =w\rightharpoonup\left(  \pi\left(  v\right)  \cdot
\gamma\left(  T\right)  +v\rightharpoonup\left(  \gamma\left(  T\right)
\right)  \right) \\
&  =\underbrace{w\rightharpoonup\left(  \pi\left(  v\right)  \cdot
\gamma\left(  T\right)  \right)  }_{\substack{=\left(  w\rightharpoonup\left(
\pi\left(  v\right)  \right)  \right)  \cdot\gamma\left(  T\right)
+\pi\left(  v\right)  \cdot\left(  w\rightharpoonup\left(  \gamma\left(
T\right)  \right)  \right)  \\\text{(by (\ref{defs.semi.g-alg.cond}) (applied
to }\otimes\mathfrak{n}\text{, }w\text{, }\pi\left(  v\right)  \text{ and
}\gamma\left(  T\right)  \\\text{instead of }A\text{, }a\text{, }u\text{ and
}v\text{), because }\otimes\mathfrak{n}\text{ is a }\left(  \mathfrak{g}%
,\mathfrak{h}\right)  \text{-semialgebra)}}}+w\rightharpoonup\left(
v\rightharpoonup\left(  \gamma\left(  T\right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Lie action of }\otimes
\mathfrak{n}\text{ is }k\text{-linear}\right) \\
&  =\left(  w\rightharpoonup\left(  \pi\left(  v\right)  \right)  \right)
\cdot\gamma\left(  T\right)  +\pi\left(  v\right)  \cdot\left(
w\rightharpoonup\left(  \gamma\left(  T\right)  \right)  \right)
+w\rightharpoonup\left(  v\rightharpoonup\left(  \gamma\left(  T\right)
\right)  \right)  .
\end{align*}
Finally,%
\begin{align*}
\gamma\left(  \underbrace{\left[  v,w\right]  }_{=-\left[  w,v\right]  }\cdot
T\right)   &  =-\underbrace{\gamma\left(  \left[  w,v\right]  \cdot T\right)
}_{\substack{=\pi\left(  \left[  w,v\right]  \right)  \cdot\gamma\left(
T\right)  +\left[  w,v\right]  \rightharpoonup\left(  \gamma\left(  T\right)
\right)  \\\text{(according to (\ref{prop.gamma.rec}), applied to}\\u=\left[
w,v\right]  \text{ and }U=T\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\gamma\text{ is }k\text{-linear}\right) \\
&  =-\left(  \pi\left(  \left[  w,v\right]  \right)  \cdot\gamma\left(
T\right)  +\left[  w,v\right]  \rightharpoonup\left(  \gamma\left(  T\right)
\right)  \right)  .
\end{align*}
Now, $s=v\cdot w-w\cdot v-\left[  v,w\right]  $ yields%
\begin{align*}
\gamma\left(  s\cdot T\right)   &  =\gamma\left(  \left(  v\cdot w-w\cdot
v-\left[  v,w\right]  \right)  \cdot T\right) \\
&  =\underbrace{\gamma\left(  v\cdot w\cdot T\right)  }_{=\pi\left(  v\right)
\cdot\left(  w\rightharpoonup\left(  \gamma\left(  T\right)  \right)  \right)
+v\rightharpoonup\left(  w\rightharpoonup\left(  \gamma\left(  T\right)
\right)  \right)  }-\underbrace{\gamma\left(  w\cdot v\cdot T\right)
}_{=\left(  w\rightharpoonup\left(  \pi\left(  v\right)  \right)  \right)
\cdot\gamma\left(  T\right)  +\pi\left(  v\right)  \cdot\left(
w\rightharpoonup\left(  \gamma\left(  T\right)  \right)  \right)
+w\rightharpoonup\left(  v\rightharpoonup\left(  \gamma\left(  T\right)
\right)  \right)  }\\
&  \ \ \ \ \ \ \ \ \ \ -\underbrace{\gamma\left(  \left[  v,w\right]  \cdot
T\right)  }_{=-\left(  \pi\left(  \left[  w,v\right]  \right)  \cdot
\gamma\left(  T\right)  +\left[  w,v\right]  \rightharpoonup\left(
\gamma\left(  T\right)  \right)  \right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\gamma\text{ is }k\text{-linear}%
\right) \\
&  =\left(  \pi\left(  v\right)  \cdot\left(  w\rightharpoonup\left(
\gamma\left(  T\right)  \right)  \right)  +v\rightharpoonup\left(
w\rightharpoonup\left(  \gamma\left(  T\right)  \right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  \left(  w\rightharpoonup\left(  \pi\left(
v\right)  \right)  \right)  \cdot\gamma\left(  T\right)  +\pi\left(  v\right)
\cdot\left(  w\rightharpoonup\left(  \gamma\left(  T\right)  \right)  \right)
+w\rightharpoonup\left(  v\rightharpoonup\left(  \gamma\left(  T\right)
\right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  -\left(  \left(  w\rightharpoonup\left(
\pi\left(  v\right)  \right)  \right)  \cdot\gamma\left(  T\right)
+w\rightharpoonup\left(  v\rightharpoonup\left(  \gamma\left(  T\right)
\right)  \right)  -v\rightharpoonup\left(  w\rightharpoonup\left(
\gamma\left(  T\right)  \right)  \right)  \right)  \right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{because all terms in the sum cancel
out}\right)  ,
\end{align*}
just as we wanted to show.

The proof of $\gamma\left(  J_{0}\cdot\left(  \otimes\mathfrak{g}\right)
\right)  =0$ is thus complete. Now, Lemma \ref{lem.l37.C} (applied to
$C=J_{0}\cdot\left(  \otimes\mathfrak{g}\right)  $) yields that $\gamma\left(
\left(  \otimes\mathfrak{g}\right)  \cdot\left(  J_{0}\cdot\left(
\otimes\mathfrak{g}\right)  \right)  \right)  =0$ (because $\gamma\left(
J_{0}\cdot\left(  \otimes\mathfrak{g}\right)  \right)  =0$). Since $\left(
\otimes\mathfrak{g}\right)  \cdot\left(  J_{0}\cdot\left(  \otimes
\mathfrak{g}\right)  \right)  =\left(  \otimes\mathfrak{g}\right)  \cdot
J_{0}\cdot\left(  \otimes\mathfrak{g}\right)  =J$ (according to Proposition
\ref{prop.l25a} \textbf{(b)}), this rewrites as $\gamma\left(  J\right)  =0$.
Thus, Proposition \ref{prop.l37.gamma(J)} is proven.
\end{vershort}

\begin{verlong}
We know that $\otimes\mathfrak{n}$ is a $\left(  \mathfrak{g},\mathfrak{h}%
\right)  $-semimodule. Therefore, (according to Definition \ref{defs.semimod})
the relation (\ref{defs.semimod.->}) with $V$ replaced by $\otimes
\mathfrak{n}$ holds. In other words, we have%
\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{h}\text{, }b\in\mathfrak{g}\text{ and }%
v\in\otimes\mathfrak{n}\right)  . \label{pf.l37.gamma(J).semimod}%
\end{equation}


Let $S_{0}$ denote the subset $\left\{  v\otimes w-w\otimes v-\left[
v,w\right]  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}%
\right\}  $ of $\otimes\mathfrak{g}$. Then, we know from the definition of
$J_{0}$ that
\begin{align*}
J_{0}  &  =\left\langle v\otimes w-w\otimes v-\left[  v,w\right]
\ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}\right\rangle \\
&  =\left\langle \underbrace{\left\{  v\otimes w-w\otimes v-\left[
v,w\right]  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}%
\right\}  }_{=S_{0}}\right\rangle =\left\langle S_{0}\right\rangle .
\end{align*}


Now, let $T\in\otimes\mathfrak{g}$ be arbitrary. Let $\varrho_{T}^{\prime
}:\otimes\mathfrak{g}\rightarrow\otimes\mathfrak{n}$ be the map defined by%
\[
\left(  \varrho_{T}^{\prime}\left(  U\right)  =\gamma\left(  U\cdot T\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }U\in\otimes\mathfrak{g}\right)  .
\]
This map $\varrho_{T}^{\prime}$ is $k$-linear (since it is the composition of
the two $k$-linear maps $\gamma$ and $\otimes\mathfrak{g}\rightarrow
\otimes\mathfrak{g},$ $U\mapsto U\cdot T$). Thus, $\operatorname*{Ker}%
\varrho_{T}^{\prime}$ is a $k$-submodule of $\otimes\mathfrak{g}$.

We are now going to prove that every $s\in S_{0}$ satisfy $s\in
\operatorname*{Ker}\varrho_{T}^{\prime}$.

In fact, let $s\in S_{0}$ be arbitrary. Since $s\in S_{0}=\left\{  v\otimes
w-w\otimes v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)  \in
\mathfrak{g}\times\mathfrak{h}\right\}  $, there exists some $\left(
v,w\right)  \in\mathfrak{g}\times\mathfrak{h}$ such that $s=v\otimes
w-w\otimes v-\left[  v,w\right]  $. So we have $v\in\mathfrak{g}$ and
$w\in\mathfrak{h}$. Note that $v\in\mathfrak{g}=\mathfrak{g}^{\otimes1}$ and
$w\in\mathfrak{h}\subseteq\mathfrak{g}=\mathfrak{g}^{\otimes1}$ yield $v\cdot
w=v\otimes w$ (by (\ref{*=(X)}), applied to $1$ and $1$ instead of $n$ and
$m$), and similarly $w\cdot v=w\otimes v$. Using these observations, we see
that%
\[
s=\underbrace{v\otimes w}_{=v\cdot w}-\underbrace{w\otimes v}_{=w\cdot
v}-\left[  v,w\right]  =v\cdot w-w\cdot v-\left[  v,w\right]  .
\]


Now, we are going to prove that $\varrho_{T}^{\prime}\left(  s\right)  =0$. In
order to do this, we will compute $\varrho_{T}^{\prime}\left(  v\cdot
w\right)  $, $\varrho_{T}^{\prime}\left(  w\cdot v\right)  $ and $\varrho
_{T}^{\prime}\left(  \left[  v,w\right]  \right)  $. First let us compute
$\gamma\left(  v\cdot T\right)  $ and $\gamma\left(  w\cdot T\right)  $.

We have $\gamma\left(  v\cdot T\right)  =\pi\left(  v\right)  \cdot
\gamma\left(  T\right)  +v\rightharpoonup\left(  \gamma\left(  T\right)
\right)  $ (according to (\ref{prop.gamma.rec}), applied to $u=v$ and $U=T$)
and $\gamma\left(  w\cdot T\right)  =w\rightharpoonup\left(  \gamma\left(
T\right)  \right)  $ (according to (\ref{prop.gamma.rec.h}), applied to $u=w$
and $U=T$).

The definition of $\varrho_{T}^{\prime}$ yields
\begin{align}
\varrho_{T}^{\prime}\left(  v\cdot w\right)   &  =\gamma\left(  v\cdot w\cdot
T\right)  =\pi\left(  v\right)  \cdot\underbrace{\gamma\left(  w\cdot
T\right)  }_{=w\rightharpoonup\left(  \gamma\left(  T\right)  \right)
}+v\rightharpoonup\left(  \underbrace{\gamma\left(  w\cdot T\right)
}_{=w\rightharpoonup\left(  \gamma\left(  T\right)  \right)  }\right)
\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{according to (\ref{prop.gamma.rec}),
applied to }u=v\text{ and }U=w\cdot T\right) \nonumber\\
&  =\pi\left(  v\right)  \cdot\left(  w\rightharpoonup\left(  \gamma\left(
T\right)  \right)  \right)  +v\rightharpoonup\left(  w\rightharpoonup\left(
\gamma\left(  T\right)  \right)  \right)  . \label{pf.l37.gamma(J).vw}%
\end{align}


The definition of $\varrho_{T}^{\prime}$ yields%
\begin{align}
\varrho_{T}^{\prime}\left(  w\cdot v\right)   &  =\gamma\left(  w\cdot v\cdot
T\right)  =w\rightharpoonup\left(  \underbrace{\gamma\left(  v\cdot T\right)
}_{=\pi\left(  v\right)  \cdot\gamma\left(  T\right)  +v\rightharpoonup\left(
\gamma\left(  T\right)  \right)  }\right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{prop.gamma.rec.h}), applied to
}u=w\text{ and }U=v\cdot T\right) \nonumber\\
&  =w\rightharpoonup\left(  \pi\left(  v\right)  \cdot\gamma\left(  T\right)
+v\rightharpoonup\left(  \gamma\left(  T\right)  \right)  \right) \nonumber\\
&  =\underbrace{w\rightharpoonup\left(  \pi\left(  v\right)  \cdot
\gamma\left(  T\right)  \right)  }_{\substack{=\left(  w\rightharpoonup\left(
\pi\left(  v\right)  \right)  \right)  \cdot\gamma\left(  T\right)
+\pi\left(  v\right)  \cdot\left(  w\rightharpoonup\left(  \gamma\left(
T\right)  \right)  \right)  \\\text{(by (\ref{defs.semi.g-alg.cond}) (applied
to }\otimes\mathfrak{n}\text{, }w\text{, }\pi\left(  v\right)  \text{ and
}\gamma\left(  T\right)  \\\text{instead of }A\text{, }a\text{, }u\text{ and
}v\text{), because }\otimes\mathfrak{n}\text{ is a }\left(  \mathfrak{g}%
,\mathfrak{h}\right)  \text{-semialgebra)}}}+w\rightharpoonup\left(
v\rightharpoonup\left(  \gamma\left(  T\right)  \right)  \right) \nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Lie action on }\otimes
\mathfrak{n}\text{ is }k\text{-linear}\right) \nonumber\\
&  =\left(  w\rightharpoonup\left(  \pi\left(  v\right)  \right)  \right)
\cdot\gamma\left(  T\right)  +\pi\left(  v\right)  \cdot\left(
w\rightharpoonup\left(  \gamma\left(  T\right)  \right)  \right)
+w\rightharpoonup\left(  v\rightharpoonup\left(  \gamma\left(  T\right)
\right)  \right)  . \label{pf.l37.gamma(J).wv}%
\end{align}
Finally, the definition of $\varrho_{T}^{\prime}$ yields%
\begin{align*}
\varrho_{T}^{\prime}\left(  \left[  v,w\right]  \right)   &  =\gamma\left(
\underbrace{\left[  v,w\right]  }_{\substack{=-\left[  w,v\right]
\\\text{(due to (\ref{[]-3}))}}}\cdot T\right)  =\gamma\left(  -\left[
w,v\right]  \cdot T\right) \\
&  =-\underbrace{\gamma\left(  \left[  w,v\right]  \cdot T\right)
}_{\substack{=\pi\left(  \left[  w,v\right]  \right)  \cdot\gamma\left(
T\right)  +\left[  w,v\right]  \rightharpoonup\left(  \gamma\left(  T\right)
\right)  \\\text{(according to (\ref{prop.gamma.rec}), applied to}\\u=\left[
w,v\right]  \text{ and }U=T\text{)}}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\gamma\text{ is }k\text{-linear}\right) \\
&  =-\left(  \pi\left(  \left[  w,v\right]  \right)  \cdot\gamma\left(
T\right)  +\left[  w,v\right]  \rightharpoonup\left(  \gamma\left(  T\right)
\right)  \right)
\end{align*}
Since%
\[
\pi\left(  \underbrace{\left[  w,v\right]  }_{\substack{=w\rightharpoonup
v\text{ (since (\ref{prop.g-is-g-module.form}) (applied to }v\text{ and
}w\\\text{instead of }w\text{ and }v\text{) yields }w\rightharpoonup v=\left[
w,v\right]  \text{)}}}\right)  =\pi\left(  w\rightharpoonup v\right)
=w\rightharpoonup\left(  \pi\left(  v\right)  \right)
\]
(since $\pi$ is an $\mathfrak{h}$-module map, while $w\in\mathfrak{h}$) and%
\begin{align*}
\left[  w,v\right]  \rightharpoonup\left(  \gamma\left(  T\right)  \right)
&  =w\rightharpoonup\left(  v\rightharpoonup\left(  \gamma\left(  T\right)
\right)  \right)  -v\rightharpoonup\left(  w\rightharpoonup\left(
\gamma\left(  T\right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.l37.gamma(J).semimod}),
applied to }w\text{, }v\text{ and }\gamma\left(  T\right)  \text{ instead of
}a\text{, }b\text{ and }v\right)  ,
\end{align*}
this rewrites as%
\begin{align}
\varrho_{T}^{\prime}\left(  \left[  v,w\right]  \right)   &  =-\left(
\underbrace{\pi\left(  \left[  w,v\right]  \right)  }_{=w\rightharpoonup
\left(  \pi\left(  v\right)  \right)  }\cdot\gamma\left(  T\right)
+\underbrace{\left[  w,v\right]  \rightharpoonup\left(  \gamma\left(
T\right)  \right)  }_{=w\rightharpoonup\left(  v\rightharpoonup\left(
\gamma\left(  T\right)  \right)  \right)  -v\rightharpoonup\left(
w\rightharpoonup\left(  \gamma\left(  T\right)  \right)  \right)  }\right)
\nonumber\\
&  =-\left(  \left(  w\rightharpoonup\left(  \pi\left(  v\right)  \right)
\right)  \cdot\gamma\left(  T\right)  +w\rightharpoonup\left(
v\rightharpoonup\left(  \gamma\left(  T\right)  \right)  \right)
-v\rightharpoonup\left(  w\rightharpoonup\left(  \gamma\left(  T\right)
\right)  \right)  \right)  . \label{pf.l37.gamma(J).[v,w]}%
\end{align}


Now, $s=v\cdot w-w\cdot v-\left[  v,w\right]  $ yields%
\begin{align*}
\varrho_{T}^{\prime}\left(  s\right)   &  =\varrho_{T}^{\prime}\left(  v\cdot
w-w\cdot v-\left[  v,w\right]  \right) \\
&  =\underbrace{\varrho_{T}^{\prime}\left(  v\cdot w\right)  }_{\substack{=\pi
\left(  v\right)  \cdot\left(  w\rightharpoonup\left(  \gamma\left(  T\right)
\right)  \right)  +v\rightharpoonup\left(  w\rightharpoonup\left(
\gamma\left(  T\right)  \right)  \right)  \\\text{(due to
(\ref{pf.l37.gamma(J).vw}))}}}-\underbrace{\varrho_{T}^{\prime}\left(  w\cdot
v\right)  }_{\substack{=\left(  w\rightharpoonup\left(  \pi\left(  v\right)
\right)  \right)  \cdot\gamma\left(  T\right)  +\pi\left(  v\right)
\cdot\left(  w\rightharpoonup\left(  \gamma\left(  T\right)  \right)  \right)
+w\rightharpoonup\left(  v\rightharpoonup\left(  \gamma\left(  T\right)
\right)  \right)  \\\text{(due to (\ref{pf.l37.gamma(J).wv}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ -\underbrace{\varrho_{T}^{\prime}\left(  \left[
v,w\right]  \right)  }_{\substack{=-\left(  \left(  w\rightharpoonup\left(
\pi\left(  v\right)  \right)  \right)  \cdot\gamma\left(  T\right)
+w\rightharpoonup\left(  v\rightharpoonup\left(  \gamma\left(  T\right)
\right)  \right)  -v\rightharpoonup\left(  w\rightharpoonup\left(
\gamma\left(  T\right)  \right)  \right)  \right)  \\\text{(due to
(\ref{pf.l37.gamma(J).[v,w]}))}}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\varrho_{T}^{\prime}\text{ is
}k\text{-linear}\right) \\
&  =\left(  \pi\left(  v\right)  \cdot\left(  w\rightharpoonup\left(
\gamma\left(  T\right)  \right)  \right)  +v\rightharpoonup\left(
w\rightharpoonup\left(  \gamma\left(  T\right)  \right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  \left(  w\rightharpoonup\left(  \pi\left(
v\right)  \right)  \right)  \cdot\gamma\left(  T\right)  +\pi\left(  v\right)
\cdot\left(  w\rightharpoonup\left(  \gamma\left(  T\right)  \right)  \right)
+w\rightharpoonup\left(  v\rightharpoonup\left(  \gamma\left(  T\right)
\right)  \right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  -\left(  \left(  w\rightharpoonup\left(
\pi\left(  v\right)  \right)  \right)  \cdot\gamma\left(  T\right)
+w\rightharpoonup\left(  v\rightharpoonup\left(  \gamma\left(  T\right)
\right)  \right)  -v\rightharpoonup\left(  w\rightharpoonup\left(
\gamma\left(  T\right)  \right)  \right)  \right)  \right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{because all terms in the sum cancel
out}\right)  ,
\end{align*}
so that $s\in\operatorname*{Ker}\varrho_{T}^{\prime}$.

We have thus proven that every $s\in S_{0}$ satisfies $s\in\operatorname*{Ker}%
\varrho_{T}^{\prime}$. Thus, $S_{0}\subseteq\operatorname*{Ker}\varrho
_{T}^{\prime}$. Since $\operatorname*{Ker}\varrho_{T}^{\prime}$ is a
$k$-module, Proposition \ref{prop.<>} \textbf{(a)} (applied to $\otimes
\mathfrak{g}$, $S_{0}$ and $\operatorname*{Ker}\varrho_{T}^{\prime}$ instead
of $M$, $S$ and $Q$) now yields $\left\langle S_{0}\right\rangle
\subseteq\operatorname*{Ker}\varrho_{T}^{\prime}$. Since $\left\langle
S_{0}\right\rangle =J_{0}$, this rewrites as $J_{0}\subseteq
\operatorname*{Ker}\varrho_{T}^{\prime} $. Thus, $\varrho_{T}^{\prime}\left(
J_{0}\right)  =0$. In other words, every $j\in J_{0}$ satisfies $\varrho
_{T}^{\prime}\left(  j\right)  =0$. Since $\varrho_{T}^{\prime}\left(
j\right)  =\gamma\left(  j\cdot T\right)  $ (by the definition of $\varrho
_{T}^{\prime}$), this rewrites as $\gamma\left(  j\cdot T\right)  =0$, and
thus $j\cdot T\in\operatorname*{Ker}\gamma$ for every $j\in J_{0}$.

We have thus shown that every $j\in J_{0}$ and $T\in\otimes\mathfrak{g}$
satisfy $j\cdot T\in\operatorname*{Ker}\gamma$. In other words, every $\left(
j,T\right)  \in J_{0}\times\left(  \otimes\mathfrak{g}\right)  $ satisfies
$j\cdot T\in\operatorname*{Ker}\gamma$. In other words, $\left\{  j\cdot
T\ \mid\ \left(  j,T\right)  \in J_{0}\times\left(  \otimes\mathfrak{g}%
\right)  \right\}  \subseteq\operatorname*{Ker}\gamma$.

Now, applying Proposition \ref{prop.<>} \textbf{(a)} to $\otimes\mathfrak{g}
$, $\left\{  j\cdot T\ \mid\ \left(  j,T\right)  \in J_{0}\times\left(
\otimes\mathfrak{g}\right)  \right\}  $ and $\operatorname*{Ker}\gamma$
instead of $M$, $S$ and $Q$, we see that $\left\langle \left\{  j\cdot
T\ \mid\ \left(  j,T\right)  \in J_{0}\times\left(  \otimes\mathfrak{g}%
\right)  \right\}  \right\rangle \subseteq\operatorname*{Ker}\gamma$ (since
$\left\{  j\cdot T\ \mid\ \left(  j,T\right)  \in J_{0}\times\left(
\otimes\mathfrak{g}\right)  \right\}  \subseteq\operatorname*{Ker}\gamma$ and
since $\operatorname*{Ker}\gamma$ is a $k$-module). Now%
\[
J_{0}\cdot\left(  \otimes\mathfrak{g}\right)  =\left\langle j\cdot
T\ \mid\ \left(  j,T\right)  \in J_{0}\times\left(  \otimes\mathfrak{g}%
\right)  \right\rangle =\left\langle \left\{  j\cdot T\ \mid\ \left(
j,T\right)  \in J_{0}\times\left(  \otimes\mathfrak{g}\right)  \right\}
\right\rangle \subseteq\operatorname*{Ker}\gamma
\]
and thus $\gamma\left(  J_{0}\cdot\left(  \otimes\mathfrak{g}\right)  \right)
=0$.

Now, Lemma \ref{lem.l37.C} (applied to $C=J_{0}\cdot\left(  \otimes
\mathfrak{g}\right)  $) yields that $\gamma\left(  \left(  \otimes
\mathfrak{g}\right)  \cdot\left(  J_{0}\cdot\left(  \otimes\mathfrak{g}%
\right)  \right)  \right)  =0$ (because $\gamma\left(  J_{0}\cdot\left(
\otimes\mathfrak{g}\right)  \right)  =0$). Since $\left(  \otimes
\mathfrak{g}\right)  \cdot\left(  J_{0}\cdot\left(  \otimes\mathfrak{g}%
\right)  \right)  =\left(  \otimes\mathfrak{g}\right)  \cdot J_{0}\cdot\left(
\otimes\mathfrak{g}\right)  =J$ (according to Proposition \ref{prop.l25a}
\textbf{(b)}), this rewrites as $\gamma\left(  J\right)  =0$. Thus,
Proposition \ref{prop.l37.gamma(J)} is proven.
\end{verlong}

\textit{Proof of Proposition \ref{prop.l37.gamma(xgh)}.} Every $u\in
\mathfrak{h}$ satisfies%
\begin{align*}
\gamma\left(  u\right)   &  =\pi\left(  u\right)  \ \ \ \ \ \ \ \ \ \ \left(
\text{according to (\ref{rmk.gamma-1}), since }u\in\mathfrak{h}\subseteq
\mathfrak{g}\right) \\
&  =0\ \ \ \ \ \ \ \ \ \ \left(  \text{since }u\in\mathfrak{h}\text{, while
}\pi\text{ is a projection with kernel }\mathfrak{h}\right)  .
\end{align*}
Thus, $\gamma\left(  \mathfrak{h}\right)  =0$. We can thus follow from Lemma
\ref{lem.l37.C} (applied to $C=\mathfrak{h}$) that $\gamma\left(  \left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  =0$ (because
$\gamma\left(  \mathfrak{h}\right)  =0$). Thus, Proposition
\ref{prop.l37.gamma(xgh)} is proven.

\subsection{The homomorphism $\overline{\gamma}$}

We now formulate our further procedure:

\begin{proposition}
\label{prop.l37.Ker-gamma}\textbf{(a)} The $\mathfrak{h}$-module homomorphism
$\gamma:\otimes\mathfrak{g}\rightarrow\otimes\mathfrak{n}$ satisfies
$J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\subseteq
\operatorname*{Ker}\gamma$. Thus, $\gamma$ induces an $\mathfrak{h}$-module
homomorphism $\overline{\gamma}:\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\rightarrow\otimes\mathfrak{n}$ which satisfies $\gamma=\overline{\gamma}%
\circ\zeta$ (where $\zeta$ denotes the canonical projection $\otimes
\mathfrak{g}\rightarrow\left(  \otimes\mathfrak{g}\right)  \diagup\left(
J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ as in
Theorem \ref{thm.l25}).\newline\textbf{(b)} The homomorphism $\overline
{\gamma}$ respects the filtration. Here, the filtration on $\left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  $ is given by $\left(  F_{n}\right)
_{n\geq0}$, and the filtration on $\otimes\mathfrak{n}$ is given by $\left(
\mathfrak{n}^{\otimes\leq n}\right)  _{n\geq0}$.\newline\textbf{(c)} The
homomorphism $\overline{\gamma}$ is an $\mathfrak{h}$-module isomorphism, and
its inverse $\overline{\gamma}^{-1}$ also respects the filtration.
\end{proposition}

We notice that the first two of the three parts of this proposition are trivial:

\textit{Proof of Proposition \ref{prop.l37.Ker-gamma} \textbf{(a)}.} Since
$\gamma$ is $k$-linear, it is clear that $\operatorname*{Ker}\gamma$ is a $k$-module.

We know from Proposition \ref{prop.l37.gamma.h} that $\gamma$ is an
$\mathfrak{h}$-module homomorphism. Further, $J\subseteq\operatorname*{Ker}%
\gamma$ (due to Proposition \ref{prop.l37.gamma(J)}) and $\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\subseteq\operatorname*{Ker}%
\gamma$ (due to Proposition \ref{prop.l37.gamma(xgh)}). Thus, $\underbrace{J}%
_{\subseteq\operatorname*{Ker}\gamma}+\underbrace{\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}}_{\subseteq\operatorname*{Ker}\gamma
}\subseteq\operatorname*{Ker}\gamma+\operatorname*{Ker}\gamma\subseteq
\operatorname*{Ker}\gamma$ (since $\operatorname*{Ker}\gamma$ is a
$k$-module). Thus, by the homomorphism theorem, we see that $\gamma$ induces
an $\mathfrak{h}$-module homomorphism $\overline{\gamma}:\left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  \rightarrow\otimes\mathfrak{n}$ which
satisfies $\gamma=\overline{\gamma}\circ\zeta$ (where $\zeta$ denotes the
canonical projection $\otimes\mathfrak{g}\rightarrow\left(  \otimes
\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  $ as in Theorem \ref{thm.l25}). This proves
Proposition \ref{prop.l37.Ker-gamma} \textbf{(a)}.

\textit{Proof of Proposition \ref{prop.l37.Ker-gamma} \textbf{(b)}.} Let
$n\in\mathbb{N}$ be arbitrary. Since $F_{n}=\zeta\left(  \mathfrak{g}%
^{\otimes\leq n}\right)  $, we have%
\begin{align*}
\overline{\gamma}\left(  F_{n}\right)   &  =\overline{\gamma}\left(
\zeta\left(  \mathfrak{g}^{\otimes\leq n}\right)  \right)
=\underbrace{\left(  \overline{\gamma}\circ\zeta\right)  }_{=\gamma}\left(
\mathfrak{g}^{\otimes\leq n}\right)  =\gamma\left(  \mathfrak{g}^{\otimes\leq
n}\right)  \subseteq\mathfrak{n}^{\otimes\leq n}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{according to Proposition
\ref{prop.l37.gamma.filt}}\right)  .
\end{align*}
In other words, the homomorphism $\overline{\gamma}$ respects the filtration.
This proves Proposition \ref{prop.l37.Ker-gamma} \textbf{(b)}.

The rest of Section \ref{sect.split} will now be devoted to proving
Proposition \ref{prop.l37.Ker-gamma} \textbf{(c)}.

\subsection{Approximating $\gamma$ by $\otimes\pi$}

As an auxiliary result, we need the following analogue of Proposition
\ref{prop.l25.alpha.mod}:

\begin{proposition}
\label{prop.l37.gamma.mod}Consider the $k$-module homomorphism $\otimes
\pi:\otimes\mathfrak{g}\rightarrow\otimes\mathfrak{n}$ induced by the
$k$-module homomorphism $\pi:\mathfrak{g}\rightarrow\mathfrak{n}$. This
homomorphism $\otimes\pi$ respects the filtration and satisfies%
\begin{equation}
\left(  \gamma-\left(  \otimes\pi\right)  \right)  \left(  \mathfrak{g}%
^{\otimes\leq n}\right)  \subseteq\mathfrak{n}^{\otimes\leq\left(  n-1\right)
}\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}\text{.}
\label{prop.l37.gamma.mod.in}%
\end{equation}

\end{proposition}

\begin{vershort}
The proof of this proposition is too similar to that of Proposition
\ref{prop.l25.alpha.mod} to be given here.
\end{vershort}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.l37.gamma.mod}.} Proposition
\ref{prop.tensor.filt.(X)f} (applied to $\pi$, $\mathfrak{g}$ and
$\mathfrak{n}$ instead of $f$, $V$ and $W$) yields that $\otimes\pi$ respects
the filtration. Thus, the only thing that remains to be done for the proof of
Proposition \ref{prop.l37.gamma.mod} is proving the relation
(\ref{prop.l37.gamma.mod.in}).

We are going to prove (\ref{prop.l37.gamma.mod.in}) by induction over $n$:

\textit{Induction base:} Every $\lambda\in\mathfrak{g}^{\otimes\leq0}$
satisfies $\lambda\in k$ (since $\mathfrak{g}^{\otimes\leq0}=\bigoplus
\limits_{i=0}^{0}\mathfrak{g}^{\otimes i}=\mathfrak{g}^{\otimes0}=k$).
Therefore, every $\lambda\in\mathfrak{g}^{\otimes\leq0}$ satisfies
$\gamma\left(  \lambda\right)  =\lambda$ (according to (\ref{rmk.gamma-0}),
since $\lambda\in k$) and $\left(  \otimes\pi\right)  \left(  \lambda\right)
=\lambda$ (by the definition of $\otimes\pi$, since $\lambda\in k$), so that
$\left(  \gamma-\left(  \otimes\pi\right)  \right)  \left(  \lambda\right)
=\underbrace{\gamma\left(  \lambda\right)  }_{=\lambda}-\underbrace{\left(
\otimes\pi\right)  \left(  \lambda\right)  }_{=\lambda}=0$ and thus
$\lambda\in\operatorname*{Ker}\left(  \gamma-\left(  \otimes\pi\right)
\right)  $. We have thus shown that every $\lambda\in\mathfrak{g}^{\otimes
\leq0}$ satisfies $\lambda\in\operatorname*{Ker}\left(  \gamma-\left(
\otimes\pi\right)  \right)  $. Hence, $\mathfrak{g}^{\otimes\leq0}%
\subseteq\operatorname*{Ker}\left(  \gamma-\left(  \otimes\pi\right)  \right)
$, so that $\left(  \gamma-\left(  \otimes\pi\right)  \right)  \left(
\mathfrak{g}^{\otimes\leq0}\right)  =0\subseteq\mathfrak{n}^{\otimes
\leq\left(  0-1\right)  }$. In other words, (\ref{prop.l37.gamma.mod.in})
holds for $n=0$. This completes the induction base.

\textit{Induction step:} Let $p\in\mathbb{N}$. Assume that
(\ref{prop.l37.gamma.mod.in}) holds for $n=p$. We now must show that
(\ref{prop.l37.gamma.mod.in}) also holds for $n=p+1$.

Since (\ref{prop.l37.gamma.mod.in}) holds for $n=p$, we have $\left(
\gamma-\left(  \otimes\pi\right)  \right)  \left(  \mathfrak{g}^{\otimes\leq
p}\right)  \subseteq\mathfrak{n}^{\otimes\leq\left(  p-1\right)  }$.

We are now going to show that $\left(  \gamma-\left(  \otimes\pi\right)
\right)  \left(  \mathfrak{g}^{\otimes\left(  p+1\right)  }\right)
\subseteq\mathfrak{n}^{\otimes\leq p}$. This will quickly yield $\left(
\gamma-\left(  \otimes\pi\right)  \right)  \left(  \mathfrak{g}^{\otimes
\leq\left(  p+1\right)  }\right)  \subseteq\mathfrak{n}^{\otimes\leq p}$,
which will bring us to the end of the induction step.

Proposition \ref{prop.left-ind2} (applied to $p+1$ and $\mathfrak{g}$ instead
of $p$ and $V$) yields $\mathfrak{g}^{\otimes\left(  p+1\right)
}=\left\langle \mathfrak{g}_{\operatorname*{lind}}^{\otimes\left(  p+1\right)
}\right\rangle $, where we are using Convention \ref{conv.left-ind2}.

Note that $\gamma-\left(  \otimes\pi\right)  $ is a $k$-linear map (since
$\gamma$ and $\otimes\pi$ are $k$-linear), and thus $\operatorname*{Ker}%
\left(  \gamma-\left(  \otimes\pi\right)  \right)  $ is a $k$-module.

Now we are going to prove that $\mathfrak{g}_{\operatorname*{lind}}%
^{\otimes\left(  p+1\right)  }\subseteq\left(  \gamma-\left(  \otimes
\pi\right)  \right)  ^{-1}\left(  \mathfrak{n}^{\otimes\leq p}\right)  $.
Indeed, let $V\in\mathfrak{g}_{\operatorname*{lind}}^{\otimes\left(
p+1\right)  }$ be arbitrary. Then, $V$ is a left-induced tensor in
$\mathfrak{g}^{\otimes\left(  p+1\right)  }$ (since $V\in\mathfrak{g}%
_{\operatorname*{lind}}^{\otimes\left(  p+1\right)  }$), and thus there exist
$u\in\mathfrak{g}$ and $U\in\mathfrak{g}^{\otimes p}$ such that $V=u\otimes
U$. Consider these $u$ and $U$. Since $u\in\mathfrak{g}=\mathfrak{g}%
^{\otimes1}$ and $U\in\mathfrak{g}^{\otimes p}$, we have $u\cdot U=u\otimes U$
(due to (\ref{*=(X)}), applied to $u$, $U$, $1$ and $p$ instead of $a$, $b$,
$n$ and $m$).

Since $U\in\mathfrak{g}^{\otimes p}\subseteq\mathfrak{g}^{\otimes\leq p}$ and
since $\gamma\left(  \mathfrak{g}^{\otimes\leq p}\right)  \subseteq
\mathfrak{n}^{\otimes\leq p}$ (because the map $\gamma$ respects the
filtration), we have $\gamma\left(  \underbrace{U}_{\in\mathfrak{g}%
^{\otimes\leq p}}\right)  \in\gamma\left(  \mathfrak{g}^{\otimes\leq
p}\right)  \subseteq\mathfrak{n}^{\otimes\leq p}$. Thus, $u\rightharpoonup
\left(  \gamma\left(  U\right)  \right)  \in\mathfrak{n}^{\otimes\leq p}$
(since $\mathfrak{n}^{\otimes\leq p}$ is a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule (which is because $\mathfrak{n}$ is a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule)). In other words,
$u\rightharpoonup\left(  \gamma\left(  U\right)  \right)  \equiv
0\operatorname{mod}\mathfrak{n}^{\otimes\leq p}$.

On the other hand, $U\in\mathfrak{g}^{\otimes p}\subseteq\mathfrak{g}%
^{\otimes\leq p}$ yields $\left(  \gamma-\left(  \otimes\pi\right)  \right)
\left(  U\right)  \in\left(  \gamma-\left(  \otimes\pi\right)  \right)
\left(  \mathfrak{g}^{\otimes\leq p}\right)  \subseteq\mathfrak{n}%
^{\otimes\leq\left(  p-1\right)  }$. Since $\left(  \gamma-\left(  \otimes
\pi\right)  \right)  \left(  U\right)  =\gamma\left(  U\right)  -\left(
\otimes\pi\right)  \left(  U\right)  $, this rewrites as $\gamma\left(
U\right)  -\left(  \otimes\pi\right)  \left(  U\right)  \in\mathfrak{n}%
^{\otimes\leq\left(  p-1\right)  }$. On the other hand, $\otimes\pi$ is a
$k$-algebra homomorphism, so that%
\[
\left(  \otimes\pi\right)  \left(  u\cdot U\right)  =\underbrace{\left(
\otimes\pi\right)  \left(  u\right)  }_{\substack{=\pi\left(  u\right)  \text{
(by the definition}\\\text{of }\otimes\pi\text{, since }u\in\mathfrak{g}%
=\mathfrak{g}^{\otimes1}\text{)}}}\cdot\left(  \otimes\pi\right)  \left(
U\right)  =\pi\left(  u\right)  \cdot\left(  \otimes\pi\right)  \left(
U\right)  .
\]
Now,%
\begin{align*}
&  \pi\left(  u\right)  \cdot\gamma\left(  U\right)  -\left(  \otimes
\pi\right)  \left(  \underbrace{V}_{=u\otimes U=u\cdot U}\right) \\
&  =\pi\left(  u\right)  \cdot\gamma\left(  U\right)  -\underbrace{\left(
\otimes\pi\right)  \left(  u\cdot U\right)  }_{=\pi\left(  u\right)
\cdot\left(  \otimes\pi\right)  \left(  U\right)  }=\pi\left(  u\right)
\cdot\gamma\left(  U\right)  -\pi\left(  u\right)  \cdot\left(  \otimes
\pi\right)  \left(  U\right) \\
&  =\underbrace{\pi\left(  u\right)  }_{\in\mathfrak{n}=\mathfrak{n}%
^{\otimes1}\subseteq\mathfrak{n}^{\otimes\leq1}}\cdot\underbrace{\left(
\gamma\left(  U\right)  -\left(  \otimes\pi\right)  \left(  U\right)  \right)
}_{\in\mathfrak{n}^{\otimes\leq\left(  p-1\right)  }}\in\mathfrak{n}%
^{\otimes\leq1}\cdot\mathfrak{n}^{\otimes\leq\left(  p-1\right)  }\\
&  \subseteq\mathfrak{n}^{\otimes\leq\left(  1+\left(  p-1\right)  \right)
}\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{according to Proposition \ref{prop.tensor.filt-is-alg-filt}
\textbf{(b)}, applied}\\
\text{to }\mathfrak{n}\text{, }1\text{ and }p-1\text{ instead of }V\text{,
}n\text{ and }m
\end{array}
\right) \\
&  =\mathfrak{n}^{\otimes\leq p}.
\end{align*}
In other words, $\pi\left(  u\right)  \cdot\gamma\left(  U\right)
\equiv\left(  \otimes\pi\right)  \left(  V\right)  \operatorname{mod}%
\mathfrak{n}^{\otimes\leq p}$.

Now, $V=u\otimes U=u\cdot U$ yields%
\begin{align*}
\gamma\left(  V\right)   &  =\gamma\left(  u\cdot U\right)  =\underbrace{\pi
\left(  u\right)  \cdot\gamma\left(  U\right)  }_{\equiv\left(  \otimes
\pi\right)  \left(  V\right)  \operatorname{mod}\mathfrak{n}^{\otimes\leq p}%
}+\underbrace{u\rightharpoonup\left(  \gamma\left(  U\right)  \right)
}_{\equiv0\operatorname{mod}\mathfrak{n}^{\otimes\leq p}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{prop.gamma.rec})}\right) \\
&  \equiv\left(  \otimes\pi\right)  \left(  V\right)  +0=\left(  \otimes
\pi\right)  \left(  V\right)  \operatorname{mod}\mathfrak{n}^{\otimes\leq p}.
\end{align*}
In other words, $\gamma\left(  V\right)  -\left(  \otimes\pi\right)  \left(
V\right)  \in\mathfrak{n}^{\otimes\leq p}$. Since $\gamma\left(  V\right)
-\left(  \otimes\pi\right)  \left(  V\right)  =\left(  \gamma-\left(
\otimes\pi\right)  \right)  \left(  V\right)  $, this rewrites as $\left(
\gamma-\left(  \otimes\pi\right)  \right)  \left(  V\right)  \in
\mathfrak{n}^{\otimes\leq p}$. In other words, $V\in\left(  \gamma-\left(
\otimes\pi\right)  \right)  ^{-1}\left(  \mathfrak{n}^{\otimes\leq p}\right)
$.

So we have proven that every $V\in\mathfrak{g}_{\operatorname*{lind}}%
^{\otimes\left(  p+1\right)  }$ satisfies $V\in\left(  \gamma-\left(
\otimes\pi\right)  \right)  ^{-1}\left(  \mathfrak{n}^{\otimes\leq p}\right)
$. In other words, $\mathfrak{g}_{\operatorname*{lind}}^{\otimes\left(
p+1\right)  }\subseteq\left(  \gamma-\left(  \otimes\pi\right)  \right)
^{-1}\left(  \mathfrak{n}^{\otimes\leq p}\right)  $. Therefore, Proposition
\ref{prop.<>} \textbf{(a)} (applied to $\mathfrak{g}^{\otimes\left(
p+1\right)  }$, $\mathfrak{g}_{\operatorname*{lind}}^{\otimes\left(
p+1\right)  }$ and $\left(  \gamma-\left(  \otimes\pi\right)  \right)
^{-1}\left(  \mathfrak{n}^{\otimes\leq p}\right)  $ instead of $M$, $S$ and
$Q$) yields $\left\langle \mathfrak{g}_{\operatorname*{lind}}^{\otimes\left(
p+1\right)  }\right\rangle \subseteq\left(  \gamma-\left(  \otimes\pi\right)
\right)  ^{-1}\left(  \mathfrak{n}^{\otimes\leq p}\right)  $ (since $\left(
\gamma-\left(  \otimes\pi\right)  \right)  ^{-1}\left(  \mathfrak{n}%
^{\otimes\leq p}\right)  $ is a $k$-module). Altogether we now have
$\mathfrak{g}^{\otimes\left(  p+1\right)  }=\left\langle \mathfrak{g}%
_{\operatorname*{lind}}^{\otimes\left(  p+1\right)  }\right\rangle
\subseteq\left(  \gamma-\left(  \otimes\pi\right)  \right)  ^{-1}\left(
\mathfrak{n}^{\otimes\leq p}\right)  $, so that $\left(  \gamma-\left(
\otimes\pi\right)  \right)  \left(  \mathfrak{g}^{\otimes\left(  p+1\right)
}\right)  \subseteq\mathfrak{n}^{\otimes\leq p}$.

Now, the definition of $\mathfrak{g}^{\otimes\leq\left(  p+1\right)  }$ is
$\mathfrak{g}^{\otimes\leq\left(  p+1\right)  }=\bigoplus\limits_{i=0}%
^{p+1}\mathfrak{g}^{\otimes i}$, while the definition of $\mathfrak{g}%
^{\otimes\leq p}=\bigoplus\limits_{i=0}^{p}\mathfrak{g}^{\otimes i}$. Thus,%
\[
\mathfrak{g}^{\otimes\leq\left(  p+1\right)  }=\bigoplus\limits_{i=0}%
^{p+1}\mathfrak{g}^{\otimes i}=\underbrace{\left(  \bigoplus\limits_{i=0}%
^{p}\mathfrak{g}^{\otimes i}\right)  }_{=\mathfrak{g}^{\otimes\leq p}}%
\oplus\mathfrak{g}^{\otimes\left(  p+1\right)  }=\mathfrak{g}^{\otimes\leq
p}\oplus\mathfrak{g}^{\otimes\left(  p+1\right)  }=\mathfrak{g}^{\otimes\leq
p}+\mathfrak{g}^{\otimes\left(  p+1\right)  }%
\]
(since direct sums are sums). Thus,%
\begin{align*}
\left(  \gamma-\left(  \otimes\pi\right)  \right)  \left(  \mathfrak{g}%
^{\otimes\leq\left(  p+1\right)  }\right)   &  =\left(  \gamma-\left(
\otimes\pi\right)  \right)  \left(  \mathfrak{g}^{\otimes\leq p}%
+\mathfrak{g}^{\otimes\left(  p+1\right)  }\right) \\
&  =\underbrace{\left(  \gamma-\left(  \otimes\pi\right)  \right)  \left(
\mathfrak{g}^{\otimes\leq p}\right)  }_{\substack{\subseteq\mathfrak{n}%
^{\otimes\leq\left(  p-1\right)  }\subseteq\mathfrak{n}^{\otimes\leq
p}\\\text{(since }\left(  \mathfrak{n}^{\otimes\leq n}\right)  _{n\geq0}\text{
is a filtration)}}}+\underbrace{\left(  \gamma-\left(  \otimes\pi\right)
\right)  \left(  \mathfrak{g}^{\otimes\left(  p+1\right)  }\right)
}_{\subseteq\mathfrak{n}^{\otimes\leq p}}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\gamma-\left(  \otimes\pi\right)
\text{ is }k\text{-linear}\right) \\
&  \subseteq\mathfrak{n}^{\otimes\leq p}+\mathfrak{n}^{\otimes\leq
p}=\mathfrak{n}^{\otimes\leq p}\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\mathfrak{n}^{\otimes\leq p}\text{ is a }k\text{-module}\right) \\
&  =\mathfrak{n}^{\otimes\leq\left(  \left(  p+1\right)  -1\right)  }.
\end{align*}
In other words, (\ref{prop.l37.gamma.mod.in}) holds for $n=p+1$. This
completes the induction step. Thus, (\ref{prop.l37.gamma.mod.in}) is proven
for all $n\in\mathbb{N}$. In other words, Proposition \ref{prop.l37.gamma.mod}
is proven.
\end{verlong}

\subsection{Finishing the proof}

Proposition \ref{prop.l37.gamma.mod} provides us with the following consequence:

\begin{corollary}
\label{cor.l37.gamma.mod}Consider the maps $\overline{\varphi}:\left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  \rightarrow\otimes N$ and $\pi\mid
_{N}:N\rightarrow\mathfrak{n}$ defined in Proposition \ref{prop.l25.Ker-alpha}%
. Also, let $\eta=\otimes\left(  \pi\mid_{N}\right)  $. Then,%
\[
\left(  \overline{\gamma}-\eta\circ\overline{\varphi}\right)  \left(
F_{n}\right)  \subseteq\mathfrak{n}^{\otimes\leq\left(  n-1\right)
}\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}\text{.}%
\]

\end{corollary}

\textit{Proof of Corollary \ref{cor.l37.gamma.mod}.} Let $x\in F_{n}$. Since
$F_{n}=\zeta\left(  \mathfrak{g}^{\otimes\leq n}\right)  $, this becomes
$x\in\zeta\left(  \mathfrak{g}^{\otimes\leq n}\right)  $, so that there exists
some $y\in\mathfrak{g}^{\otimes\leq n}$ such that $x=\zeta\left(  y\right)  $.
Consider this $y$. It satisfies%
\begin{align*}
&  \left(  \overline{\gamma}-\eta\circ\overline{\varphi}\right)  \left(
\underbrace{x}_{=\zeta\left(  y\right)  }\right) \\
&  =\left(  \overline{\gamma}-\eta\circ\overline{\varphi}\right)  \left(
\zeta\left(  y\right)  \right)  =\underbrace{\overline{\gamma}\left(
\zeta\left(  y\right)  \right)  }_{=\left(  \overline{\gamma}\circ
\zeta\right)  \left(  y\right)  }-\underbrace{\left(  \eta\circ\overline
{\varphi}\right)  \left(  \zeta\left(  y\right)  \right)  }_{=\left(
\eta\circ\overline{\varphi}\circ\zeta\right)  \left(  y\right)  }\\
&  =\underbrace{\left(  \overline{\gamma}\circ\zeta\right)  }_{=\gamma}\left(
y\right)  -\left(  \eta\circ\underbrace{\overline{\varphi}\circ\zeta
}_{=\varphi}\right)  \left(  y\right)  =\gamma\left(  y\right)  -\left(
\eta\circ\varphi\right)  \left(  y\right) \\
&  =\underbrace{\left(  \gamma\left(  y\right)  -\left(  \otimes\pi\right)
\left(  y\right)  \right)  }_{=\left(  \gamma-\left(  \otimes\pi\right)
\right)  \left(  y\right)  }-\underbrace{\left(  \left(  \eta\circ
\varphi\right)  \left(  y\right)  -\left(  \otimes\pi\right)  \left(
y\right)  \right)  }_{=\left(  \eta\circ\varphi-\left(  \otimes\pi\right)
\right)  \left(  y\right)  }\\
&  =\left(  \gamma-\left(  \otimes\pi\right)  \right)  \left(  \underbrace{y}%
_{\in\mathfrak{g}^{\otimes\leq n}}\right)  -\left(  \eta\circ\varphi-\left(
\otimes\pi\right)  \right)  \left(  \underbrace{y}_{\in\mathfrak{g}%
^{\otimes\leq n}}\right) \\
&  \in\underbrace{\left(  \gamma-\left(  \otimes\pi\right)  \right)  \left(
\mathfrak{g}^{\otimes\leq n}\right)  }_{\substack{\subseteq\mathfrak{n}%
^{\otimes\leq\left(  n-1\right)  }\\\text{(according to
(\ref{prop.l37.gamma.mod.in}))}}}-\underbrace{\left(  \eta\circ\varphi-\left(
\otimes\pi\right)  \right)  \left(  \mathfrak{g}^{\otimes\leq n}\right)
}_{\substack{\subseteq\mathfrak{n}^{\otimes\leq\left(  n-1\right)
}\\\text{(according to (\ref{pf.l25.ha-Xp.lowers}), applied}\\\text{to
}n\text{ instead of }p\text{)}}}\\
&  \subseteq\mathfrak{n}^{\otimes\leq\left(  n-1\right)  }+\mathfrak{n}%
^{\otimes\leq\left(  n-1\right)  }\subseteq\mathfrak{n}^{\otimes\leq\left(
n-1\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\mathfrak{n}%
^{\otimes\leq\left(  n-1\right)  }\text{ is a }k\text{-module}\right)  .
\end{align*}


We have thus shown that $\left(  \overline{\gamma}-\eta\circ\overline{\varphi
}\right)  \left(  x\right)  \in\mathfrak{n}^{\otimes\leq\left(  n-1\right)  }$
for every $x\in F_{n}$. In other words, $\left(  \overline{\gamma}-\eta
\circ\overline{\varphi}\right)  \left(  F_{n}\right)  \subseteq\mathfrak{n}%
^{\otimes\leq\left(  n-1\right)  }$. This proves Corollary
\ref{cor.l37.gamma.mod}.

A further corollary from this corollary:

\begin{corollary}
\label{cor.l37.gamma.mod.cor2}Consider the maps $\overline{\varphi}:\left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  \rightarrow\otimes N$ and $\pi\mid
_{N}:N\rightarrow\mathfrak{n}$ defined in Proposition \ref{prop.l25.Ker-alpha}%
. Also, let $\eta=\otimes\left(  \pi\mid_{N}\right)  $. Then, $\eta
\circ\overline{\varphi}$ is a $k$-module isomorphism and satisfies%
\[
\left(  \operatorname*{id}-\overline{\gamma}\circ\left(  \eta\circ
\overline{\varphi}\right)  ^{-1}\right)  \left(  \mathfrak{n}^{\otimes\leq
n}\right)  \subseteq\mathfrak{n}^{\otimes\leq\left(  n-1\right)
}\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}\text{.}%
\]

\end{corollary}

\textit{Proof of Corollary \ref{cor.l37.gamma.mod.cor2}.} Since $\eta$ and
$\overline{\varphi}$ are $k$-module isomorphisms, their composition $\eta
\circ\overline{\varphi}$ must also be a $k$-module isomorphism.

Besides, $\eta=\otimes\left(  \pi\mid_{N}\right)  $, so that $\eta
^{-1}=\left(  \otimes\left(  \pi\mid_{N}\right)  \right)  ^{-1}=\otimes\left(
\left(  \pi\mid_{N}\right)  ^{-1}\right)  $, and thus $\eta^{-1}$ respects the
filtration (by Proposition \ref{prop.tensor.filt.(X)f}). On the other hand,
$\overline{\varphi}^{-1}$ respects the filtration (according to Proposition
\ref{prop.l25.Ker-alpha} \textbf{(d)}). Thus, Proposition
\ref{prop.gr.functor} \textbf{(b)} (applied to $\otimes\mathfrak{n}$, $\left(
\mathfrak{n}^{\otimes\leq n}\right)  _{n\geq0}$, $\otimes N$, $\left(
N^{\otimes\leq n}\right)  _{n\geq0}$, $\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
$, $\left(  F_{n}\right)  _{n\geq0}$, $\eta^{-1}$ and $\overline{\varphi}%
^{-1}$ instead of $U$, $\left(  U_{n}\right)  _{n\geq0}$, $V$, $\left(
V_{n}\right)  _{n\geq0}$, $W$, $\left(  W_{n}\right)  _{n\geq0}$, $f$ and $g$)
yields that the composition $\overline{\varphi}^{-1}\circ\eta^{-1}$ also
respects the filtration. Since $\overline{\varphi}^{-1}\circ\eta^{-1}=\left(
\eta\circ\overline{\varphi}\right)  ^{-1}$, this rewrites as follows: The
homomorphism $\left(  \eta\circ\overline{\varphi}\right)  ^{-1}$ respects the
filtration. In other words, $\left(  \eta\circ\overline{\varphi}\right)
^{-1}\left(  \mathfrak{n}^{\otimes\leq n}\right)  \subseteq F_{n}$ for every
$n\in\mathbb{N}$.

Now, every $n\in\mathbb{N}$ satisfies%
\begin{align*}
&  \left(  \underbrace{\operatorname*{id}}_{=\left(  \eta\circ\overline
{\varphi}\right)  \circ\left(  \eta\circ\overline{\varphi}\right)  ^{-1}%
}-\overline{\gamma}\circ\left(  \eta\circ\overline{\varphi}\right)
^{-1}\right)  \left(  \mathfrak{n}^{\otimes\leq n}\right) \\
&  =\left(  \underbrace{\left(  \eta\circ\overline{\varphi}\right)
\circ\left(  \eta\circ\overline{\varphi}\right)  ^{-1}-\overline{\gamma}%
\circ\left(  \eta\circ\overline{\varphi}\right)  ^{-1}}_{=\left(  \eta
\circ\overline{\varphi}-\overline{\gamma}\right)  \circ\left(  \eta
\circ\overline{\varphi}\right)  ^{-1}}\right)  \left(  \mathfrak{n}%
^{\otimes\leq n}\right) \\
&  =\left(  \left(  \eta\circ\overline{\varphi}-\overline{\gamma}\right)
\circ\left(  \eta\circ\overline{\varphi}\right)  ^{-1}\right)  \left(
\mathfrak{n}^{\otimes\leq n}\right)  =\left(  \eta\circ\overline{\varphi
}-\overline{\gamma}\right)  \left(  \underbrace{\left(  \eta\circ
\overline{\varphi}\right)  ^{-1}\left(  \mathfrak{n}^{\otimes\leq n}\right)
}_{\subseteq F_{n}}\right) \\
&  \subseteq\underbrace{\left(  \eta\circ\overline{\varphi}-\overline{\gamma
}\right)  }_{=-\left(  \overline{\gamma}-\eta\circ\overline{\varphi}\right)
}\left(  F_{n}\right)  =-\underbrace{\left(  \overline{\gamma}-\eta
\circ\overline{\varphi}\right)  \left(  F_{n}\right)  }_{\subseteq
\mathfrak{n}^{\otimes\leq\left(  n-1\right)  }\text{ (by Corollary
\ref{cor.l37.gamma.mod})}}\\
&  \subseteq-\mathfrak{n}^{\otimes\leq\left(  n-1\right)  }\subseteq
\mathfrak{n}^{\otimes\leq\left(  n-1\right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\mathfrak{n}^{\otimes\leq\left(  n-1\right)  }\text{ is a
}k\text{-module}\right)  .
\end{align*}
This proves Corollary \ref{cor.l37.gamma.mod.cor2}.

Now finally we can come to the \textit{proof of Proposition
\ref{prop.l37.Ker-gamma}:}

We have already proven parts \textbf{(a)} and \textbf{(b)} of Proposition
\ref{prop.l37.Ker-gamma}. Now to part \textbf{(c)}:

In the proof of Corollary \ref{cor.l37.gamma.mod.cor2}, we showed that the map
$\eta\circ\overline{\varphi}$ respects the filtration.

We now know that:

\begin{itemize}
\item The filtration on $\otimes\mathfrak{n}$ is $\left(  \mathfrak{n}%
^{\otimes\leq n}\right)  _{n\geq0}$, and $\mathfrak{n}^{\otimes\leq\left(
-1\right)  }=0$.

\item The map $\left(  \eta\circ\overline{\varphi}\right)  ^{-1}$ is a
$k$-module isomorphism (because it has an inverse, namely $\eta\circ
\overline{\varphi}$).

\item The map $\left(  \eta\circ\overline{\varphi}\right)  ^{-1}$ respects the
filtration (this was proven in the proof of Corollary
\ref{cor.l37.gamma.mod.cor2}).

\item The map $\left(  \left(  \eta\circ\overline{\varphi}\right)
^{-1}\right)  ^{-1}$ respects the filtration (because $\left(  \left(
\eta\circ\overline{\varphi}\right)  ^{-1}\right)  ^{-1}=\eta\circ
\overline{\varphi}$, while we know that the map $\eta\circ\overline{\varphi}$
respects the filtration).

\item We have $\left(  \operatorname*{id}-\overline{\gamma}\circ\left(
\eta\circ\overline{\varphi}\right)  ^{-1}\right)  \left(  \mathfrak{n}%
^{\otimes\leq n}\right)  \subseteq\mathfrak{n}^{\otimes\leq\left(  n-1\right)
}$ for every $n\in\mathbb{N}$ (according to Corollary
\ref{cor.l37.gamma.mod.cor2}).
\end{itemize}

Thus, we can apply Corollary \ref{cor.filtr.iso.cor} to $\otimes\mathfrak{n}$,
$\left(  \mathfrak{n}^{\otimes\leq n}\right)  _{n\geq0}$, $\left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  $, $\left(  F_{n}\right)  _{n\geq0}$,
$\left(  \eta\circ\overline{\varphi}\right)  ^{-1}$ and $\overline{\gamma}$
instead of $V$, $\left(  V_{n}\right)  _{n\geq0}$, $W$, $\left(  W_{n}\right)
_{n\geq0}$, $f$ and $g$. This yields that the $k$-module homomorphism
$\overline{\gamma}$ is an isomorphism (according to Corollary
\ref{cor.filtr.iso.cor} \textbf{(a)}) and that each of the maps $\overline
{\gamma}$ and $\overline{\gamma}^{-1}$ respects the filtration (according to
Corollary \ref{cor.filtr.iso.cor} \textbf{(b)}).

We now know that $\overline{\gamma}$ is a $k$-module isomorphism, but we also
know that $\overline{\gamma}$ is an $\mathfrak{h}$-module homomorphism (due to
Proposition \ref{prop.l37.gamma.h}). Thus, $\overline{\gamma}$ is an
$\mathfrak{h}$-module isomorphism (by an application of Proposition
\ref{prop.liemodiso}). This proves Proposition \ref{prop.l37.Ker-gamma}
\textbf{(c)}, and thus completes the proof of Proposition
\ref{prop.l37.Ker-gamma}.

We can now easily obtain Theorem \ref{thm.l37ba}:

\textit{Proof of Theorem \ref{thm.l37ba}.} \textbf{(a)} Theorem
\ref{thm.l37ba} \textbf{(a)} is identic with Theorem \ref{thm.l25}
\textbf{(a)}. Thus, Theorem \ref{thm.l37ba} \textbf{(a)} is already proven (as
we have proven Theorem \ref{thm.l25} \textbf{(a)} in Section \ref{sect.L25}).

\textbf{(b)} We already know from Theorem \ref{thm.l25} \textbf{(b)} that
$\left(  F_{n}\right)  _{n\geq0}$ is an $\mathfrak{h}$-module filtration of
$\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $.

Now, let $n\in\mathbb{N}$. Since $\overline{\gamma}$ respects the filtration,
we have $\overline{\gamma}\left(  F_{n}\right)  \subseteq\mathfrak{n}%
^{\otimes\leq n}$. Since the homomorphism $\overline{\gamma}^{-1}$ respects
the filtration, we have $\overline{\gamma}^{-1}\left(  \mathfrak{n}%
^{\otimes\leq n}\right)  \subseteq F_{n}$. Thus, $F_{n}\supseteq
\overline{\gamma}^{-1}\left(  \mathfrak{n}^{\otimes\leq n}\right)  $, so that
$\overline{\gamma}\left(  F_{n}\right)  \supseteq\overline{\gamma}\left(
\overline{\gamma}^{-1}\left(  \mathfrak{n}^{\otimes\leq n}\right)  \right)
=\mathfrak{n}^{\otimes\leq n}$ (since $\overline{\gamma}$ is an isomorphism).
Combining this with $\overline{\gamma}\left(  F_{n}\right)  \subseteq
\mathfrak{n}^{\otimes\leq n}$, we obtain $\overline{\gamma}\left(
F_{n}\right)  =\mathfrak{n}^{\otimes\leq n}$. Since $\overline{\gamma}$ is an
$\mathfrak{h}$-module isomorphism, it thus follows that $\overline{\gamma}$
induces an $\mathfrak{h}$-module isomorphism $F_{n}\rightarrow\mathfrak{n}%
^{\otimes\leq n}$. Hence, $F_{n}\cong\mathfrak{n}^{\otimes\leq n}$ as
$\mathfrak{h}$-modules. This completes the proof of Theorem \ref{thm.l37ba}
\textbf{(b)}.

\textbf{(c)} The map $\overline{\gamma}$ is an $\mathfrak{h}$-module
isomorphism $\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \rightarrow
\otimes\mathfrak{n}$, and we know that for every $n\in\mathbb{N}$, the image
of $F_{n}$ under this isomorphism is $\mathfrak{n}^{\otimes\leq n}$ (because
$\overline{\gamma}\left(  F_{n}\right)  =\mathfrak{n}^{\otimes\leq n}$). Thus,
Theorem \ref{thm.l37ba} \textbf{(c)} must hold.

\begin{vershort}
\textbf{(d)} Theorem \ref{thm.l37ba} \textbf{(c)} shows that the
$\mathfrak{h}$-module $\left(  \otimes\mathfrak{g}\right)  \diagup\left(
J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ with its
filtration $\left(  F_{n}\right)  _{n\geq0}$ is isomorphic to the
$\mathfrak{h}$-module $\otimes\mathfrak{n}$ with its filtration $\left(
\mathfrak{n}^{\otimes\leq n}\right)  _{n\geq0}$. Thus, in order to prove that
the filtration $\left(  F_{n}\right)  _{n\geq0}$ is $\mathfrak{h}$-split, it
is enough to show that the filtration $\left(  \mathfrak{n}^{\otimes\leq
n}\right)  _{n\geq0}$ is $\mathfrak{h}$-split. But this is clear, since the
latter filtration comes from the grading $\left(  \mathfrak{n}^{\otimes
n}\right)  _{n\in\mathbb{N}}$ of the $\mathfrak{h}$-module $\otimes
\mathfrak{n}$. This proves Theorem \ref{thm.l37ba} \textbf{(d)}.
\end{vershort}

\begin{verlong}
\textbf{(d)} Since the map $\overline{\gamma}$ is an $\mathfrak{h}$-module
isomorphism, its inverse $\overline{\gamma}^{-1}$ must be an $\mathfrak{h}%
$-module isomorphism as well.

For every $n\in\mathbb{N}$, define an $\mathfrak{h}$-submodule $W_{n}$ of
$\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ by $W_{n}=\overline{\gamma
}^{-1}\left(  \mathfrak{n}^{\otimes n}\right)  $ (here, we use that
$\overline{\gamma}^{-1}\left(  \mathfrak{n}^{\otimes n}\right)  $ is an
$\mathfrak{h}$-submodule of $\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
$, because $\overline{\gamma}^{-1}$ is an $\mathfrak{h}$-module isomorphism).

We have $\otimes\mathfrak{n}=\bigoplus\limits_{n\in\mathbb{N}}\mathfrak{n}%
^{\otimes n}$ and thus%
\begin{align*}
\overline{\gamma}^{-1}\left(  \otimes\mathfrak{n}\right)   &  =\overline
{\gamma}^{-1}\left(  \bigoplus\limits_{n\in\mathbb{N}}\mathfrak{n}^{\otimes
n}\right)  =\bigoplus\limits_{n\in\mathbb{N}}\underbrace{\overline{\gamma
}^{-1}\left(  \mathfrak{n}^{\otimes n}\right)  }_{=W_{n}}%
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\overline{\gamma}^{-1}\text{ is an
}\mathfrak{h}\text{-module isomorphism}\right) \\
&  =\bigoplus\limits_{n\in\mathbb{N}}W_{n}.
\end{align*}
Now, since $\overline{\gamma}^{-1}$ is an isomorphism, we have $\left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  =\overline{\gamma}^{-1}\left(
\otimes\mathfrak{n}\right)  =\bigoplus\limits_{n\in\mathbb{N}}W_{n}$.

Every $n\in\mathbb{N}$ satisfies $\overline{\gamma}\left(  F_{n}\right)
=\mathfrak{n}^{\otimes\leq n}$ (as we saw during the proof of Theorem
\ref{thm.l37ba} \textbf{(b)}). Renaming $n$ as $p$ in this result, we obtain:
Every $p\in\mathbb{N}$ satisfies $\overline{\gamma}\left(  F_{p}\right)
=\mathfrak{n}^{\otimes\leq p}$. Now, every $p\in\mathbb{N}$ satisfies%
\begin{align*}
F_{p}  &  =\overline{\gamma}^{-1}\left(  \underbrace{\overline{\gamma}\left(
F_{p}\right)  }_{=\mathfrak{n}^{\otimes\leq p}}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\overline{\gamma}\text{ is an
isomorphism}\right) \\
&  =\overline{\gamma}^{-1}\left(  \underbrace{\mathfrak{n}^{\otimes\leq p}%
}_{=\bigoplus\limits_{n=0}^{p}\mathfrak{n}^{\otimes n}}\right)  =\overline
{\gamma}^{-1}\left(  \bigoplus\limits_{n=0}^{p}\mathfrak{n}^{\otimes n}\right)
\\
&  =\bigoplus\limits_{n=0}^{p}\underbrace{\overline{\gamma}^{-1}\left(
\mathfrak{n}^{\otimes n}\right)  }_{=W_{n}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\overline{\gamma}^{-1}\text{ is an }\mathfrak{h}\text{-module
isomorphism}\right) \\
&  =\bigoplus\limits_{n=0}^{p}W_{n}.
\end{align*}


Thus we have found a family $\left(  W_{n}\right)  _{n\in\mathbb{N}}$ of
$\mathfrak{h}$-submodules of $\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
$ such that $\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  =\bigoplus
\limits_{n\in\mathbb{N}}W_{n}$ and such that every $p\in\mathbb{N}$ satisfies
$F_{p}=\bigoplus\limits_{n=0}^{p}W_{n}$.

Now, Proposition \ref{prop.split.filtr-g} (applied to $\mathfrak{h}$, $\left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  $ and $\left(  F_{n}\right)  _{n\geq0}$
instead of $\mathfrak{g}$, $V$ and $\left(  V_{n}\right)  _{n\geq0}$) yields
that the filtration $\left(  F_{n}\right)  _{n\geq0}$ is $\mathfrak{h}$-split
if and only if there exists a family $\left(  W_{n}\right)  _{n\in\mathbb{N}}$
of $\mathfrak{h}$-submodules of $\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
$ such that $\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  =\bigoplus
\limits_{n\in\mathbb{N}}W_{n}$ and such that every $p\in\mathbb{N}$ satisfies
$F_{p}=\bigoplus\limits_{n=0}^{p}W_{n}$. Since we know that there exists a
family $\left(  W_{n}\right)  _{n\in\mathbb{N}}$ of $\mathfrak{h}$-submodules
of $\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ such that $\left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  =\bigoplus\limits_{n\in\mathbb{N}}W_{n}$
and such that every $p\in\mathbb{N}$ satisfies $F_{p}=\bigoplus\limits_{n=0}%
^{p}W_{n}$, we can thus conclude that the filtration $\left(  F_{n}\right)
_{n\geq0}$ is $\mathfrak{h}$-split. This proves Theorem \ref{thm.l37ba}
\textbf{(d)}.
\end{verlong}

Hence, the whole Theorem \ref{thm.l37ba} is proven.

\section{\label{sect.pbw}The Poincar\'{e}-Birkhoff-Witt theorem}

\subsection{The symmetric powers of a module}

We recall the definition of the $n$\textit{-th symmetric power} of a $k$-module:

\begin{definition}
\label{defs.symm-alg}Let $k$ be a commutative ring. Let $V$ be a $k$-module.
Let $n\in\mathbb{N}$.\newline Let $K_{n}\left(  V\right)  $ be the
$k$-submodule%
\[
\left\langle v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\sigma\left(
1\right)  }\otimes v_{\sigma\left(  2\right)  }\otimes...\otimes
v_{\sigma\left(  n\right)  }\ \mid\ \left(  \left(  v_{1},v_{2},...,v_{n}%
\right)  ,\sigma\right)  \in V^{n}\times S_{n}\right\rangle
\]
of the $k$-module $V^{\otimes n}$ (where we are using Convention
\ref{conv.<>}, and are denoting the $n$-th symmetric group by $S_{n}%
$).\newline The factor $k$-module $V^{\otimes n}\diagup K_{n}\left(  V\right)
$ is called the $n$\textit{-th symmetric power} of the $k$-module $V$ and will
be denoted by $\operatorname*{Sym}^{n}V$. We denote by $\operatorname*{sym}%
\nolimits_{V,n}$ the canonical projection $V^{\otimes n}\rightarrow V^{\otimes
n}\diagup K_{n}\left(  V\right)  =\operatorname{Sym}^{n}V$. Clearly, this map
$\operatorname*{sym}\nolimits_{V,n}$ is a surjective $k$-module homomorphism.
\end{definition}

Here is an alternative description of the module $K_{n}\left(  V\right)  $
from this definition:

\begin{proposition}
\label{rmk.symm-alg}Let $k$ be a commutative ring. Let $V$ be a $k$-module.
Let $n\in\mathbb{N}$.\newline Then,%
\[
K_{n}\left(  V\right)  =\sum_{i=1}^{n-1}\left\langle v_{1}\otimes v_{2}%
\otimes...\otimes v_{n}-v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau
_{i}\left(  2\right)  }\otimes...\otimes v_{\tau_{i}\left(  n\right)  }%
\ \mid\ \left(  v_{1},v_{2},...,v_{n}\right)  \in V^{n}\right\rangle ,
\]
where $\tau_{i}$ denotes the transposition $\left(  i,i+1\right)  \in S_{n}$.
\end{proposition}

\begin{vershort}
In fact, this proposition can be concluded from the definition of
$K_{n}\left(  V\right)  $ and the fact that the transpositions $\tau_{1}$,
$\tau_{2}$, $...$, $\tau_{n-1}$ generate the symmetric group $S_{n}$. The
details of this proof are classical and can be found in any serious treatise
on symmetric powers.
\end{vershort}

\begin{verlong}
\textit{Proof of Proposition \ref{rmk.symm-alg}.} Let $T$ denote the subset%
\[
\left\{  v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\sigma\left(  1\right)
}\otimes v_{\sigma\left(  2\right)  }\otimes...\otimes v_{\sigma\left(
n\right)  }\ \mid\ \left(  \left(  v_{1},v_{2},...,v_{n}\right)
,\sigma\right)  \in V^{n}\times S_{n}\right\}
\]
of $V^{\otimes n}$. Then,%
\begin{align*}
\left\langle T\right\rangle  &  =\left\langle \left\{  v_{1}\otimes
v_{2}\otimes...\otimes v_{n}-v_{\sigma\left(  1\right)  }\otimes
v_{\sigma\left(  2\right)  }\otimes...\otimes v_{\sigma\left(  n\right)
}\ \mid\ \left(  \left(  v_{1},v_{2},...,v_{n}\right)  ,\sigma\right)  \in
V^{n}\times S_{n}\right\}  \right\rangle \\
&  =\left\langle v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\sigma\left(
1\right)  }\otimes v_{\sigma\left(  2\right)  }\otimes...\otimes
v_{\sigma\left(  n\right)  }\ \mid\ \left(  \left(  v_{1},v_{2},...,v_{n}%
\right)  ,\sigma\right)  \in V^{n}\times S_{n}\right\rangle \\
&  =K_{n}\left(  V\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Definition
\ref{defs.symm-alg}}\right)  .
\end{align*}
On the other hand,%
\begin{align}
&  v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\sigma\left(  1\right)
}\otimes v_{\sigma\left(  2\right)  }\otimes...\otimes v_{\sigma\left(
n\right)  }\in T\label{pf.symm-alg.1}\\
&  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(  \left(  v_{1},v_{2}%
,...,v_{n}\right)  ,\sigma\right)  \in V^{n}\times S_{n}\nonumber
\end{align}
(since%
\[
T=\left\{  v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\sigma\left(
1\right)  }\otimes v_{\sigma\left(  2\right)  }\otimes...\otimes
v_{\sigma\left(  n\right)  }\ \mid\ \left(  \left(  v_{1},v_{2},...,v_{n}%
\right)  ,\sigma\right)  \in V^{n}\times S_{n}\right\}
\]
).

On the other hand, let $Z$ denote the $k$-submodule%
\[
\sum_{i=1}^{n-1}\left\langle v_{1}\otimes v_{2}\otimes...\otimes v_{n}%
-v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }%
\otimes...\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  v_{1}%
,v_{2},...,v_{n}\right)  \in V^{n}\right\rangle
\]
of $V^{\otimes n}$. Then,%
\begin{equation}
\left\langle v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\tau_{\mathbf{I}%
}\left(  1\right)  }\otimes v_{\tau_{\mathbf{I}}\left(  2\right)  }%
\otimes...\otimes v_{\tau_{\mathbf{I}}\left(  n\right)  }\ \mid\ \left(
v_{1},v_{2},...,v_{n}\right)  \in V^{n}\right\rangle \subseteq Z
\label{pf.symm-alg.2}%
\end{equation}
for every $\mathbf{I}\in\left\{  1,2,...,n-1\right\}  $. Now,%
\begin{align*}
&  \left\{  w_{1}\otimes w_{2}\otimes...\otimes w_{n}-w_{\tau_{\mathbf{I}%
}\left(  1\right)  }\otimes w_{\tau_{\mathbf{I}}\left(  2\right)  }%
\otimes...\otimes w_{\tau_{\mathbf{I}}\left(  n\right)  }\ \mid\ \left(
w_{1},w_{2},...,w_{n}\right)  \in V^{n}\right\} \\
&  =\left\{  v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\tau_{\mathbf{I}%
}\left(  1\right)  }\otimes v_{\tau_{\mathbf{I}}\left(  2\right)  }%
\otimes...\otimes v_{\tau_{\mathbf{I}}\left(  n\right)  }\ \mid\ \left(
v_{1},v_{2},...,v_{n}\right)  \in V^{n}\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we renamed }\left(  w_{1}%
,w_{2},...,w_{n}\right)  \text{ as }\left(  v_{1},v_{2},...,v_{n}\right)
\right) \\
&  \subseteq\left\langle \left\{  v_{1}\otimes v_{2}\otimes...\otimes
v_{n}-v_{\tau_{\mathbf{I}}\left(  1\right)  }\otimes v_{\tau_{\mathbf{I}%
}\left(  2\right)  }\otimes...\otimes v_{\tau_{\mathbf{I}}\left(  n\right)
}\ \mid\ \left(  v_{1},v_{2},...,v_{n}\right)  \in V^{n}\right\}
\right\rangle \\
&  =\left\langle v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\tau
_{\mathbf{I}}\left(  1\right)  }\otimes v_{\tau_{\mathbf{I}}\left(  2\right)
}\otimes...\otimes v_{\tau_{\mathbf{I}}\left(  n\right)  }\ \mid\ \left(
v_{1},v_{2},...,v_{n}\right)  \in V^{n}\right\rangle \subseteq Z
\end{align*}
for every $\mathbf{I}\in\left\{  1,2,...,n-1\right\}  $. Thus,%
\begin{align}
&  w_{1}\otimes w_{2}\otimes...\otimes w_{n}-w_{\tau_{\mathbf{I}}\left(
1\right)  }\otimes w_{\tau_{\mathbf{I}}\left(  2\right)  }\otimes...\otimes
w_{\tau_{\mathbf{I}}\left(  n\right)  }\in Z\nonumber\\
&  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(  w_{1},w_{2},...,w_{n}\right)
\in V^{n}\text{ and every }\mathbf{I}\in\left\{  1,2,...,n-1\right\}  .
\label{pf.symm-alg.3}%
\end{align}


We are now going to show that $Z=\left\langle T\right\rangle $.

First, let us prove that $Z\subseteq\left\langle T\right\rangle $. In fact,
every $i\in\left\{  1,2,...,n-1\right\}  $ satisfies%
\[
\left\{  v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\tau_{i}\left(
1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }\otimes...\otimes v_{\tau
_{i}\left(  n\right)  }\ \mid\ \left(  v_{1},v_{2},...,v_{n}\right)  \in
V^{n}\right\}  \subseteq\left\langle T\right\rangle
\]
(since every $\left(  v_{1},v_{2},...,v_{n}\right)  \in V^{n}$ satisfies%
\begin{align*}
&  v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\tau_{i}\left(  1\right)
}\otimes v_{\tau_{i}\left(  2\right)  }\otimes...\otimes v_{\tau_{i}\left(
n\right)  }\\
&  \in T\ \ \ \ \ \ \ \ \ \ \left(  \text{due to (\ref{pf.symm-alg.1}),
applied to }\left(  \left(  v_{1},v_{2},...,v_{n}\right)  ,\tau_{i}\right)
\text{ instead of }\ \left(  \left(  v_{1},v_{2},...,v_{n}\right)
,\sigma\right)  \right) \\
&  \subseteq\left\langle T\right\rangle
\end{align*}
). Now,%
\begin{align*}
Z  &  =\sum_{i=1}^{n-1}\underbrace{\left\langle v_{1}\otimes v_{2}%
\otimes...\otimes v_{n}-v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau
_{i}\left(  2\right)  }\otimes...\otimes v_{\tau_{i}\left(  n\right)  }%
\ \mid\ \left(  v_{1},v_{2},...,v_{n}\right)  \in V^{n}\right\rangle
}_{\subseteq\left\langle T\right\rangle }\\
&  \subseteq\sum_{i=1}^{n-1}\left\langle T\right\rangle \subseteq\left\langle
T\right\rangle \ \ \ \ \ \ \ \ \ \ \left(  \text{since }\left\langle
T\right\rangle \text{ is a }k\text{-module}\right)  .
\end{align*}


Now, let us show that $\left\langle T\right\rangle \subseteq Z$. To that aim,
we will show that $T\subseteq Z$.

In fact, let $\left(  \left(  v_{1},v_{2},...,v_{n}\right)  ,\sigma\right)
\in V^{n}\times S_{n}$ be arbitrary. Then, $\left(  v_{1},v_{2},...,v_{n}%
\right)  \in V^{n}$ and $\sigma\in S_{n}$. Now, it is known that every element
of the symmetric group $S_{n}$ can be written as a product of some
transpositions from the set $\left\{  \tau_{1},\tau_{2},...,\tau
_{n-1}\right\}  $. Applying this to the element $\sigma\in S_{n}$, we conclude
that $\sigma$ can be written as a product of some transpositions from the set
$\left\{  \tau_{1},\tau_{2},...,\tau_{n-1}\right\}  $. In other words, there
exists a natural number $m\in\mathbb{N}$ and a sequence $\left(  i_{1}%
,i_{2},...,i_{m}\right)  \in\left\{  1,2,...,n-1\right\}  ^{m}$ such that
$\sigma=\tau_{i_{1}}\tau_{i_{2}}...\tau_{i_{m}}$. Consider this $m$ and this
$\left(  i_{1},i_{2},...,i_{m}\right)  $. For every $j\in\left\{
0,1,...,m\right\}  $, let $\sigma_{j}$ denote the permutation $\tau_{i_{1}%
}\tau_{i_{2}}...\tau_{i_{j}}\in S_{n}$. Then, $\sigma_{0}=\tau_{i_{1}}%
\tau_{i_{2}}...\tau_{i_{0}}=\left(  \text{empty product}\right)
=\operatorname{id}$ and $\sigma_{m}=\tau_{i_{1}}\tau_{i_{2}}...\tau_{i_{m}%
}=\sigma$. Moreover, every $j\in\left\{  1,2,...,m\right\}  $ satisfies
$v_{\sigma_{j-1}\left(  1\right)  }\otimes v_{\sigma_{j-1}\left(  2\right)
}\otimes...\otimes v_{\sigma_{j-1}\left(  n\right)  }-v_{\sigma_{j}\left(
1\right)  }\otimes v_{\sigma_{j}\left(  2\right)  }\otimes...\otimes
v_{\sigma_{j}\left(  n\right)  }\in Z$.\ \ \ \ \footnote{\textit{Proof.} Let
$j\in\left\{  1,2,...,m\right\}  $ be arbitrary. Then,
\[
\underbrace{\sigma_{j-1}}_{\substack{=\tau_{i_{1}}\tau_{i_{2}}...\tau
_{i_{j-1}}\\\text{(by the formula }\sigma_{j}=\tau_{i_{1}}\tau_{i_{2}}%
...\tau_{i_{j}}\text{,}\\\text{applied to }j-1\text{ instead of }j\text{)}%
}}\tau_{i_{j}}=\tau_{i_{1}}\tau_{i_{2}}...\tau_{i_{j-1}}\tau_{i_{j}}%
=\tau_{i_{1}}\tau_{i_{2}}...\tau_{i_{j}}=\sigma_{j},
\]
so that $\sigma_{j}=\sigma_{j-1}\tau_{i_{j}}$. Denote $i_{j}$ by $\mathbf{I}$.
Then, $\sigma_{j}=\sigma_{j-1}\tau_{i_{j}}$ rewrites as $\sigma_{j}%
=\sigma_{j-1}\tau_{\mathbf{I}}$.
\par
Now, define an $n$-tuple $\left(  w_{1},w_{2},...,w_{n}\right)  \in V^{n}$ by
$\left(  w_{\rho}=v_{\sigma_{j-1}\left(  \rho\right)  }\text{ for every }%
\rho\in\left\{  1,2,...,n\right\}  \right)  $. Then, $\left(  w_{1}%
,w_{2},...,w_{n}\right)  =\left(  v_{\sigma_{j-1}\left(  1\right)  }%
,v_{\sigma_{j-1}\left(  2\right)  },...,v_{\sigma_{j-1}\left(  n\right)
}\right)  $, so that $w_{1}\otimes w_{2}\otimes...\otimes w_{n}=v_{\sigma
_{j-1}\left(  1\right)  }\otimes v_{\sigma_{j-1}\left(  2\right)  }%
\otimes...\otimes v_{\sigma_{j-1}\left(  n\right)  }$. On the other hand,
every $\xi\in\left\{  1,2,...,n\right\}  $ satisfies
\begin{align*}
w_{\tau_{\mathbf{I}}\left(  \xi\right)  }  &  =v_{\sigma_{j-1}\left(
\tau_{\mathbf{I}}\left(  \xi\right)  \right)  }\ \ \ \ \ \ \ \ \ \ \left(
\text{by the formula }w_{\rho}=v_{\sigma_{j-1}\left(  \rho\right)  }\text{,
applied to }\rho=\tau_{\mathbf{I}}\left(  \xi\right)  \right) \\
&  =v_{\sigma_{j}\left(  \xi\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{since
}\sigma_{j-1}\left(  \tau_{\mathbf{I}}\left(  \xi\right)  \right)
=\underbrace{\left(  \sigma_{j-1}\tau_{\mathbf{I}}\right)  }_{=\sigma_{j}%
}\left(  \xi\right)  =\sigma_{j}\left(  \xi\right)  \right)  .
\end{align*}
Thus, $\left(  w_{\tau_{\mathbf{I}}\left(  1\right)  },w_{\tau_{\mathbf{I}%
}\left(  2\right)  },...,w_{\tau_{\mathbf{I}}\left(  n\right)  }\right)
=\left(  v_{\sigma_{j}\left(  1\right)  },v_{\sigma_{j}\left(  2\right)
},...,v_{\sigma_{j}\left(  n\right)  }\right)  $, so that $w_{\tau
_{\mathbf{I}}\left(  1\right)  }\otimes w_{\tau_{\mathbf{I}}\left(  2\right)
}\otimes...\otimes w_{\tau_{\mathbf{I}}\left(  n\right)  }=v_{\sigma
_{j}\left(  1\right)  }\otimes v_{\sigma_{j}\left(  2\right)  }\otimes
...\otimes v_{\sigma_{j}\left(  n\right)  }$.
\par
Now,%
\begin{align*}
&  \underbrace{v_{\sigma_{j-1}\left(  1\right)  }\otimes v_{\sigma
_{j-1}\left(  2\right)  }\otimes...\otimes v_{\sigma_{j-1}\left(  n\right)  }%
}_{=w_{1}\otimes w_{2}\otimes...\otimes w_{n}}-\underbrace{v_{\sigma
_{j}\left(  1\right)  }\otimes v_{\sigma_{j}\left(  2\right)  }\otimes
...\otimes v_{\sigma_{j}\left(  n\right)  }}_{=w_{\tau_{\mathbf{I}}\left(
1\right)  }\otimes w_{\tau_{\mathbf{I}}\left(  2\right)  }\otimes...\otimes
w_{\tau_{\mathbf{I}}\left(  n\right)  }}\\
&  =w_{1}\otimes w_{2}\otimes...\otimes w_{n}-w_{\tau_{\mathbf{I}}\left(
1\right)  }\otimes w_{\tau_{\mathbf{I}}\left(  2\right)  }\otimes...\otimes
w_{\tau_{\mathbf{I}}\left(  n\right)  }\in Z\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.symm-alg.3})}\right)  ,
\end{align*}
qed.} Thus,%
\[
\sum_{j=1}^{m}\underbrace{\left(  v_{\sigma_{j-1}\left(  1\right)  }\otimes
v_{\sigma_{j-1}\left(  2\right)  }\otimes...\otimes v_{\sigma_{j-1}\left(
n\right)  }-v_{\sigma_{j}\left(  1\right)  }\otimes v_{\sigma_{j}\left(
2\right)  }\otimes...\otimes v_{\sigma_{j}\left(  n\right)  }\right)  }_{\in
Z}\in\sum_{j=1}^{m}Z\subseteq Z
\]
(since $Z$ is a $k$-module). Since%
\begin{align}
&  \sum_{j=1}^{m}\left(  v_{\sigma_{j-1}\left(  1\right)  }\otimes
v_{\sigma_{j-1}\left(  2\right)  }\otimes...\otimes v_{\sigma_{j-1}\left(
n\right)  }-v_{\sigma_{j}\left(  1\right)  }\otimes v_{\sigma_{j}\left(
2\right)  }\otimes...\otimes v_{\sigma_{j}\left(  n\right)  }\right)
\nonumber\\
&  =v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\sigma\left(  1\right)
}\otimes v_{\sigma\left(  2\right)  }\otimes...\otimes v_{\sigma\left(
n\right)  } \label{pf.symm-alg.4}%
\end{align}
\footnote{\textit{Proof of (\ref{pf.symm-alg.4}).} We distinguish between two
cases: the case when $m>0$, and the case when $m=0$.
\par
In the case when $m>0$, we have%
\begin{align*}
&  \sum_{j=1}^{m}\left(  v_{\sigma_{j-1}\left(  1\right)  }\otimes
v_{\sigma_{j-1}\left(  2\right)  }\otimes...\otimes v_{\sigma_{j-1}\left(
n\right)  }-v_{\sigma_{j}\left(  1\right)  }\otimes v_{\sigma_{j}\left(
2\right)  }\otimes...\otimes v_{\sigma_{j}\left(  n\right)  }\right) \\
&  =\sum_{j=1}^{m}v_{\sigma_{j-1}\left(  1\right)  }\otimes v_{\sigma
_{j-1}\left(  2\right)  }\otimes...\otimes v_{\sigma_{j-1}\left(  n\right)
}-\sum_{j=1}^{m}v_{\sigma_{j}\left(  1\right)  }\otimes v_{\sigma_{j}\left(
2\right)  }\otimes...\otimes v_{\sigma_{j}\left(  n\right)  }\\
&  =\underbrace{\sum_{j=0}^{m-1}v_{\sigma_{j}\left(  1\right)  }\otimes
v_{\sigma_{j}\left(  2\right)  }\otimes...\otimes v_{\sigma_{j}\left(
n\right)  }}_{=v_{\sigma_{0}\left(  1\right)  }\otimes v_{\sigma_{0}\left(
2\right)  }\otimes...\otimes v_{\sigma_{0}\left(  n\right)  }+\sum
\limits_{j=1}^{m-1}v_{\sigma_{j}\left(  1\right)  }\otimes v_{\sigma
_{j}\left(  2\right)  }\otimes...\otimes v_{\sigma_{j}\left(  n\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ -\underbrace{\sum_{j=1}^{m}v_{\sigma_{j}\left(
1\right)  }\otimes v_{\sigma_{j}\left(  2\right)  }\otimes...\otimes
v_{\sigma_{j}\left(  n\right)  }}_{=\sum\limits_{j=1}^{m-1}v_{\sigma
_{j}\left(  1\right)  }\otimes v_{\sigma_{j}\left(  2\right)  }\otimes
...\otimes v_{\sigma_{j}\left(  n\right)  }+v_{\sigma_{m}\left(  1\right)
}\otimes v_{\sigma_{m}\left(  2\right)  }\otimes...\otimes v_{\sigma
_{m}\left(  n\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }j\text{ for
}j-1\text{ in the first sum}\right) \\
&  =\left(  v_{\sigma_{0}\left(  1\right)  }\otimes v_{\sigma_{0}\left(
2\right)  }\otimes...\otimes v_{\sigma_{0}\left(  n\right)  }+\sum
\limits_{j=1}^{m-1}v_{\sigma_{j}\left(  1\right)  }\otimes v_{\sigma
_{j}\left(  2\right)  }\otimes...\otimes v_{\sigma_{j}\left(  n\right)
}\right) \\
&  \ \ \ \ \ \ \ \ \ \ -\left(  \sum\limits_{j=1}^{m-1}v_{\sigma_{j}\left(
1\right)  }\otimes v_{\sigma_{j}\left(  2\right)  }\otimes...\otimes
v_{\sigma_{j}\left(  n\right)  }+v_{\sigma_{m}\left(  1\right)  }\otimes
v_{\sigma_{m}\left(  2\right)  }\otimes...\otimes v_{\sigma_{m}\left(
n\right)  }\right) \\
&  =\underbrace{v_{\sigma_{0}\left(  1\right)  }\otimes v_{\sigma_{0}\left(
2\right)  }\otimes...\otimes v_{\sigma_{0}\left(  n\right)  }}%
_{\substack{=v_{1}\otimes v_{2}\otimes...\otimes v_{n}\\\text{(since }%
\sigma_{0}=\operatorname{id}\text{ yields}\\v_{\sigma_{0}\left(  1\right)
}\otimes v_{\sigma_{0}\left(  2\right)  }\otimes...\otimes v_{\sigma
_{0}\left(  n\right)  }=v_{\operatorname{id}\left(  1\right)  }\otimes
v_{\operatorname{id}\left(  2\right)  }\otimes...\otimes v_{\operatorname{id}%
\left(  n\right)  }\\=v_{1}\otimes v_{2}\otimes...\otimes v_{n}\text{)}%
}}-\underbrace{v_{\sigma_{m}\left(  1\right)  }\otimes v_{\sigma_{m}\left(
2\right)  }\otimes...\otimes v_{\sigma_{m}\left(  n\right)  }}%
_{\substack{=v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(  2\right)
}\otimes...\otimes v_{\sigma\left(  n\right)  }\\\text{(since }\sigma
_{m}=\sigma\text{)}}}\\
&  =v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\sigma\left(  1\right)
}\otimes v_{\sigma\left(  2\right)  }\otimes...\otimes v_{\sigma\left(
n\right)  },
\end{align*}
so that (\ref{pf.symm-alg.4}) is proven in the case when $m>0$.
\par
In the case when $m=0$, we have%
\begin{align*}
&  \sum_{j=1}^{m}\left(  v_{\sigma_{j-1}\left(  1\right)  }\otimes
v_{\sigma_{j-1}\left(  2\right)  }\otimes...\otimes v_{\sigma_{j-1}\left(
n\right)  }-v_{\sigma_{j}\left(  1\right)  }\otimes v_{\sigma_{j}\left(
2\right)  }\otimes...\otimes v_{\sigma_{j}\left(  n\right)  }\right) \\
&  =\left(  \text{empty sum}\right)  =0\\
&  =\underbrace{v_{\operatorname{id}\left(  1\right)  }\otimes
v_{\operatorname{id}\left(  2\right)  }\otimes...\otimes v_{\operatorname{id}%
\left(  n\right)  }}_{=v_{1}\otimes v_{2}\otimes...\otimes v_{n}%
}-\underbrace{v_{\operatorname{id}\left(  1\right)  }\otimes
v_{\operatorname{id}\left(  2\right)  }\otimes...\otimes v_{\operatorname{id}%
\left(  n\right)  }}_{\substack{=v_{\sigma\left(  1\right)  }\otimes
v_{\sigma\left(  2\right)  }\otimes...\otimes v_{\sigma\left(  n\right)
}\\\text{(since }m=0\text{ leads to }\sigma_{m}=\sigma_{0}\text{, so
that}\\\operatorname{id}=\sigma_{0}=\sigma_{m}=\sigma\text{)}}}\\
&  =v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\sigma\left(  1\right)
}\otimes v_{\sigma\left(  2\right)  }\otimes...\otimes v_{\sigma\left(
n\right)  },
\end{align*}
so that (\ref{pf.symm-alg.4}) is proven in the case when $m=0$.
\par
Thus, (\ref{pf.symm-alg.4}) is proven in both possible cases. This completes
the proof of (\ref{pf.symm-alg.4}).}, this rewrites as%
\[
v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\sigma\left(  1\right)  }\otimes
v_{\sigma\left(  2\right)  }\otimes...\otimes v_{\sigma\left(  n\right)  }\in
Z.
\]


We have thus shown that every $\left(  \left(  v_{1},v_{2},...,v_{n}\right)
,\sigma\right)  \in V^{n}\times S_{n}$ satisfies $v_{1}\otimes v_{2}%
\otimes...\otimes v_{n}-v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(
2\right)  }\otimes...\otimes v_{\sigma\left(  n\right)  }\in Z$. Thus,%
\[
\left\{  v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\sigma\left(  1\right)
}\otimes v_{\sigma\left(  2\right)  }\otimes...\otimes v_{\sigma\left(
n\right)  }\ \mid\ \left(  \left(  v_{1},v_{2},...,v_{n}\right)
,\sigma\right)  \in V^{n}\times S_{n}\right\}  \subseteq Z.
\]
Since $\left\{  v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\sigma\left(
1\right)  }\otimes v_{\sigma\left(  2\right)  }\otimes...\otimes
v_{\sigma\left(  n\right)  }\ \mid\ \left(  \left(  v_{1},v_{2},...,v_{n}%
\right)  ,\sigma\right)  \in V^{n}\times S_{n}\right\}  =T$, this rewrites as
$T\subseteq Z$. Proposition \ref{prop.<>} \textbf{(a)} (applied to $V^{\otimes
n}$, $T$ and $Z$ instead of $M$, $S$ and $Q$) thus yields that $\left\langle
T\right\rangle \subseteq Z$. Combined with $Z\subseteq\left\langle
T\right\rangle $, this yields $Z=\left\langle T\right\rangle $.

We now have%
\begin{align*}
&  \sum_{i=1}^{n-1}\left\langle v_{1}\otimes v_{2}\otimes...\otimes
v_{n}-v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)
}\otimes...\otimes v_{\tau_{i}\left(  n\right)  }\ \mid\ \left(  v_{1}%
,v_{2},...,v_{n}\right)  \in V^{n}\right\rangle \\
&  =Z=\left\langle T\right\rangle =K_{n}\left(  V\right)  .
\end{align*}
This proves Proposition \ref{rmk.symm-alg}.

Just like the $n$-th tensor power, $\operatorname*{Sym}^{n}V$ becomes a
$\mathfrak{g}$-module when $V$ is a $\mathfrak{g}$-module:
\end{verlong}

\begin{definition}
\label{defs.symm-alg.g}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $V$ be a $\mathfrak{g}$-module. Let $n\in\mathbb{N}%
$.\newline Consider the $\mathfrak{g}$-module $V^{\otimes n}$ (the $n$-th
tensor power of the $\mathfrak{g}$-module $V$). Then, the $k$-module
$K_{n}\left(  V\right)  $ introduced in Definition \ref{defs.symm-alg} is a
$\mathfrak{g}$-submodule of the $\mathfrak{g}$-module $V^{\otimes n}$
(according to Proposition \ref{prop.symm-alg.g} below). Thus, the factor
$k$-module $V^{\otimes n}\diagup K_{n}\left(  V\right)  $ becomes a
$\mathfrak{g}$-module. Since $V^{\otimes n}\diagup K_{n}\left(  V\right)
=\operatorname*{Sym}^{n}V$, this means that the $k$-module
$\operatorname*{Sym}^{n}V$ becomes a $\mathfrak{g}$-module. Whenever we will
speak of "the $\mathfrak{g}$-module $\operatorname*{Sym}^{n}V$", we are going
to mean the $\mathfrak{g}$-module $\operatorname*{Sym}^{n}V$ just
defined.\newline It is clear that the map $\operatorname*{sym}\nolimits_{V,n}$
(being the canonical projection $V^{\otimes n}\rightarrow V^{\otimes n}\diagup
K_{n}\left(  V\right)  $) is a $\mathfrak{g}$-module homomorphism.
\end{definition}

Here, we are using the following standard fact:

\begin{proposition}
\label{prop.symm-alg.g}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $V$ be a $\mathfrak{g}$-module. Let $n\in\mathbb{N}%
$.\newline Consider the $\mathfrak{g}$-module $V^{\otimes n}$ (the $n$-th
tensor power of the $\mathfrak{g}$-module $V$). Then, the $k$-module
$K_{n}\left(  V\right)  $ introduced in Definition \ref{defs.symm-alg} is a
$\mathfrak{g}$-submodule of the $\mathfrak{g}$-module $V^{\otimes n}$.
\end{proposition}

\begin{verlong}
\textit{Proof of Proposition \ref{prop.symm-alg.g}.} Let $T$ denote the subset%
\[
\left\{  v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\sigma\left(  1\right)
}\otimes v_{\sigma\left(  2\right)  }\otimes...\otimes v_{\sigma\left(
n\right)  }\ \mid\ \left(  \left(  v_{1},v_{2},...,v_{n}\right)
,\sigma\right)  \in V^{n}\times S_{n}\right\}
\]
of $V^{\otimes n}$. Then, clearly,%
\begin{equation}
w_{1}\otimes w_{2}\otimes...\otimes w_{n}-w_{\sigma\left(  1\right)  }\otimes
w_{\sigma\left(  2\right)  }\otimes...\otimes w_{\sigma\left(  n\right)  }\in
T \label{pf.symm-alg.g.0}%
\end{equation}
for every $\left(  \left(  w_{1},w_{2},...,w_{n}\right)  ,\sigma\right)  \in
V^{n}\times S_{n}$. We will now show that%
\begin{equation}
a\rightharpoonup t\in K_{n}\left(  V\right)  \ \ \ \ \ \ \ \ \ \ \text{for
every }a\in\mathfrak{g}\text{ and }t\in T. \label{pf.symm-alg.g.1}%
\end{equation}


\textit{Proof of (\ref{pf.symm-alg.g.1}).} We will be using Convention
\ref{conv.replace} during the following proof of (\ref{pf.symm-alg.g.1}).

Also, let us introduce another notation: If $\alpha$ and $\beta$ are two
integers such that $\alpha\leq\beta+1$, and if $p_{\alpha}$, $p_{\alpha+1}$,
$...$, $p_{\beta}$ are some elements of $V$, then $\overrightarrow{\bigotimes
\limits_{j=\alpha}^{\beta}}p_{j}$ will mean the tensor $p_{\alpha}\otimes
p_{\alpha+1}\otimes...\otimes p_{\beta}\in V^{\otimes\left(  \beta
-\alpha+1\right)  }$. (This tensor means $1$ if $\alpha=\beta+1$, and it means
$p_{\alpha}$ if $\alpha=\beta$.)

We notice that every $i\in\left\{  1,2,...,n\right\}  $, every $\left(
v_{1},v_{2},...,v_{n}\right)  \in V^{n}$ and every $w_{i}\in V$ satisfy
\begin{equation}
v_{1}\otimes v_{2}\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$w_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
}\otimes...\otimes v_{n}=\left(  \overrightarrow{\bigotimes\limits_{j=1}%
^{i-1}}v_{j}\right)  \otimes w_{i}\otimes\left(  \overrightarrow{\bigotimes
\limits_{j=i+1}^{n}}v_{j}\right)  . \label{pf.symm-alg.g.a}%
\end{equation}
\footnote{This is because Convention \ref{conv.replace} (applied to $V_{j}=V$
and $W_{i}=V$) yields%
\begin{align*}
v_{1}\otimes v_{2}\otimes...\otimes\ \text{%
\begin{tabular}
[c]{|c|}\hline
$w_{i}$\\\hline
$v_{i}$\\\hline
\end{tabular}
\ }\otimes...\otimes v_{n}  &  =\underbrace{v_{1}\otimes v_{2}\otimes
...\otimes v_{i-1}}_{\substack{\text{tensor product of the}\\\text{first
}i-1\text{ vectors }v_{\ell}}}\otimes w_{i}\otimes\underbrace{v_{i+1}\otimes
v_{i+2}\otimes...\otimes v_{n}}_{\substack{\text{tensor product of
the}\\\text{last }n-i\text{ vectors }v_{\ell}}}\\
&  =\underbrace{v_{1}\otimes v_{2}\otimes...\otimes v_{i-1}}%
_{=\overrightarrow{\bigotimes\limits_{j=1}^{i-1}}v_{j}}\otimes w_{i}%
\otimes\underbrace{v_{i+1}\otimes v_{i+2}\otimes...\otimes v_{n}%
}_{=\overrightarrow{\bigotimes\limits_{j=i+1}^{n}}v_{j}}\\
&  =\left(  \overrightarrow{\bigotimes\limits_{j=1}^{i-1}}v_{j}\right)
\otimes w_{i}\otimes\left(  \overrightarrow{\bigotimes\limits_{j=i+1}^{n}%
}v_{j}\right)  .
\end{align*}
}

Let $a\in\mathfrak{g}$ and $t\in T$ be arbitrary. Then,%
\[
t\in T=\left\{  v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\sigma\left(
1\right)  }\otimes v_{\sigma\left(  2\right)  }\otimes...\otimes
v_{\sigma\left(  n\right)  }\ \mid\ \left(  \left(  v_{1},v_{2},...,v_{n}%
\right)  ,\sigma\right)  \in V^{n}\times S_{n}\right\}  .
\]
Hence, there exists some $\left(  \left(  u_{1},u_{2},...,u_{n}\right)
,\sigma\right)  \in V^{n}\times S_{n}$ such that $t=u_{1}\otimes u_{2}%
\otimes...\otimes u_{n}-u_{\sigma\left(  1\right)  }\otimes u_{\sigma\left(
2\right)  }\otimes...\otimes u_{\sigma\left(  n\right)  }$. Consider this
$\left(  \left(  u_{1},u_{2},...,u_{n}\right)  ,\sigma\right)  $. Then,
$\left(  u_{1},u_{2},...,u_{n}\right)  \in V^{n}$ and $\sigma\in S_{n}$ (since
$\left(  \left(  u_{1},u_{2},...,u_{n}\right)  ,\sigma\right)  \in V^{n}\times
S_{n}$).

For every $i\in\left\{  1,2,...,n\right\}  $, let us define a tensor $M_{i}\in
V^{\otimes n}$ by%
\begin{equation}
M_{i}=u_{1}\otimes u_{2}\otimes...\otimes\
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup u_{i}$\\\hline
$u_{i}$\\\hline
\end{tabular}
\ \otimes...\otimes u_{n}. \label{pf.symm-alg.g.M}%
\end{equation}


For every $i\in\left\{  1,2,...,n\right\}  $, let us define a tensor $N_{i}\in
V^{\otimes n}$ by%
\begin{equation}
N_{i}=u_{\sigma\left(  1\right)  }\otimes u_{\sigma\left(  2\right)  }%
\otimes...\otimes\
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup u_{\sigma\left(  i\right)  }$\\\hline
$u_{\sigma\left(  i\right)  }$\\\hline
\end{tabular}
\ \otimes...\otimes u_{\sigma\left(  n\right)  }. \label{pf.symm-alg.g.N}%
\end{equation}


Proposition \ref{prop.lie(X)mult.exp} (applied to $V_{i}=V$ and $v_{i}=u_{i}$)
yields%
\[
a\rightharpoonup\left(  u_{1}\otimes u_{2}\otimes...\otimes u_{n}\right)
=\underbrace{\sum\limits_{i=1}^{n}}_{=\sum\limits_{i\in\left\{
1,2,...,n\right\}  }}\underbrace{u_{1}\otimes u_{2}\otimes...\otimes\
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup u_{i}$\\\hline
$u_{i}$\\\hline
\end{tabular}
\ \otimes...\otimes u_{n}}_{=M_{i}}=\sum\limits_{i\in\left\{
1,2,...,n\right\}  }M_{i}.
\]
On the other hand, Proposition \ref{prop.lie(X)mult.exp} (applied to $V_{i}=V$
and $v_{i}=u_{\sigma\left(  i\right)  }$) yields%
\begin{align*}
&  a\rightharpoonup\left(  u_{\sigma\left(  1\right)  }\otimes u_{\sigma
\left(  2\right)  }\otimes...\otimes u_{\sigma\left(  n\right)  }\right) \\
&  =\underbrace{\sum\limits_{i=1}^{n}}_{=\sum\limits_{i\in\left\{
1,2,...,n\right\}  }}\underbrace{u_{\sigma\left(  1\right)  }\otimes
u_{\sigma\left(  2\right)  }\otimes...\otimes\
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup u_{\sigma\left(  i\right)  }$\\\hline
$u_{\sigma\left(  i\right)  }$\\\hline
\end{tabular}
\ \otimes...\otimes u_{\sigma\left(  n\right)  }}_{=N_{i}}=\sum_{i\in\left\{
1,2,...,n\right\}  }N_{i}=\sum\limits_{i\in\left\{  1,2,...,n\right\}
}N_{\sigma^{-1}\left(  i\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, we substituted }\sigma^{-1}\left(  i\right)  \text{ for }i\text{
in the sum, since}\\
\sigma:\left\{  1,2,...,n\right\}  \rightarrow\left\{  1,2,...,n\right\}
\text{ is a bijection (because }\sigma\in S_{n}\text{)}%
\end{array}
\right)  .
\end{align*}
Since $t=u_{1}\otimes u_{2}\otimes...\otimes u_{n}-u_{\sigma\left(  1\right)
}\otimes u_{\sigma\left(  2\right)  }\otimes...\otimes u_{\sigma\left(
n\right)  }$, we have%
\begin{align*}
a\rightharpoonup t  &  =a\rightharpoonup\left(  u_{1}\otimes u_{2}%
\otimes...\otimes u_{n}-u_{\sigma\left(  1\right)  }\otimes u_{\sigma\left(
2\right)  }\otimes...\otimes u_{\sigma\left(  n\right)  }\right) \\
&  =\underbrace{a\rightharpoonup\left(  u_{1}\otimes u_{2}\otimes...\otimes
u_{n}\right)  }_{=\sum\limits_{i\in\left\{  1,2,...,n\right\}  }M_{i}%
}-\underbrace{a\rightharpoonup\left(  u_{\sigma\left(  1\right)  }\otimes
u_{\sigma\left(  2\right)  }\otimes...\otimes u_{\sigma\left(  n\right)
}\right)  }_{=\sum\limits_{i\in\left\{  1,2,...,n\right\}  }N_{\sigma
^{-1}\left(  i\right)  }}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{since the Lie action of }V^{\otimes
n}\text{ is }k\text{-bilinear}\right) \\
&  =\sum\limits_{i\in\left\{  1,2,...,n\right\}  }M_{i}-\sum\limits_{i\in
\left\{  1,2,...,n\right\}  }N_{\sigma^{-1}\left(  i\right)  }=\sum
\limits_{i\in\left\{  1,2,...,n\right\}  }\left(  M_{i}-N_{\sigma^{-1}\left(
i\right)  }\right)  .
\end{align*}


Now, every $i\in\left\{  1,2,...,n\right\}  $ satisfies%
\begin{equation}
M_{i}-N_{\sigma^{-1}\left(  i\right)  }\in T. \label{pf.symm-alg.g.2}%
\end{equation}


\textit{Proof of (\ref{pf.symm-alg.g.2}).} Let $i\in\left\{
1,2,...,n\right\}  $ be arbitrary. Let $r=\sigma^{-1}\left(  i\right)  $. This
is well-defined since $\sigma$ is a bijection (because $\sigma\in S_{n}$).
Clearly, $\sigma\left(  r\right)  =i$.

For every $j\in\left\{  1,2,...,n\right\}  $, define an element $w_{j}$ of $V$
by $w_{j}=\left\{
\begin{array}
[c]{c}%
u_{j}\text{, if }j\neq i;\\
a\rightharpoonup u_{j}\text{, if }j=i
\end{array}
\right.  $. Thus we have defined an $n$-tuple $\left(  w_{1},w_{2}%
,...,w_{n}\right)  \in V^{n}$. Every $j\in\left\{  1,2,...,n\right\}  $
satisfying $j\neq i$ satisfies $w_{j}=\left\{
\begin{array}
[c]{c}%
u_{j}\text{, if }j\neq i;\\
a\rightharpoonup u_{j}\text{, if }j=i
\end{array}
\right.  =u_{j}$ (since $j\neq i$). In other words,
\begin{equation}
\text{every }j\in\left\{  1,2,...,n\right\}  \text{ satisfying }j\neq i\text{
satisfies }u_{j}=w_{j}. \label{pf.symm-alg.g.3}%
\end{equation}
Hence,%
\begin{equation}
\text{every }j\in\left\{  1,2,...,n\right\}  \text{ satisfying }j\neq r\text{
satisfies }u_{\sigma\left(  j\right)  }=w_{\sigma\left(  j\right)  }
\label{pf.symm-alg.g.4}%
\end{equation}
(by (\ref{pf.symm-alg.g.3}) (applied to $\sigma\left(  j\right)  $ instead of
$j$), since $\sigma\left(  j\right)  \neq i$ (because $j\neq r=\sigma
^{-1}\left(  i\right)  $ and because $\sigma$ is a bijection)).

Also, the definition of $w_{i}$ yields $w_{i}=\left\{
\begin{array}
[c]{c}%
u_{i}\text{, if }i\neq i;\\
a\rightharpoonup u_{i}\text{, if }i=i
\end{array}
\right.  =a\rightharpoonup u_{i}$ (since $i=i$).

Now,%
\begin{align*}
&  M_{i}=u_{1}\otimes u_{2}\otimes...\otimes\
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup u_{i}$\\\hline
$u_{i}$\\\hline
\end{tabular}
\ \otimes...\otimes u_{n}\\
&  =\left(  \overrightarrow{\bigotimes\limits_{j=1}^{i-1}}\underbrace{u_{j}%
}_{\substack{=w_{j}\\\text{(by (\ref{pf.symm-alg.g.3}),}\\\text{since
}j<i\text{ yields }j\neq i\text{)}}}\right)  \otimes\underbrace{\left(
a\rightharpoonup u_{i}\right)  }_{=w_{i}}\otimes\left(
\overrightarrow{\bigotimes\limits_{j=i+1}^{n}}\underbrace{u_{j}}%
_{\substack{=w_{j}\\\text{(by (\ref{pf.symm-alg.g.3}),}\\\text{since
}j>i\text{ yields }j\neq i\text{)}}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{due to (\ref{pf.symm-alg.g.a}), applied
to }u_{j}\text{ and }a\rightharpoonup u_{i}\text{ instead of }v_{j}\text{ and
}w_{i}\right) \\
&  =\underbrace{\left(  \overrightarrow{\bigotimes\limits_{j=1}^{i-1}}%
w_{j}\right)  }_{=w_{1}\otimes w_{2}\otimes...\otimes w_{i-1}}\otimes
w_{i}\otimes\underbrace{\left(  \overrightarrow{\bigotimes\limits_{j=i+1}^{n}%
}w_{j}\right)  }_{=w_{i+1}\otimes w_{i+2}\otimes...\otimes w_{n}}\\
&  =\left(  w_{1}\otimes w_{2}\otimes...\otimes w_{i-1}\right)  \otimes
w_{i}\otimes\left(  w_{i+1}\otimes w_{i+2}\otimes...\otimes w_{n}\right)
=w_{1}\otimes w_{2}\otimes...\otimes w_{n}%
\end{align*}
and%
\begin{align*}
&  N_{\sigma^{-1}\left(  i\right)  }=N_{r}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }\sigma^{-1}\left(  i\right)  =r\right) \\
&  =u_{\sigma\left(  1\right)  }\otimes u_{\sigma\left(  2\right)  }%
\otimes...\otimes\
\begin{tabular}
[c]{|c|}\hline
$a\rightharpoonup u_{\sigma\left(  r\right)  }$\\\hline
$u_{\sigma\left(  r\right)  }$\\\hline
\end{tabular}
\ \ \ \ \ \otimes...\otimes u_{\sigma\left(  n\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.symm-alg.g.N}), applied to
}r\text{ instead of }i\right) \\
&  =\left(  \overrightarrow{\bigotimes\limits_{j=1}^{r-1}}%
\underbrace{u_{\sigma\left(  j\right)  }}_{\substack{=w_{\sigma\left(
j\right)  }\\\text{(by (\ref{pf.symm-alg.g.4})}\\\text{(since }j<r\text{
yields }j\neq r\text{))}}}\right)  \otimes\underbrace{\left(  a\rightharpoonup
u_{\sigma\left(  r\right)  }\right)  }_{=a\rightharpoonup u_{i}\text{ (since
}\sigma\left(  r\right)  =i\text{)}}\otimes\left(  \overrightarrow{\bigotimes
\limits_{j=r+1}^{n}}\underbrace{u_{\sigma\left(  j\right)  }}%
_{\substack{=w_{\sigma\left(  j\right)  }\\\text{(by (\ref{pf.symm-alg.g.4}%
)}\\\text{(since }j>r\text{ yields }j\neq r\text{))}}}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{due to (\ref{pf.symm-alg.g.a}), applied
to }r\text{, }u_{\sigma\left(  j\right)  }\text{ and }a\rightharpoonup
u_{\sigma\left(  r\right)  }\text{ instead of }i\text{, }v_{j}\text{ and
}w_{i}\right) \\
&  =\underbrace{\left(  \overrightarrow{\bigotimes\limits_{j=1}^{r-1}%
}w_{\sigma\left(  j\right)  }\right)  }_{=w_{\sigma\left(  1\right)  }\otimes
w_{\sigma\left(  2\right)  }\otimes...\otimes w_{\sigma\left(  r-1\right)  }%
}\otimes\underbrace{\left(  a\rightharpoonup u_{i}\right)  }_{\substack{=w_{i}%
=w_{\sigma\left(  r\right)  }\\\text{(since }i=\sigma\left(  r\right)
\text{)}}}\otimes\underbrace{\left(  \overrightarrow{\bigotimes\limits_{j=r+1}%
^{n}}w_{\sigma\left(  j\right)  }\right)  }_{=w_{\sigma\left(  r+1\right)
}\otimes w_{\sigma\left(  r+2\right)  }\otimes...\otimes w_{\sigma\left(
n\right)  }}\\
&  =\left(  w_{\sigma\left(  1\right)  }\otimes w_{\sigma\left(  2\right)
}\otimes...\otimes w_{\sigma\left(  r-1\right)  }\right)  \otimes
w_{\sigma\left(  r\right)  }\otimes\left(  w_{\sigma\left(  r+1\right)
}\otimes w_{\sigma\left(  r+2\right)  }\otimes...\otimes w_{\sigma\left(
n\right)  }\right) \\
&  =w_{\sigma\left(  1\right)  }\otimes w_{\sigma\left(  2\right)  }%
\otimes...\otimes w_{\sigma\left(  n\right)  }.
\end{align*}
Hence,%
\begin{align*}
&  \underbrace{M_{i}}_{=w_{1}\otimes w_{2}\otimes...\otimes w_{n}%
}-\underbrace{N_{\sigma^{-1}\left(  i\right)  }}_{=w_{\sigma\left(  1\right)
}\otimes w_{\sigma\left(  2\right)  }\otimes...\otimes w_{\sigma\left(
n\right)  }}\\
&  =w_{1}\otimes w_{2}\otimes...\otimes w_{n}-w_{\sigma\left(  1\right)
}\otimes w_{\sigma\left(  2\right)  }\otimes...\otimes w_{\sigma\left(
n\right)  }\in T\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.symm-alg.g.0}%
)}\right)  ,
\end{align*}
and thus (\ref{pf.symm-alg.g.2}) is proven.

Let us now continue proving (\ref{pf.symm-alg.g.1}): We have%
\[
a\rightharpoonup t=\sum\limits_{i\in\left\{  1,2,...,n\right\}  }%
\underbrace{\left(  M_{i}-N_{\sigma^{-1}\left(  i\right)  }\right)  }_{\in
T\text{ (by (\ref{pf.symm-alg.g.2}))}}\in\left\langle T\right\rangle .
\]
This proves (\ref{pf.symm-alg.g.1}).

Now, let $a\in\mathfrak{g}$ be arbitrary. Define a map $\partial
_{a}:V^{\otimes n}\rightarrow V^{\otimes n}$ by%
\[
\left(  \partial_{a}\left(  x\right)  =a\rightharpoonup x\text{ for every
}x\in V^{\otimes n}\right)  .
\]
Then, $\partial_{a}$ is a $k$-linear map (since the Lie action of $V^{\otimes
n}$ is $k$-bilinear). Proposition \ref{prop.<>} \textbf{(b)} (applied to
$V^{\otimes n}$, $T$, $V^{\otimes n}$ and $\partial_{a}$ instead of $M$, $S$,
$R$ and $f$) thus yields that $\partial_{a}\left(  \left\langle T\right\rangle
\right)  =\left\langle \partial_{a}\left(  T\right)  \right\rangle $. But
since%
\begin{align*}
\partial_{a}\left(  T\right)   &  =\left\{  \underbrace{\partial_{a}\left(
x\right)  }_{=a\rightharpoonup x}\ \mid\ x\in T\right\}  =\left\{
a\rightharpoonup x\ \mid\ x\in T\right\}  =\left\{  a\rightharpoonup
t\ \mid\ t\in T\right\} \\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{here, we substituted }t\text{ for
}x\right) \\
&  \subseteq K_{n}\left(  V\right)  \ \ \ \ \ \ \ \ \ \ \left(  \text{due to
(\ref{pf.symm-alg.g.1})}\right)  ,
\end{align*}
Proposition \ref{prop.<>} \textbf{(a)} (applied to $V^{\otimes n}$, $T$ and
$K_{n}\left(  V\right)  $ instead of $M$, $S$ and $Q$) yields $\left\langle
\partial_{a}\left(  T\right)  \right\rangle \subseteq K_{n}\left(  V\right)  $.

Now, let $x\in K_{n}\left(  V\right)  $ be arbitrary. Then,%
\begin{align*}
x  &  \in K_{n}\left(  V\right) \\
&  =\left\langle v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\sigma\left(
1\right)  }\otimes v_{\sigma\left(  2\right)  }\otimes...\otimes
v_{\sigma\left(  n\right)  }\ \mid\ \left(  \left(  v_{1},v_{2},...,v_{n}%
\right)  ,\sigma\right)  \in V^{n}\times S_{n}\right\rangle \\
&  =\left\langle \underbrace{\left\{  v_{1}\otimes v_{2}\otimes...\otimes
v_{n}-v_{\sigma\left(  1\right)  }\otimes v_{\sigma\left(  2\right)  }%
\otimes...\otimes v_{\sigma\left(  n\right)  }\ \mid\ \left(  \left(
v_{1},v_{2},...,v_{n}\right)  ,\sigma\right)  \in V^{n}\times S_{n}\right\}
}_{=T}\right\rangle \\
&  =\left\langle T\right\rangle ,
\end{align*}
so that $\partial_{a}\left(  x\right)  \in\partial_{a}\left(  \left\langle
T\right\rangle \right)  =\left\langle \partial_{a}\left(  T\right)
\right\rangle \subseteq K_{n}\left(  V\right)  $. Since $\partial_{a}\left(
x\right)  =a\rightharpoonup x$, this rewrites as $a\rightharpoonup x\in
K_{n}\left(  V\right)  $.

We have thus proven that $a\rightharpoonup x\in K_{n}\left(  V\right)  $ for
every $a\in\mathfrak{g}$ and every $x\in K_{n}\left(  V\right)  $. In other
words, $K_{n}\left(  V\right)  $ is a $\mathfrak{g}$-submodule of $V^{\otimes
n}$. This proves Proposition \ref{prop.symm-alg.g}.
\end{verlong}

\subsection{The PBW map}

The next proposition sets the stage for the Poincar\'{e}-Birkhoff-Witt theorem:

\begin{proposition}
\label{prop.PBWmap}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Consider the universal enveloping algebra $U\left(
\mathfrak{g}\right)  $ defined in Definition \ref{defs.U}.\newline Let
$\psi:\otimes\mathfrak{g}\rightarrow U\left(  \mathfrak{g}\right)  $ be the
canonical projection (which is well-defined since $U\left(  \mathfrak{g}%
\right)  $ is defined as a factor algebra of $\otimes\mathfrak{g}$). Clearly,
$\psi$ is a surjective $k$-algebra homomorphism.\newline For every
$n\in\mathbb{N}$, let $U_{\leq n}\left(  \mathfrak{g}\right)  $ be the
$k$-submodule $\psi\left(  \mathfrak{g}^{\otimes\leq n}\right)  $ of $U\left(
\mathfrak{g}\right)  $.\newline\textbf{(a)} Then, $\left(  U_{\leq n}\left(
\mathfrak{g}\right)  \right)  _{n\geq0}$ is a $\mathfrak{g}$-module filtration
of the $\mathfrak{g}$-module $U\left(  \mathfrak{g}\right)  $%
.\ \ \ \ {\footnotemark}\newline\textbf{(b)} The $k$-module homomorphism
$\psi:\otimes\mathfrak{g}\rightarrow U\left(  \mathfrak{g}\right)  $ respects
the filtration. Thus, for every $n\in\mathbb{N}$, we have a $k$-module
homomorphism $\operatorname*{gr}\nolimits_{n}\psi:\operatorname*{gr}%
\nolimits_{n}\left(  \otimes\mathfrak{g}\right)  \rightarrow\operatorname*{gr}%
\nolimits_{n}\left(  U\left(  \mathfrak{g}\right)  \right)  $. Composing this
homomorphism with the canonical $k$-module isomorphism $\operatorname{grad}%
_{\mathfrak{g},n}:\mathfrak{g}^{\otimes n}\rightarrow\operatorname*{gr}%
\nolimits_{n}\left(  \otimes\mathfrak{g}\right)  $, we obtain a $k$-module
homomorphism $\operatorname*{gr}\nolimits_{n}\psi\circ\operatorname{grad}%
_{\mathfrak{g},n}:\mathfrak{g}^{\otimes n}\rightarrow\operatorname*{gr}%
\nolimits_{n}\left(  U\left(  \mathfrak{g}\right)  \right)  $. Explicitly this
homomorphism looks as follows:%
\begin{align*}
&  \left(  \operatorname*{gr}\nolimits_{n}\psi\circ\operatorname{grad}%
_{\mathfrak{g},n}\right)  \left(  v_{1}\otimes v_{2}\otimes...\otimes
v_{n}\right) \\
&  =\left(  \text{the residue class of }\overline{v_{1}\otimes v_{2}%
\otimes...\otimes v_{n}}\in U_{\leq n}\left(  \mathfrak{g}\right)  \text{
modulo }U_{\leq\left(  n-1\right)  }\left(  \mathfrak{g}\right)  \right)
\end{align*}
for every $\left(  v_{1},v_{2},...,v_{n}\right)  \in\mathfrak{g}^{n}$%
.\newline\textbf{(c)} For every $n\in\mathbb{N}$, there exists a unique
$k$-module homomorphism $\mathbf{P}:\operatorname*{Sym}\nolimits^{n}%
\mathfrak{g}\rightarrow\operatorname*{gr}\nolimits_{n}\left(  U\left(
\mathfrak{g}\right)  \right)  $ such that $\operatorname*{gr}\nolimits_{n}%
\psi\circ\operatorname{grad}_{\mathfrak{g},n}=\mathbf{P}\circ
\operatorname*{sym}\nolimits_{\mathfrak{g},n}$. This homomorphism $\mathbf{P}$
is surjective.
\end{proposition}

\begin{vershort}
\footnotetext{Let us remind ourselves that here (and in the following), "the
$\mathfrak{g}$-module $U\left(  \mathfrak{g}\right)  $" is to be understood
according to Remark \ref{rmk.g-mod-U(g).short}. In other words, "the
$\mathfrak{g}$-module $U\left(  \mathfrak{g}\right)  $" means the
$\mathfrak{g}$-module obtained by applying Definition \ref{defs.lietensor} to
$V=\mathfrak{g}$ and setting $U\left(  \mathfrak{g}\right)  =\left(
\otimes\mathfrak{g}\right)  \diagup I_{\mathfrak{g}}$, not the $\mathfrak{g}%
$-module structure given by (\ref{rmk.g-mod-U(g).short.bad}).}
\end{vershort}

\begin{verlong}
\footnotetext{Let us remind ourselves that here (and in the following), "the
$\mathfrak{g}$-module $U\left(  \mathfrak{g}\right)  $" is to be understood
according to Remark \ref{rmk.g-mod-U(g).long}. In other words, "the
$\mathfrak{g}$-module $U\left(  \mathfrak{g}\right)  $" means the
$\mathfrak{g}$-module obtained by applying Definition \ref{defs.lietensor} to
$V=\mathfrak{g}$ and setting $U\left(  \mathfrak{g}\right)  =\left(
\otimes\mathfrak{g}\right)  \diagup I_{\mathfrak{g}}$, not the $\mathfrak{g}%
$-module structure given by (\ref{rmk.g-mod-U(g).long.bad}).}
\end{verlong}

We are not going to prove this proposition in detail, as every text on the
Poincar\'{e}-Birkhoff-Witt theorem does it. Here is a rough outline:

\textit{Proof of Proposition \ref{prop.PBWmap} (sketched).} \textbf{(a)}
Clear, since $\psi$ is surjective and a $\mathfrak{g}$-module homomorphism.

\textbf{(b)} Clear from the definition of $\psi$.

\textbf{(c)} Since $\operatorname*{sym}\nolimits_{\mathfrak{g},n}$ is
surjective, it is enough to prove that $\operatorname{Ker}\operatorname*{sym}%
\nolimits_{\mathfrak{g},n}\subseteq\operatorname{Ker}\left(
\operatorname*{gr}\nolimits_{n}\psi\circ\operatorname{grad}_{\mathfrak{g}%
,n}\right)  $ (because when this is proven, Proposition \ref{prop.PBWmap}
\textbf{(c)} follows from the homomorphism theorem). But $\operatorname{Ker}%
\operatorname*{sym}\nolimits_{\mathfrak{g},n}=K_{n}\left(  \mathfrak{g}%
\right)  $, so we must check that $K_{n}\left(  \mathfrak{g}\right)
\subseteq\operatorname{Ker}\left(  \operatorname*{gr}\nolimits_{n}\psi
\circ\operatorname{grad}_{\mathfrak{g},n}\right)  $. Due to Proposition
\ref{rmk.symm-alg} (applied to $V=\mathfrak{g}$), this only requires showing
that%
\[
v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\tau_{i}\left(  1\right)
}\otimes v_{\tau_{i}\left(  2\right)  }\otimes...\otimes v_{\tau_{i}\left(
n\right)  }\in\operatorname{Ker}\left(  \operatorname*{gr}\nolimits_{n}%
\psi\circ\operatorname{grad}_{\mathfrak{g},n}\right)
\]
for every $\left(  v_{1},v_{2},...,v_{n}\right)  \in\mathfrak{g}^{n}$ and
every $i\in\left\{  1,2,...,n-1\right\}  $. In other words, we have to show
that%
\[
\left(  \operatorname*{gr}\nolimits_{n}\psi\circ\operatorname{grad}%
_{\mathfrak{g},n}\right)  \left(  v_{1}\otimes v_{2}\otimes...\otimes
v_{n}-v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)
}\otimes...\otimes v_{\tau_{i}\left(  n\right)  }\right)  =0
\]
for every $\left(  v_{1},v_{2},...,v_{n}\right)  \in\mathfrak{g}^{n}$ and
every $i\in\left\{  1,2,...,n-1\right\}  $. But this is easy: From
\textbf{(b)}, we have%
\begin{align*}
&  \left(  \operatorname*{gr}\nolimits_{n}\psi\circ\operatorname{grad}%
_{\mathfrak{g},n}\right)  \left(  v_{1}\otimes v_{2}\otimes...\otimes
v_{n}-v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)
}\otimes...\otimes v_{\tau_{i}\left(  n\right)  }\right) \\
&  =\left(  \text{the residue class of }\overline{v_{1}\otimes v_{2}%
\otimes...\otimes v_{n}}\in U_{\leq n}\left(  \mathfrak{g}\right)  \text{
modulo }U_{\leq\left(  n-1\right)  }\left(  \mathfrak{g}\right)  \right) \\
&  -\left(  \text{the residue class of }\overline{v_{\tau_{i}\left(  1\right)
}\otimes v_{\tau_{i}\left(  2\right)  }\otimes...\otimes v_{\tau_{i}\left(
n\right)  }}\in U_{\leq n}\left(  \mathfrak{g}\right)  \text{ modulo }%
U_{\leq\left(  n-1\right)  }\left(  \mathfrak{g}\right)  \right) \\
&  =\left(  \text{the residue class of }\overline{v_{1}\otimes v_{2}%
\otimes...\otimes v_{n}-v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau
_{i}\left(  2\right)  }\otimes...\otimes v_{\tau_{i}\left(  n\right)  }}\in
U_{\leq n}\left(  \mathfrak{g}\right)  \text{ modulo }U_{\leq\left(
n-1\right)  }\left(  \mathfrak{g}\right)  \right)  ,
\end{align*}
so we must prove that $\overline{v_{1}\otimes v_{2}\otimes...\otimes
v_{n}-v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)
}\otimes...\otimes v_{\tau_{i}\left(  n\right)  }}\in U_{\leq\left(
n-1\right)  }\left(  \mathfrak{g}\right)  $. But since $\tau_{i}$ is the
transposition $\left(  i,i+1\right)  $, we have%
\begin{align*}
&  \overline{v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{\tau_{i}\left(
1\right)  }\otimes v_{\tau_{i}\left(  2\right)  }\otimes...\otimes v_{\tau
_{i}\left(  n\right)  }}\\
&  =\overline{v_{1}\otimes v_{2}\otimes...\otimes v_{n}-v_{1}\otimes
v_{2}\otimes...\otimes v_{i-1}\otimes v_{i+1}\otimes v_{i}\otimes
v_{i+2}\otimes v_{i+3}\otimes...\otimes v_{n}}\\
&  =\overline{v_{1}\otimes v_{2}\otimes...\otimes v_{i-1}\otimes\left(
v_{i}\otimes v_{i+1}-v_{i+1}\otimes v_{i}\right)  \otimes v_{i+2}\otimes
v_{i+3}\otimes...\otimes v_{n}}\\
&  =\overline{v_{1}\otimes v_{2}\otimes...\otimes v_{i-1}}\cdot
\underbrace{\overline{v_{i}\otimes v_{i+1}-v_{i+1}\otimes v_{i}}%
}_{\substack{=\overline{\left[  v_{i},v_{i+1}\right]  }\\\text{(since we are
in }U\left(  \mathfrak{g}\right)  \text{)}}}\cdot\overline{v_{i+2}\otimes
v_{i+3}\otimes...\otimes v_{n}}\\
&  =\overline{v_{1}\otimes v_{2}\otimes...\otimes v_{i-1}}\cdot\overline
{\left[  v_{i},v_{i+1}\right]  }\cdot\overline{v_{i+2}\otimes v_{i+3}%
\otimes...\otimes v_{n}}\\
&  =\overline{v_{1}\otimes v_{2}\otimes...\otimes v_{i-1}\otimes\left[
v_{i},v_{i+1}\right]  \otimes v_{i+2}\otimes v_{i+3}\otimes...\otimes v_{n}%
}\in U_{\leq\left(  n-1\right)  }\left(  \mathfrak{g}\right)  .
\end{align*}
This completes the proof of Proposition \ref{prop.PBWmap}.

\begin{definition}
\label{defs.PBWmap}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $n\in\mathbb{N}$.\newline\textbf{(a)} According to
Proposition \ref{prop.PBWmap} \textbf{(c)}, there exists a unique $k$-module
homomorphism $\mathbf{P}:\operatorname*{Sym}\nolimits^{n}\mathfrak{g}%
\rightarrow\operatorname*{gr}\nolimits_{n}\left(  U\left(  \mathfrak{g}%
\right)  \right)  $ such that $\operatorname*{gr}\nolimits_{n}\psi
\circ\operatorname{grad}_{\mathfrak{g},n}=\mathbf{P}\circ\operatorname*{sym}%
\nolimits_{\mathfrak{g},n}$ (where the notations used are those of Proposition
\ref{prop.PBWmap}). This homomorphism will be denoted by $\operatorname*{PBW}%
\nolimits_{\mathfrak{g},n}$ and called the $n$\textit{-PBW homomorphism} of
the Lie algebra $\mathfrak{g}$.\newline\textbf{(b)} We say that the Lie
algebra $\mathfrak{g}$ \textit{satisfies the }$n$\textit{-PBW condition} if
the $n$-PBW homomorphism $\operatorname*{PBW}\nolimits_{\mathfrak{g},n}$ is a
$k$-module isomorphism.
\end{definition}

\begin{remark}
The terminology "$n$-PBW isomorphism" and "$n$-PBW condition" is peculiar to
the author. Most people only speak of the "PBW isomorphism" (which is the
direct sum of the $n$-PBW isomorphisms over all $n\in\mathbb{N}$) and of the
"PBW condition" (which is the conjunction of the $n$-PBW conditions over all
$n\in\mathbb{N}$), as it is most often enough to consider all $n$ together.
However, in considering restricted Lie algebras in characteristic $p$ it is
often necessary to consider only the $n$-PBW conditions for $n<p$, so we
prefer to have a way to refer to the individual $n$-PBW conditions
separately.\newline Note that "PBW" is an abbreviation for
Poincar\'{e}-Birkhoff-Witt, and "BW" is an abbreviation for "Birkhoff-Witt".
Some older literature writes "BW" instead of "PBW", as Poincar\'{e}'s
discovery of universal enveloping algebras was forgotten for a long time.
\end{remark}

\subsection{The Poincar\'{e}-Birkhoff-Witt theorem}

There are several interrelated (but still different and non-equivalent) facts
referred to as Poincar\'{e}-Birkhoff-Witt theorems in literature. All of them
are similar in that each of them has a condition (such as: $k$ is a
$\mathbb{Q}$-algebra, or: $k$ is a field, or: $\mathfrak{g}$ is a free
$k$-module), under which they claim that the universal enveloping algebra
$U\left(  \mathfrak{g}\right)  $ of a $k$-Lie algebra $\mathfrak{g}$ is
"similar" in a certain way to the symmetric algebra $\operatorname*{Sym}%
\mathfrak{g}$ of the $k$-module $\mathfrak{g}$ as a $k$-module, as a
$\mathfrak{g}$-module, or as a $k$-algebra. What this "similar" means depends
on which of the various Poincar\'{e}-Birkhoff-Witt theorems we consider. There
is usually no isomorphism of $k$-algebras, but there is often an isomorphism
of $k$-modules, sometimes one of $\mathfrak{g}$-modules, and often one between
the associated graded $k$-algebras. Here we are interested in the latter. The
fact we are going to use is the following one:

\begin{proposition}
\label{prop.pbw}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie
algebra.\newline\textbf{(a)} Assume that the Lie algebra $\mathfrak{g}$
satisfies the $n$-PBW condition for every $n\in\mathbb{N}$. Then, the
associated graded $k$-algebra $\operatorname*{gr}\left(  U\left(
\mathfrak{g}\right)  \right)  $ is isomorphic to the symmetric algebra
$\operatorname*{Sym}\mathfrak{g}=\bigoplus\limits_{p\in\mathbb{N}%
}\operatorname{Sym}^{p}\mathfrak{g}$ as a $k$-algebra. (Here, the
\textit{associated graded }$k$\textit{-algebra }$\operatorname*{gr}\left(
U\left(  \mathfrak{g}\right)  \right)  $ is defined as the direct sum
$\bigoplus\limits_{p\in\mathbb{N}}\operatorname*{gr}_{p}\left(  U\left(
\mathfrak{g}\right)  \right)  $, with multiplication defined by%
\begin{align*}
\left(  \overline{u_{p}}\right)  _{p\in\mathbb{N}}\cdot\left(  \overline
{v_{p}}\right)  _{p\in\mathbb{N}}  &  =\left(  \overline{\sum_{i=0}^{p}%
u_{i}\cdot v_{p-i}}\right)  _{p\in\mathbb{N}}\\
&  \ \ \ \ \ \ \ \ \ \ \text{for every }\left(  \overline{u_{p}}\right)
_{p\in\mathbb{N}}\in\bigoplus\limits_{p\in\mathbb{N}}\operatorname{gr}%
_{p}\left(  U\left(  \mathfrak{g}\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \text{ and }\left(  \overline{v_{p}}\right)
_{p\in\mathbb{N}}\in\bigoplus\limits_{p\in\mathbb{N}}\operatorname{gr}%
_{p}\left(  U\left(  \mathfrak{g}\right)  \right)  ,
\end{align*}
where $\overline{u_{p}}$, $\overline{v_{p}}$ and $\overline{\sum
\limits_{i=0}^{p}u_{i}\cdot v_{p-i}}$ are to be understood as residue classes
of certain elements of $U_{\leq p}\left(  \mathfrak{g}\right)  $ modulo
$U_{\leq\left(  p-1\right)  }\left(  \mathfrak{g}\right)  $.)$\newline%
$\textbf{(b)} Let $n\in\mathbb{N}$. Assume that the Lie algebra $\mathfrak{g}$
satisfies the $n$-PBW condition. Then, $\operatorname*{Ker}\left(
\operatorname*{gr}\nolimits_{n}\psi\right)  =\operatorname*{grad}%
\nolimits_{\mathfrak{g},n}\left(  K_{n}\left(  \mathfrak{g}\right)  \right)
$. Here, $\psi$ is defined as in Proposition \ref{prop.PBWmap}, and
$K_{n}\left(  \mathfrak{g}\right)  $ is defined as in Definition
\ref{defs.symm-alg} (applied to $V=\mathfrak{g}$).
\end{proposition}

\textit{Proof of Proposition \ref{prop.pbw}.} \textbf{(a)} We assumed that the
Lie algebra $\mathfrak{g}$ satisfies the $n$-PBW condition for every
$n\in\mathbb{N}$. Thus, for every $n\in\mathbb{N}$, the map
$\operatorname{PBW}_{\mathfrak{g},n}:\operatorname*{Sym}\nolimits^{n}%
\mathfrak{g}\rightarrow\operatorname*{gr}\nolimits_{n}\left(  U\left(
\mathfrak{g}\right)  \right)  $ is a $k$-module isomorphism. The direct sum
$\bigoplus\limits_{p\in\mathbb{N}}\operatorname{PBW}_{\mathfrak{g},p}$ of
these maps is thus a $k$-module isomorphism from $\bigoplus\limits_{p\in
\mathbb{N}}\operatorname{Sym}^{p}\mathfrak{g}=\operatorname{Sym}\mathfrak{g}$
to $\bigoplus\limits_{p\in\mathbb{N}}\operatorname*{gr}_{p}\left(  U\left(
\mathfrak{g}\right)  \right)  =\operatorname*{gr}\left(  U\left(
\mathfrak{g}\right)  \right)  $. It only remains to show that this is a
$k$-algebra isomorphism. This is straightforward and left to the reader
(especially given that we will not be using this fact anyway).

\textbf{(b)} We assumed that the Lie algebra $\mathfrak{g}$ satisfies the
$n$-PBW condition. This means that the map $\operatorname{PBW}_{\mathfrak{g}%
,n}:\operatorname*{Sym}\nolimits^{n}\mathfrak{g}\rightarrow\operatorname*{gr}%
\nolimits_{n}\left(  U\left(  \mathfrak{g}\right)  \right)  $ is a $k$-module
isomorphism. Thus, $\operatorname{Ker}\left(  \operatorname{PBW}%
_{\mathfrak{g},n}\circ\operatorname*{sym}\nolimits_{\mathfrak{g},n}\right)
=\operatorname{Ker}\operatorname*{sym}\nolimits_{\mathfrak{g},n}=K_{n}\left(
\mathfrak{g}\right)  $ (since $\operatorname*{sym}\nolimits_{\mathfrak{g},n}$
is the projection of $\mathfrak{g}^{\otimes n}$ onto $\mathfrak{g}^{\otimes
n}\diagup K_{n}\left(  \mathfrak{g}\right)  $). But the definition of
$\operatorname{PBW}_{\mathfrak{g},n}$ yields $\operatorname*{gr}%
\nolimits_{n}\psi\circ\operatorname{grad}_{\mathfrak{g},n}=\operatorname{PBW}%
_{\mathfrak{g},n}\circ\operatorname*{sym}\nolimits_{\mathfrak{g},n}$. Thus,
$\operatorname{Ker}\left(  \operatorname{PBW}_{\mathfrak{g},n}\circ
\operatorname*{sym}\nolimits_{\mathfrak{g},n}\right)  =K_{n}\left(
\mathfrak{g}\right)  $ becomes $\operatorname{Ker}\left(  \operatorname*{gr}%
\nolimits_{n}\psi\circ\operatorname{grad}_{\mathfrak{g},n}\right)
=K_{n}\left(  \mathfrak{g}\right)  $. But $\operatorname{Ker}\left(
\operatorname*{gr}\nolimits_{n}\psi\circ\operatorname{grad}_{\mathfrak{g}%
,n}\right)  =\operatorname{grad}_{\mathfrak{g},n}^{-1}\left(
\operatorname{Ker}\left(  \operatorname*{gr}\nolimits_{n}\psi\right)  \right)
$ (since $\operatorname{grad}_{\mathfrak{g},n}$ is a $k$-module isomorphism).
Thus we get%
\[
K_{n}\left(  \mathfrak{g}\right)  =\operatorname{Ker}\left(
\operatorname*{gr}\nolimits_{n}\psi\circ\operatorname{grad}_{\mathfrak{g}%
,n}\right)  =\operatorname{grad}_{\mathfrak{g},n}^{-1}\left(
\operatorname{Ker}\left(  \operatorname*{gr}\nolimits_{n}\psi\right)  \right)
,
\]
so that $\operatorname*{Ker}\left(  \operatorname*{gr}\nolimits_{n}%
\psi\right)  =\operatorname*{grad}\nolimits_{\mathfrak{g},n}\left(
K_{n}\left(  \mathfrak{g}\right)  \right)  $ (again since $\operatorname{grad}%
_{\mathfrak{g},n}$ is an isomorphism). This proves Proposition \ref{prop.pbw}
\textbf{(b)}.

To get anything useful out of Proposition \ref{prop.pbw}, we need to know some
simple conditions under which the $n$-PBW condition is guaranteed to hold (the
$n$-PBW condition itself is rather hard to check in most cases, particularly
if we want to check it \textit{for all }$n\in\mathbb{N}$\textit{ at once}).
These are the essence of the Poincar\'{e}-Birkhoff-Witt
theorems:{\footnotemark}\footnotetext{Note that Theorem \ref{thm.pbw} (at
least part \textbf{(a)}, but probably some of the other parts as well) depends
on the axiom of choice (or at least its proof does). So do some of its
consequences which we are going to use.}

\begin{theorem}
\label{thm.pbw}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie
algebra. Let $n\in\mathbb{N}$.\newline\textbf{(a)} If $\mathfrak{g}$ is a free
$k$-module, then $\mathfrak{g}$ satisfies the $n$-PBW condition.\newline%
\textbf{(b)} If $k$ is a $\mathbb{Q}$-algebra, then $\mathfrak{g}$ satisfies
the $n$-PBW condition.\newline\textbf{(c) }If $\mathfrak{g}$ is a projective
$k$-module, then $\mathfrak{g}$ satisfies the $n$-PBW condition.\newline%
\textbf{(d)} If the (additive) abelian group $\mathfrak{g}$ is torsion-free,
then $\mathfrak{g}$ satisfies the $n$-PBW condition.\newline\textbf{(e)} If
$k$ is a Dedekind domain, then $\mathfrak{g}$ satisfies the $n$-PBW
condition.\newline\textbf{(f)} If the $k$-module $\mathfrak{g}$ is the direct
sum of cyclic modules (where a module is said to be \textit{cyclic} if it is
generated by one element), then $\mathfrak{g}$ satisfies the $n$-PBW
condition.\newline\textbf{(g)} If $\mathfrak{g}$ is a flat $k$-module, then
$\mathfrak{g}$ satisfies the $n$-PBW condition.
\end{theorem}

Calling this theorem "Poincar\'{e}-Birkhoff-Witt theorem" is an anachronism,
although a rather convenient one. The first to discover anything related to
this result was apparently Poincar\'{e} in 1900; it was a weak version of
Theorem \ref{thm.pbw} which required $k$ to be a field of characteristic $0$
and claimed that if $\left(  v_{1},v_{2},...,v_{m}\right)  $ is a basis of the
$k$-vector space $\mathfrak{g}$, then $\left(  \overline{v_{i_{1}}\otimes
v_{i_{2}}\otimes...\otimes v_{i_{n}}}\right)  _{\substack{n\in\mathbb{N}%
;\ \left(  i_{1},i_{2},...,i_{n}\right)  \in\left\{  1,2,...,m\right\}
^{n};\\i_{1}\leq i_{2}\leq...\leq i_{n}}}$ is a basis of the $k$-vector space
$U\left(  \mathfrak{g}\right)  $ (see Theorem \ref{thm.pbw.basis} for why this
is weaker than Theorem \ref{thm.pbw}). This can be shown to be equivalent to
the claim that $\mathfrak{g}$ satisfies the $n$-PBW condition for every
$n\in\mathbb{N}$. Whether Poincar\'{e}'s proof of this is correct is still a
matter of controversy. More historical details, along with a modernized
version of Poincar\'{e}'s original alleged proof, can be found in
\cite{ton-that}.

A comprehensive proof of Theorem \ref{thm.pbw} can be found in Higgins's paper
\cite[Theorems 6 and 7 and Corollary 2]{higgins} (where he proves all parts
except for \textbf{(d)} and \textbf{(g)}, but \textbf{(d)} can be derived from
\textbf{(b)} by tensoring with $\mathbb{Q}$, and \textbf{(g)} can be derived
from his Theorem 8 combined with Lazard's theorem that any flat module is a
direct limit of free modules). However, its parts \textbf{(a)} and
\textbf{(b)} are proven more frequently in different sources: Theorem
\ref{thm.pbw} \textbf{(a)} is shown in most references which consider the
Poincar\'{e}-Birkhoff-Witt theorem: for example, \cite[Theorem 1.3.1]{abe},
\cite[Theorem 8.2.2]{neisendorfer}, \cite[\S 17.3]{humphreys}, \cite[\S 2.1]%
{dixmier}, \cite[\S 1.9]{sternberg}, \cite[Theorem 5.15]{milnor-moore},
\cite[\S 2.7, Th\'{e}or\`{e}me 1]{bourbaki-1}, \cite[Theorem 3.3.1]%
{varadarajan}, \cite[Part I, Chapter III, Theorem 4.3]{serre}, \cite[Theorem
2]{booher}, \cite[Theorem 6.5]{lingle} give proofs (and while most of these
sources superficially require $k$ to be a field, the only condition they
actually use is that $\mathfrak{g}$ be a free $k$-module). \cite[Part 1,
Chapter 1, \S 1.3.7]{deligne} proves Theorem \ref{thm.pbw} \textbf{(b)} as
well, and so do \cite[\S 2.5]{petracci} and \cite[\S 5.5]{meinrenken}. Theorem
\ref{thm.pbw} \textbf{(e)} and \textbf{(f)} were also shown by Pierre Cartier
in \cite{cartier}.

In the case when $\mathfrak{g}$ is a free $k$-module, the following is also
known as "the Poincar\'{e}-Birkhoff-Witt theorem", as it is an equivalent
version of Theorem \ref{thm.pbw} \textbf{(a)}:

\begin{theorem}
\label{thm.pbw.basis}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Assume that the $k$-module $\mathfrak{g}$ has a basis
$\left(  e_{i}\right)  _{i\in I}$, where $I$ is a totally ordered set. Then,
$\left(  \overline{e_{i_{1}}\otimes e_{i_{2}}\otimes...\otimes e_{i_{n}}%
}\right)  _{\substack{n\in\mathbb{N};\ \left(  i_{1},i_{2},...,i_{n}\right)
\in I^{n};\\i_{1}\leq i_{2}\leq...\leq i_{n}}}$ is a basis of the $k$-module
$U\left(  \mathfrak{g}\right)  $.
\end{theorem}

\begin{condition}
\label{conv.Un}Here and in the following, we use the notation $U_{\leq
n}\left(  \mathfrak{g}\right)  $ for every $n\in\mathbb{N}$ as defined in
Proposition \ref{prop.PBWmap}. This means that for every $n\in\mathbb{N}$, we
denote by $U_{\leq n}\left(  \mathfrak{g}\right)  $ the $k$-submodule
$\psi\left(  \mathfrak{g}^{\otimes\leq n}\right)  $ of $U\left(
\mathfrak{g}\right)  $.
\end{condition}

Theorem \ref{thm.pbw.basis} appears, for instance, in literature as
\cite[Theorem 5.1.1]{beidar}, \cite[Theorem 3.1]{bergman-diamond},
\cite[Section V.2, Theorem 3]{jacobson}, \cite[Theorem 5]{booher},
\cite[Chapter XIII, Theorem 3.1]{cartan-eilenberg}, \cite[\S 2.7, Corollaire
3]{bourbaki-1}, \cite[Chapter III, Theorem 3.8]{knapp}\footnote{Note that
\cite[Chapter III, Theorem 3.8]{knapp} requires $k$ to be $\mathbb{C}$, but
this requirement is neither necessary for the theorem nor used in the proof.
The proof works just as well for the general case.} or \cite[Theorem
3.2.2]{varadarajan}. It can be easily derived from Theorem \ref{thm.pbw}
\textbf{(a)}, but the other direction is more standard: Almost all proofs of
Theorem \ref{thm.pbw} \textbf{(a)} proceed by deriving Theorem
\ref{thm.pbw.basis} first (\cite[\S 2.1]{dixmier}, \cite{humphreys} and
\cite[\S 1.9]{sternberg} are very explicit about doing so - for example,
Theorem \ref{thm.pbw.basis} is \cite[Lemma 2.1.8]{dixmier} and \cite[\S 17.3,
Corollary C]{humphreys}). Also, one of the most translucent proofs for Theorem
\ref{thm.pbw.basis} is given in \cite{garrett-pbw} and in \cite[\S 7.1]%
{procesi}.

\begin{remark}
Some sources prove Theorem \ref{thm.pbw} \textbf{(a)} only in the case when
$I$ is well-ordered (and not just totally ordered). However, once Theorem
\ref{thm.pbw} \textbf{(a)} is proven in this case, we can easily see that
Theorem \ref{thm.pbw} \textbf{(a)} holds for any totally ordered set $I$. Here
is why: The only difficult part of Theorem \ref{thm.pbw} \textbf{(a)} is the
linear independency of the family $\left(  \overline{e_{i_{1}}\otimes
e_{i_{2}}\otimes...\otimes e_{i_{n}}}\right)  _{\substack{n\in\mathbb{N}%
;\ \left(  i_{1},i_{2},...,i_{n}\right)  \in I^{n};\\i_{1}\leq i_{2}%
\leq...\leq i_{n}}}$. To prove this independency, it is enough to show that
the family $\left(  \overline{e_{i_{1}}\otimes e_{i_{2}}\otimes...\otimes
e_{i_{n}}}\right)  _{\substack{n\in\mathbb{N};\ \left(  i_{1},i_{2}%
,...,i_{n}\right)  \in J^{n};\\i_{1}\leq i_{2}\leq...\leq i_{n}}}$ is linearly
independent for every finite subset $J$ of $I$. But for every finite subset
$J$ of $I$, we can extend the ordering on $J$ to a well-ordering of $I$
(without changing the order of the elements of $J$), and apply the proof from
the well-ordered case.\newline This all requires the axiom of choice, but then
again, in all applications of Poincar\'{e}-Birkhoff-Witt I have seen, the
basis is countable and thus a well-ordering is easy to find.
\end{remark}

We note that a part of Theorem \ref{thm.pbw.basis} holds more generally:

\begin{proposition}
\label{prop.pbw.gen}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $\left(  e_{i}\right)  _{i\in I}$ be a generating set of
the $k$-module $\mathfrak{g}$, where $I$ is a totally ordered set. Then,
$\left(  \overline{e_{i_{1}}\otimes e_{i_{2}}\otimes...\otimes e_{i_{n}}%
}\right)  _{\substack{n\in\mathbb{N};\ \left(  i_{1},i_{2},...,i_{n}\right)
\in I^{n};\\i_{1}\leq i_{2}\leq...\leq i_{n}}}$ is a generating set of the
$k$-module $U\left(  \mathfrak{g}\right)  $.
\end{proposition}

This is actually the easier part of Theorem \ref{thm.pbw.basis}, and is proven
by induction in almost every text on Lie algebras. The same argument shows the
following strengthening of this proposition:

\begin{proposition}
\label{prop.pbw.gen.m}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $\left(  e_{i}\right)  _{i\in I}$ be a generating set of
the $k$-module $\mathfrak{g}$, where $I$ is a totally ordered set. Let
$m\in\mathbb{N}$. Then, the family $\left(  \overline{e_{i_{1}}\otimes
e_{i_{2}}\otimes...\otimes e_{i_{n}}}\right)  _{\substack{n\in\mathbb{N}%
;\ \left(  i_{1},i_{2},...,i_{n}\right)  \in I^{n};\\i_{1}\leq i_{2}%
\leq...\leq i_{n};\ n\leq m}}$ is a generating set of the $k$-module $U_{\leq
m}\left(  \mathfrak{g}\right)  $.
\end{proposition}

This proposition appears, e. g., in \cite[Lemma 2.1.6]{dixmier} and \cite[Part
I, Chapter III, Lemma 4.4]{serre} (although in an unnecessarily restrictive version).

We notice the following slight strengthening of Theorem \ref{thm.pbw.basis}:

\begin{corollary}
\label{cor.pbw.basis}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Assume that the $k$-module $\mathfrak{g}$ has a basis
$\left(  e_{i}\right)  _{i\in I}$, where $I$ is a totally ordered set. Let
$m\in\mathbb{N}$. Then, the family $\left(  \overline{e_{i_{1}}\otimes
e_{i_{2}}\otimes...\otimes e_{i_{n}}}\right)  _{\substack{n\in\mathbb{N}%
;\ \left(  i_{1},i_{2},...,i_{n}\right)  \in I^{n};\\i_{1}\leq i_{2}%
\leq...\leq i_{n};\ n\leq m}}$ is a basis of the $k$-module $U_{\leq m}\left(
\mathfrak{g}\right)  $.
\end{corollary}

\textit{Proof of Corollary \ref{cor.pbw.basis}.} The family $\left(
\overline{e_{i_{1}}\otimes e_{i_{2}}\otimes...\otimes e_{i_{n}}}\right)
_{\substack{n\in\mathbb{N};\ \left(  i_{1},i_{2},...,i_{n}\right)  \in
I^{n};\\i_{1}\leq i_{2}\leq...\leq i_{n};\ n\leq m}}$ is a generating set of
the $k$-module $U_{\leq m}\left(  \mathfrak{g}\right)  $ (by Proposition
\ref{prop.pbw.gen.m}) and linearly independent (by Theorem \ref{thm.pbw.basis}%
). Therefore it is a basis of the $k$-module $U_{\leq m}\left(  \mathfrak{g}%
\right)  $. This proves Corollary \ref{cor.pbw.basis}.

Note that while most literature does not explicitly mention Corollary
\ref{cor.pbw.basis}, it often tacitly uses it (for example, when deriving
Theorem \ref{thm.pbw} from Theorem \ref{thm.pbw.basis}).

Our next results are concerned with the case when $\mathfrak{h}$ is a free
$k$-module and satisfies $\mathfrak{g}=\mathfrak{h}\oplus N$ for some free
$k$-module $N$. This requirement is harsh in comparison to what we have
required in previous sections, but it still encompasses the situation when $k$
is a field, and besides is satisfied for many standard cases such as $\left(
\mathfrak{g},\mathfrak{h}\right)  =\left(  \mathfrak{gl}_{n}k,\mathfrak{sl}%
_{n}k\right)  $ even if $k$ is just a commutative ring with $1$.

We are going to prove the following consequence of Poincar\'{e}-Birkhoff-Witt:

\begin{proposition}
\label{prop.pbw.U(g)h}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie algebra. Let $m\in\mathbb{N}$.\newline Let $\mathfrak{h}$ be a Lie
subalgebra of $\mathfrak{g}$ such that $\mathfrak{h}$ is a free $k$-module and
such that there exists a free $k$-submodule $N$ of $\mathfrak{g}$ such that
$\mathfrak{g}=\mathfrak{h}\oplus N$.\newline Then, $U_{\leq m}\left(
\mathfrak{g}\right)  \cap\left(  U\left(  \mathfrak{g}\right)  \cdot
\mathfrak{h}\right)  =U_{\leq\left(  m-1\right)  }\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}$. (Here, we are using the notation of Definition
\ref{conv.Un}, and we are abbreviating the $k$-submodule $U\left(
\mathfrak{g}\right)  \cdot\psi\left(  \mathfrak{h}\right)  $ of $U\left(
\mathfrak{g}\right)  $ by $U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}$.)
\end{proposition}

Before we sketch a proof of Proposition \ref{prop.pbw.U(g)h}, let us recall a
basic from linear algebra:

\begin{lemma}
\label{lem.linalg.int}Let $k$ be a commutative ring. Let $V$ be a free
$k$-module with basis $\left(  \mathbf{e}_{\kappa}\right)  _{\kappa\in K}$,
where $K$ is a set. Let $X$ and $Y$ be two subsets of $K$. Then,%
\[
\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in X\right\rangle
\cap\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in Y\right\rangle
=\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in X\cap Y\right\rangle .
\]
(Here we are using Convention \ref{conv.<>}.)
\end{lemma}

\textit{Proof of Proposition \ref{prop.pbw.U(g)h}}. Let $\left(  e_{i}\right)
_{i\in P}$ be a basis of the $k$-module $\mathfrak{h}$, and let $\left(
e_{i}\right)  _{i\in Q}$ be a basis of the $k$-module $N$. Assume WLOG that
the sets $P$ and $Q$ are disjoint. Let $I=P\cup Q$. Choose a well-ordering on
$I$ such that every element of $Q$ is smaller than any element of $P$. Then,
$\left(  e_{i}\right)  _{i\in I}$ is a basis of the $k$-module $\mathfrak{h}%
\oplus N=\mathfrak{g}$. Proposition \ref{thm.pbw.basis} thus yields that
$\left(  \overline{e_{i_{1}}\otimes e_{i_{2}}\otimes...\otimes e_{i_{n}}%
}\right)  _{\substack{n\in\mathbb{N};\ \left(  i_{1},i_{2},...,i_{n}\right)
\in I^{n};\\i_{1}\leq i_{2}\leq...\leq i_{n}}}$ is a basis of the $k$-module
$U\left(  \mathfrak{g}\right)  $.

Let $I^{\ast}$ be the disjoint union of the sets $I^{n}$ for all
$n\in\mathbb{N}$. In other words, let $I^{\ast}$ be the set of all finite
sequences of elements of $I$. In particular, the empty sequence (i. e., the
only element of $I^{0}$) is an element of $I^{\ast}$.

For every two elements $\kappa\in I^{\ast}$ and $\kappa^{\prime}\in I^{\ast}$,
we define an element $\kappa\cdot\kappa^{\prime}\in I^{\ast}$ as follows:
Write $\kappa$ in the form $\kappa=\left(  i_{1},i_{2},...,i_{n}\right)  $ and
write $\kappa^{\prime}$ in the form $\kappa^{\prime}=\left(  j_{1}%
,j_{2},...,j_{m}\right)  $; then set $\kappa\cdot\kappa^{\prime}=\left(
i_{1},i_{2},...,i_{n},j_{1},j_{2},....,j_{m}\right)  $.

Let $K$ be the set%
\begin{align*}
&  \left\{  \left(  i_{1},i_{2},...,i_{n}\right)  \ \mid\ \left(  i_{1}%
,i_{2},...,i_{n}\right)  \in I^{\ast};\ i_{1}\leq i_{2}\leq...\leq
i_{n}\right\} \\
&  =\bigcup_{n\in\mathbb{N}}\left\{  \left(  i_{1},i_{2},...,i_{n}\right)
\ \mid\ \left(  i_{1},i_{2},...,i_{n}\right)  \in I^{n};\ i_{1}\leq i_{2}%
\leq...\leq i_{n}\right\}  .
\end{align*}
In other words, $K$ is the set of all increasing finite sequences of elements
of $I$.

Let us note that every $q\in K\cap Q^{\ast}$ and $p\in K\cap P^{\ast}$ satisfy
$q\cdot p\in K$. (This is because we have chosen a well-ordering on $I$ such
that every element of $Q$ is smaller than any element of $P$.)

For every $\kappa\in I^{\ast}$, define an element $\mathbf{e}_{\kappa}$ of
$U\left(  \mathfrak{g}\right)  $ by%
\[
\mathbf{e}_{\kappa}=\overline{e_{i_{1}}\otimes e_{i_{2}}\otimes...\otimes
e_{i_{n}}},\ \ \ \ \ \ \ \ \ \ \text{where }\left(  i_{1},i_{2},...,i_{n}%
\right)  \text{ is such that }\kappa=\left(  i_{1},i_{2},...,i_{n}\right)  .
\]


Then, $\left(  \mathbf{e}_{\kappa}\right)  _{\kappa\in K}$ is a basis of the
$k$-module $U\left(  \mathfrak{g}\right)  $ (since this is just another way to
state our knowledge that $\left(  \overline{e_{i_{1}}\otimes e_{i_{2}}%
\otimes...\otimes e_{i_{n}}}\right)  _{\substack{n\in\mathbb{N};\ \left(
i_{1},i_{2},...,i_{n}\right)  \in I^{n};\\i_{1}\leq i_{2}\leq...\leq i_{n}}}$
is a basis of the $k$-module $U\left(  \mathfrak{g}\right)  $). Thus,
$U\left(  \mathfrak{g}\right)  =\left\langle \mathbf{e}_{\kappa}\ \mid
\ \kappa\in K\right\rangle $.

Let $X$ be the subset $\left\{  \left(  i_{1},i_{2},...,i_{n}\right)  \in
K\ \mid\ i_{n}\in P\right\}  $ of $K$.

Let $Y$ be the subset $\left\{  \left(  i_{1},i_{2},...,i_{n}\right)  \in
K\ \mid\ n\leq m\right\}  $ of $K$.

Clearly, $X\cap Y$ is the subset $\left\{  \left(  i_{1},i_{2},...,i_{n}%
\right)  \in K\ \mid\ i_{n}\in P;\ n\leq m\right\}  $ of $K$.

\begin{noncompile}
Let $Z$ be the subset $\left\{  \left(  i_{1},i_{2},...,i_{n}\right)  \in
K\ \mid\ n\leq m-1\right\}  $ of $K$.
\end{noncompile}

We notice that $\left(  \mathbf{e}_{\kappa}\right)  _{\kappa\in Y}$ is a basis
of the $k$-module $U_{\leq m}\left(  \mathfrak{g}\right)  $ (since this is
just another way to state our knowledge that the family $\left(
\overline{e_{i_{1}}\otimes e_{i_{2}}\otimes...\otimes e_{i_{n}}}\right)
_{\substack{n\in\mathbb{N};\ \left(  i_{1},i_{2},...,i_{n}\right)  \in
I^{n};\\i_{1}\leq i_{2}\leq...\leq i_{n};\ n\leq m}}$ is a basis of the
$k$-module $U_{\leq m}\left(  \mathfrak{g}\right)  $).

\begin{noncompile}
Applying this very fact to $m-1$ instead of $m$, we see that $\left(
\mathbf{e}_{\kappa}\right)  _{\kappa\in Z}$ is a basis of the $k$-module
$U_{\leq\left(  m-1\right)  }\left(  \mathfrak{g}\right)  $.
\end{noncompile}

The definition of $\mathbf{e}_{\kappa}$ readily yields%
\begin{equation}
\mathbf{e}_{\kappa}\cdot\mathbf{e}_{\kappa^{\prime}}=\mathbf{e}_{\kappa
\cdot\kappa^{\prime}}\ \ \ \ \ \ \ \ \ \ \text{for every }\kappa\in I^{\ast
}\text{ and }\kappa^{\prime}\in I^{\ast}. \label{pf.pbw.U(g)h.0}%
\end{equation}
Also note that%
\begin{equation}
\mathbf{e}_{\left(  i\right)  }=e_{i}\ \ \ \ \ \ \ \ \ \ \text{for every }i\in
I, \label{pf.pbw.U(g)h.-1}%
\end{equation}
where $e_{i}$ means $\overline{e_{i}}=\psi\left(  e_{i}\right)  \in U\left(
\mathfrak{g}\right)  $ on the right hand side (this is a slight abuse of
notation, but legitimate in view of the Poincar\'{e}-Birkhoff-Witt theorem).

Now let us show that $U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}%
=\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in X\right\rangle $. In fact,
\begin{align}
\underbrace{U\left(  \mathfrak{g}\right)  }_{=\left\langle \mathbf{e}_{\kappa
}\ \mid\ \kappa\in K\right\rangle }\cdot\underbrace{\mathfrak{h}%
}_{\substack{=\left\langle e_{i}\ \mid\ i\in P\right\rangle \\\text{(since
}\left(  e_{i}\right)  _{i\in P}\text{ is a basis of }\mathfrak{h}\text{)}}}
&  =\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in K\right\rangle
\cdot\left\langle e_{i}\ \mid\ i\in P\right\rangle \nonumber\\
&  =\left\langle \mathbf{e}_{\kappa}e_{i}\ \mid\ \left(  \kappa,i\right)  \in
K\times P\right\rangle . \label{pf.pbw.U(g)h.3a}%
\end{align}
Hence, in order to prove that $U\left(  \mathfrak{g}\right)  \cdot
\mathfrak{h}=\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in X\right\rangle
$, it is enough to show that $\left\langle \mathbf{e}_{\kappa}e_{i}%
\ \mid\ \left(  \kappa,i\right)  \in K\times P\right\rangle =\left\langle
\mathbf{e}_{\kappa}\ \mid\ \kappa\in X\right\rangle $. In order to do so, we
must prove that%
\begin{equation}
\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in X\right\rangle
\subseteq\left\langle \mathbf{e}_{\kappa}e_{i}\ \mid\ \left(  \kappa,i\right)
\in K\times P\right\rangle \label{pf.pbw.U(g)h.1}%
\end{equation}
and%
\begin{equation}
\left\langle \mathbf{e}_{\kappa}e_{i}\ \mid\ \left(  \kappa,i\right)  \in
K\times P\right\rangle \subseteq\left\langle \mathbf{e}_{\kappa}\ \mid
\ \kappa\in X\right\rangle . \label{pf.pbw.U(g)h.2}%
\end{equation}


\textit{Proof of (\ref{pf.pbw.U(g)h.1}).} Every $\kappa\in X$ can be written
in the form $\left(  i_{1},i_{2},...,i_{n}\right)  $ for some $n\in\mathbb{N}$
and $\left(  i_{1},i_{2},...,i_{n}\right)  \in K$ satisfying $i_{n}\in P$.
Thus,%
\begin{align*}
\mathbf{e}_{\kappa}  &  =\mathbf{e}_{\left(  i_{1},i_{2},...,i_{n}\right)
}=\mathbf{e}_{\left(  i_{1},i_{2},...,i_{n-1}\right)  \cdot\left(
i_{n}\right)  }=\underbrace{\mathbf{e}_{\left(  i_{1},i_{2},...,i_{n-1}%
\right)  }}_{\in U\left(  \mathfrak{g}\right)  }\underbrace{\mathbf{e}%
_{\left(  i_{n}\right)  }}_{=e_{i_{n}}\text{ (by (\ref{pf.pbw.U(g)h.-1}))}%
}\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.pbw.U(g)h.0})}\right) \\
&  \in U\left(  \mathfrak{g}\right)  \underbrace{e_{i_{n}}}_{\in
\mathfrak{h}\text{ (since }i_{n}\in P\text{)}}\subseteq U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}.
\end{align*}
We have thus shown that every $\kappa\in X$ satisfies $\mathbf{e}_{\kappa}\in
U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}$. Therefore,%
\begin{equation}
\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in X\right\rangle \subseteq
U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}. \label{pf.pbw.U(g)h.1a}%
\end{equation}
Combined with (\ref{pf.pbw.U(g)h.3a}), this proves (\ref{pf.pbw.U(g)h.1}).

\textit{Proof of (\ref{pf.pbw.U(g)h.2}).} Let $\left(  \kappa,i\right)  \in
K\times P$ be arbitrary. We only have to prove that $\mathbf{e}_{\kappa}%
e_{i}\in\left\langle \mathbf{e}_{\kappa^{\prime}}\ \mid\ \kappa^{\prime}\in
X\right\rangle $ (because once this is shown for every $\left(  \kappa
,i\right)  \in K\times P$, it will become clear that $\left\langle
\mathbf{e}_{\kappa}e_{i}\ \mid\ \left(  \kappa,i\right)  \in K\times
P\right\rangle \subseteq\left\langle \mathbf{e}_{\kappa^{\prime}}%
\ \mid\ \kappa^{\prime}\in X\right\rangle =\left\langle \mathbf{e}_{\kappa
}\ \mid\ \kappa\in X\right\rangle $, and this will prove (\ref{pf.pbw.U(g)h.2})).

Since $\kappa\in K$, we can write $\kappa$ in the form $\left(  i_{1}%
,i_{2},...,i_{n}\right)  $ for some $n\in\mathbb{N}$ and $\left(  i_{1}%
,i_{2},...,i_{n}\right)  \in I^{n}$ satisfying $i_{1}\leq i_{2}\leq...\leq
i_{n}$. Let $i_{n+1}=i$. Let $\nu$ be the smallest integer in $\left\{
1,2,...,n+1\right\}  $ such that $i_{\nu}\in P$ (such a $\nu$ exists since
$i_{n+1}=i\in P$). Then, $i_{1}$, $i_{2}$, $...$, $i_{\nu-1}$ lie in $Q$
whereas $i_{\nu}$, $i_{\nu+1}$, $...$, $i_{n}$ lie in $P$ (this is because
$i_{1}\leq i_{2}\leq...\leq i_{n}$ and because every element of $Q$ is smaller
than any element of $P$). Since $i_{n+1}=i$ also lies in $P$, we conclude that
all of the elements $i_{\nu}$, $i_{\nu+1}$, $...$, $i_{n+1}$ lie in $P$. This
yields $\left(  i_{\nu},i_{\nu+1},...,i_{n+1}\right)  \in P^{\ast}$.

Since $i_{1}$, $i_{2}$, $...$, $i_{\nu-1}$ lie in $Q$, we have $\left(
i_{1},i_{2},...,i_{\nu-1}\right)  \in Q^{\ast}$. Combined with $\left(
i_{1},i_{2},...,i_{\nu-1}\right)  \in K$ (since $i_{1}\leq i_{2}\leq...\leq
i_{\nu-1}$), this yields $\left(  i_{1},i_{2},...,i_{\nu-1}\right)  \in K\cap
Q^{\ast}$.

Now we have%
\begin{align*}
\mathbf{e}_{\kappa}e_{i}  &  =\mathbf{e}_{\left(  i_{1},i_{2},...,i_{n}%
\right)  }e_{i}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\kappa=\left(
i_{1},i_{2},...,i_{n}\right)  \right) \\
&  =\mathbf{e}_{\left(  i_{1},i_{2},...,i_{n}\right)  }\underbrace{e_{i_{n+1}%
}}_{=\mathbf{e}_{\left(  i_{n+1}\right)  }\text{ (by (\ref{pf.pbw.U(g)h.-1}%
))}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }i=i_{n+1}\right) \\
&  =\mathbf{e}_{\left(  i_{1},i_{2},...,i_{n}\right)  }\mathbf{e}_{\left(
i_{n+1}\right)  }=\mathbf{e}_{\left(  i_{1},i_{2},...,i_{n}\right)
\cdot\left(  i_{n+1}\right)  }\ \ \ \ \ \ \ \ \ \ \left(  \text{by
(\ref{pf.pbw.U(g)h.0})}\right) \\
&  =\mathbf{e}_{\left(  i_{1},i_{2},...,i_{n+1}\right)  }=\mathbf{e}_{\left(
i_{1},i_{2},...,i_{\nu-1}\right)  \cdot\left(  i_{\nu},i_{\nu+1}%
,...,i_{n+1}\right)  }=\mathbf{e}_{\left(  i_{1},i_{2},...,i_{\nu-1}\right)
}\mathbf{e}_{\left(  i_{\nu},i_{\nu+1},...,i_{n+1}\right)  }%
\ \ \ \ \ \ \ \ \ \ \left(  \text{by (\ref{pf.pbw.U(g)h.0})}\right)  .
\end{align*}
But we can identify the universal enveloping algebra $U\left(  \mathfrak{h}%
\right)  $ with a $k$-submodule of $U\left(  \mathfrak{g}\right)  $ - namely,
with the $k$-submodule $\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in
P^{\ast}\right\rangle $ of $U\left(  \mathfrak{g}\right)  $. The element
$\mathbf{e}_{\left(  i_{\nu},i_{\nu+1},...,i_{n+1}\right)  }$ of $U\left(
\mathfrak{g}\right)  $ lies in this submodule $U\left(  \mathfrak{h}\right)  $
(because $\left(  i_{\nu},i_{\nu+1},...,i_{n+1}\right)  \in P^{\ast}$).

Applying Proposition \ref{prop.pbw.gen} to $\mathfrak{h}$ and $\left(
e_{i}\right)  _{i\in P}$ instead of $\mathfrak{g}$ and $\left(  e_{i}\right)
_{i\in I}$, we see that $\left(  \overline{e_{i_{1}}\otimes e_{i_{2}}%
\otimes...\otimes e_{i_{n}}}\right)  _{\substack{n\in\mathbb{N};\ \left(
i_{1},i_{2},...,i_{n}\right)  \in P^{n};\\i_{1}\leq i_{2}\leq...\leq i_{n}}}$
is a generating set of the $k$-module $U\left(  \mathfrak{h}\right)  $. In
other words,%
\begin{align*}
U\left(  \mathfrak{h}\right)   &  =\left\langle \underbrace{\overline
{e_{i_{1}}\otimes e_{i_{2}}\otimes...\otimes e_{i_{n}}}}_{=\mathbf{e}_{\left(
i_{1},i_{2},...,i_{n}\right)  }}\ \mid\ n\in\mathbb{N};\ \left(  i_{1}%
,i_{2},...,i_{n}\right)  \in P^{n};\ \underbrace{i_{1}\leq i_{2}\leq...\leq
i_{n}}_{\substack{\text{this is equivalent to}\\\left(  i_{1},i_{2}%
,...,i_{n}\right)  \in K}}\right\rangle \\
&  =\left\langle \mathbf{e}_{\left(  i_{1},i_{2},...,i_{n}\right)  }%
\ \mid\ n\in\mathbb{N};\ \left(  i_{1},i_{2},...,i_{n}\right)  \in
P^{n};\ \left(  i_{1},i_{2},...,i_{n}\right)  \in K\right\rangle \\
&  =\left\langle \mathbf{e}_{\left(  i_{1},i_{2},...,i_{n}\right)  }%
\ \mid\ n\in\mathbb{N};\ \left(  i_{1},i_{2},...,i_{n}\right)  \in K\cap
P^{n}\right\rangle =\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in K\cap
P^{\ast}\right\rangle .
\end{align*}
Therefore, $\mathbf{e}_{\left(  i_{\nu},i_{\nu+1},...,i_{n+1}\right)  }%
=\sum\limits_{\lambda\in K\cap P^{\ast}}\rho_{\lambda}\mathbf{e}_{\lambda}$
for some scalars $\rho_{\lambda}\in k$ (because $\mathbf{e}_{\left(  i_{\nu
},i_{\nu+1},...,i_{n+1}\right)  }\in U\left(  \mathfrak{h}\right)  $).
Consider these scalars $\rho_{\lambda}$. It is easily seen that $\rho_{\left(
{}\right)  }=0$, where $\left(  {}\right)  $ denotes the empty sequence (in
fact, if the map $\varepsilon_{U\left(  \mathfrak{g}\right)  }:U\left(
\mathfrak{g}\right)  \rightarrow k$ is defined as in Proposition
\ref{prop.Uhopf.easy}, then Proposition \ref{prop.Uhopf.easy} \textbf{(b)}
yields that $\varepsilon_{U\left(  \mathfrak{g}\right)  }\left(
\mathbf{e}_{\left(  i_{\nu},i_{\nu+1},...,i_{n+1}\right)  }\right)  =0$
(because $\left(  i_{\nu},i_{\nu+1},...,i_{n+1}\right)  $ is not the empty
sequence) on one hand but $\varepsilon_{U\left(  \mathfrak{g}\right)  }\left(
\sum\limits_{\lambda\in K\cap P^{\ast}}\rho_{\lambda}\mathbf{e}_{\lambda
}\right)  =\rho_{\left(  {}\right)  }$ on the other, so that we obtain
$0=\rho_{\left(  {}\right)  }$). Therefore, $\mathbf{e}_{\left(  i_{\nu
},i_{\nu+1},...,i_{n+1}\right)  }=\sum\limits_{\lambda\in K\cap P^{\ast}}%
\rho_{\lambda}\mathbf{e}_{\lambda}$ can be rewritten as%
\[
\mathbf{e}_{\left(  i_{\nu},i_{\nu+1},...,i_{n+1}\right)  }=\sum
\limits_{\substack{\lambda\in K\cap P^{\ast};\\\lambda\neq\left(  {}\right)
}}\rho_{\lambda}\mathbf{e}_{\lambda}%
\]
(here we removed the $\lambda=\left(  {}\right)  $ addend, since
$\rho_{\left(  {}\right)  }=0$).

Now,%
\begin{align*}
\mathbf{e}_{\kappa}e_{i}  &  =\mathbf{e}_{\left(  i_{1},i_{2},...,i_{\nu
-1}\right)  }\underbrace{\mathbf{e}_{\left(  i_{\nu},i_{\nu+1},...,i_{n+1}%
\right)  }}_{=\sum\limits_{\substack{\lambda\in K\cap P^{\ast};\\\lambda
\neq\left(  {}\right)  }}\rho_{\lambda}\mathbf{e}_{\lambda}}=\sum
\limits_{\substack{\lambda\in K\cap P^{\ast};\\\lambda\neq\left(  {}\right)
}}\rho_{\lambda}\underbrace{\mathbf{e}_{\left(  i_{1},i_{2},...,i_{\nu
-1}\right)  }\mathbf{e}_{\lambda}}_{\substack{=\mathbf{e}_{\left(  i_{1}%
,i_{2},...,i_{\nu-1}\right)  \cdot\lambda}\\\text{(by (\ref{pf.pbw.U(g)h.0}%
))}}}\\
&  =\sum\limits_{\substack{\lambda\in K\cap P^{\ast};\\\lambda\neq\left(
{}\right)  }}\rho_{\lambda}\mathbf{e}_{\left(  i_{1},i_{2},...,i_{\nu
-1}\right)  \cdot\lambda}.
\end{align*}
Now, every $\lambda\in K\cap P^{\ast}$ satisfies $\left(  i_{1},i_{2}%
,...,i_{\nu-1}\right)  \cdot\lambda\in K$ (because $\left(  i_{1}%
,i_{2},...,i_{\nu-1}\right)  \in K\cap Q^{\ast}$ and $\lambda\in K\cap
P^{\ast}$, and because every $q\in K\cap Q^{\ast}$ and $p\in K\cap P^{\ast}$
satisfy $q\cdot p\in K$). Therefore, every $\lambda\in K\cap P^{\ast}$ such
that $\lambda\neq\left(  {}\right)  $ satisfies $\left(  i_{1},i_{2}%
,...,i_{\nu-1}\right)  \cdot\lambda\in X$ (because $\left(  i_{1}%
,i_{2},...,i_{\nu-1}\right)  \cdot\lambda\in K$ on the one hand, but on the
other hand the last element of the sequence $\left(  i_{1},i_{2},...,i_{\nu
-1}\right)  \cdot\lambda$ lies in $P$\ \ \ \ {\footnotemark}\footnotetext{This
is because $\lambda\neq\left(  {}\right)  $ and $\lambda\in P^{\ast}$.}).
Thus,%
\[
\mathbf{e}_{\kappa}e_{i}=\sum\limits_{\substack{\lambda\in K\cap P^{\ast
};\\\lambda\neq\left(  {}\right)  }}\rho_{\lambda}\mathbf{e}_{\left(
i_{1},i_{2},...,i_{\nu-1}\right)  \cdot\lambda}\in\left\langle \mathbf{e}%
_{\kappa^{\prime}}\ \mid\ \kappa^{\prime}\in X\right\rangle
\]
(since $\left(  i_{1},i_{2},...,i_{\nu-1}\right)  \cdot\lambda\in X$ for every
$\lambda\in K\cap P^{\ast}$ such that $\lambda\neq\left(  {}\right)  $). This
proves (\ref{pf.pbw.U(g)h.2}).

We could also prove $U_{\leq\left(  m-1\right)  }\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}=\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in X\cap
Y\right\rangle $ by a similar argument, but this would be a slight overkill.
Instead we only need the weaker result that $\left\langle \mathbf{e}_{\kappa
}\ \mid\ \kappa\in X\cap Y\right\rangle \subseteq U_{\leq\left(  m-1\right)
}\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}$, which can be seen exactly
the same way as we have shown (\ref{pf.pbw.U(g)h.1a}).

Now, combining $U_{\leq\left(  m-1\right)  }\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}\subseteq\left(  U\left(  \mathfrak{g}\right)  \cdot
\mathfrak{h}\right)  \cap U_{\leq m}\left(  \mathfrak{g}\right)  $ (this is
because $\underbrace{U_{\leq\left(  m-1\right)  }\left(  \mathfrak{g}\right)
}_{\subseteq U\left(  \mathfrak{g}\right)  }\cdot\mathfrak{h}\subseteq
U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}$ and $U_{\leq\left(
m-1\right)  }\left(  \mathfrak{g}\right)  \cdot\underbrace{\mathfrak{h}%
}_{\subseteq\mathfrak{g}}\subseteq U_{\leq\left(  m-1\right)  }\left(
\mathfrak{g}\right)  \cdot\mathfrak{g}\subseteq U_{\leq m}\left(
\mathfrak{g}\right)  $) with
\begin{align*}
\underbrace{\left(  U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)
}_{=\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in X\right\rangle }%
\cap\underbrace{U_{\leq m}\left(  \mathfrak{g}\right)  }_{=\left\langle
\mathbf{e}_{\kappa}\ \mid\ \kappa\in Y\right\rangle }  &  =\left\langle
\mathbf{e}_{\kappa}\ \mid\ \kappa\in X\right\rangle \cap\left\langle
\mathbf{e}_{\kappa}\ \mid\ \kappa\in Y\right\rangle \\
&  =\left\langle \mathbf{e}_{\kappa}\ \mid\ \kappa\in X\cap Y\right\rangle
\ \ \ \ \ \ \ \ \ \ \left(  \text{by Lemma \ref{lem.linalg.int}}\right) \\
&  \subseteq U_{\leq\left(  m-1\right)  }\left(  \mathfrak{g}\right)
\cdot\mathfrak{h},
\end{align*}
we obtain $\left(  U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\cap U_{\leq m}\left(  \mathfrak{g}\right)  =U_{\leq\left(  m-1\right)
}\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}$. In other words, Proposition
\ref{prop.pbw.U(g)h} is proven.

\subsection{The kernel of $\operatorname*{gr}\nolimits_{n}\tau
:\operatorname*{gr}\nolimits_{n}\left(  \left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\right)  \rightarrow\operatorname*{gr}\nolimits_{n}\left(  U\left(
\mathfrak{g}\right)  \diagup\left(  U\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}\right)  \right)  $}

Next we prove a result that generalizes Lemma 4.3 of \cite{calaque1}:

\begin{theorem}
\label{thm.l42}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie
algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Let
$n\in\mathbb{N}$. Assume that the inclusion $\mathfrak{h}\hookrightarrow
\mathfrak{g}$ splits \textit{as a }$k$\textit{-module inclusion} (but not
necessarily as an $\mathfrak{h}$-module inclusion). This means that there
exists a $k$-submodule $N$ of $\mathfrak{g}$ such that $\mathfrak{g}%
=\mathfrak{h}\oplus N$. \newline Let us work with the notations introduced in
Theorem \ref{thm.l25} and in Definition \ref{defs.U}. Let also $\psi$ denote
the canonical projection from the $k$-algebra $\otimes\mathfrak{g}$ to the
factor algebra $\left(  \otimes\mathfrak{g}\right)  \diagup I_{\mathfrak{g}%
}=U\left(  \mathfrak{g}\right)  $. Clearly, $\psi$ is a $k$-algebra
homomorphism.\newline Let us abbreviate the $k$-submodule $U\left(
\mathfrak{g}\right)  \cdot\psi\left(  \mathfrak{h}\right)  $ of $U\left(
\mathfrak{g}\right)  $ by $U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}%
$.\newline Let $\rho$ be the canonical $k$-module projection $U\left(
\mathfrak{g}\right)  \rightarrow U\left(  \mathfrak{g}\right)  \diagup\left(
U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $.\newline\textbf{(a)}
There exists one and only one map $\theta:\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\rightarrow U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ for which the diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{5pc}
%\xymatrix{
%\otimes\mathfrak{g} \ar@{->>}[r]^{\psi} \ar@{->>}[d]_{\zeta} & U\left
%(\mathfrak{g}\right) \ar@{->>}[d]^{\rho} \\
%\left(\otimes\mathfrak{g}\right) \diagup\left(J + \left(\otimes\mathfrak
%{g}\right)\cdot\mathfrak{h}\right) \ar[r]_{\theta} & U\left(\mathfrak{g}%
%\right) \diagup\left(U\left(\mathfrak{g}\right)\cdot\mathfrak{h}\right)
%}} }%
%BeginExpansion
\xymatrixcolsep{5pc}
\xymatrix{
\otimes\mathfrak{g} \ar@{->>}[r]^{\psi} \ar@{->>}[d]_{\zeta} & U\left
(\mathfrak{g}\right) \ar@{->>}[d]^{\rho} \\
\left(\otimes\mathfrak{g}\right) \diagup\left(J + \left(\otimes\mathfrak
{g}\right)\cdot\mathfrak{h}\right) \ar[r]_{\theta} & U\left(\mathfrak{g}%
\right) \diagup\left(U\left(\mathfrak{g}\right)\cdot\mathfrak{h}\right)
}
%EndExpansion
\label{thm.l42.diag}%
\end{equation}
commutes. Denote this map $\theta$ by $\tau$.\newline\textbf{(b)} This $\tau$
is a surjective $\mathfrak{h}$-module homomorphism. Also, it satisfies
$\tau\circ\zeta=\rho\circ\psi$. In other words, the diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{5pc}
%\xymatrix{
%\otimes\mathfrak{g} \ar@{->>}[r]^{\psi} \ar@{->>}[d]_{\zeta} & U\left
%(\mathfrak{g}\right) \ar@{->>}[d]^{\rho} \\
%\left(\otimes\mathfrak{g}\right) \diagup\left(J + \left(\otimes\mathfrak
%{g}\right)\cdot\mathfrak{h}\right) \ar[r]_{\tau} & U\left(\mathfrak{g}%
%\right) \diagup\left(U\left(\mathfrak{g}\right)\cdot\mathfrak{h}\right)
%}} }%
%BeginExpansion
\xymatrixcolsep{5pc}
\xymatrix{
\otimes\mathfrak{g} \ar@{->>}[r]^{\psi} \ar@{->>}[d]_{\zeta} & U\left
(\mathfrak{g}\right) \ar@{->>}[d]^{\rho} \\
\left(\otimes\mathfrak{g}\right) \diagup\left(J + \left(\otimes\mathfrak
{g}\right)\cdot\mathfrak{h}\right) \ar[r]_{\tau} & U\left(\mathfrak{g}%
\right) \diagup\left(U\left(\mathfrak{g}\right)\cdot\mathfrak{h}\right)
}
%EndExpansion
\label{thm.l42.diag.real}%
\end{equation}
commutes.\newline\textbf{(c)} Every of the four corners of the commutative
square (\ref{thm.l42.diag.real}) is endowed with a filtration - namely as
follows:\newline- The filtration on $\otimes\mathfrak{g}$ is the degree
filtration $\left(  \mathfrak{g}^{\otimes\leq n}\right)  _{n\geq0}$.\newline-
The filtration on $\left(  \otimes\mathfrak{g}\right)  \diagup\left(
J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ is the
filtration $\left(  F_{n}\right)  _{n\geq0}$ defined in Theorem \ref{thm.l25}
\textbf{(b)} by $F_{n}=\zeta\left(  \mathfrak{g}^{\otimes\leq n}\right)
$.\newline- The filtration on $U\left(  \mathfrak{g}\right)  $ is the
filtration $\left(  U_{\leq n}\left(  \mathfrak{g}\right)  \right)  _{n\geq0}$
defined in Convention \ref{conv.Un} by $U_{\leq n}\left(  \mathfrak{g}\right)
=\psi\left(  \mathfrak{g}^{\otimes\leq n}\right)  $.\newline- The filtration
on $U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}\right)  $ is the filtration $\left(  W_{n}\right)  _{n\geq
0}$ defined by $W_{n}=\left(  \tau\circ\zeta\right)  \left(  \mathfrak{g}%
^{\otimes\leq n}\right)  $.\newline Then, $W_{n}=\left(  \rho\circ\psi\right)
\left(  \mathfrak{g}^{\otimes\leq n}\right)  =\rho\left(  U_{\leq n}\left(
\mathfrak{g}\right)  \right)  =\tau\left(  F_{n}\right)  $. Also, the maps
$\psi$, $\rho$, $\zeta$ and $\tau$ all respect the filtration. Therefore, for
every $n\in\mathbb{N}$, we can apply the functor $\operatorname*{gr}%
\nolimits_{n}$ to the diagram (\ref{thm.l42.diag.real}), and obtain the
commutative diagram%
\begin{equation}%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{5pc}
%\xymatrix{
%\operatorname*{gr}_n\left(\otimes\mathfrak{g}\right) \ar@{->>}%
%[r]^{\operatorname*{gr}_n\psi} \ar@{->>}[d]_{\operatorname*{gr}_n\zeta}
%& \operatorname*{gr}_n\left(U\left(\mathfrak{g}\right)\right) \ar@
%{->>}[d]^{\operatorname*{gr}_n\rho} \\
%\operatorname*{gr}_n\left(\left(\otimes\mathfrak{g}\right) \diagup
%\left(J + \left(\otimes\mathfrak{g}\right)\cdot\mathfrak{h}\right)\right
%) \ar[r]_{\operatorname*{gr}_n\tau} & \operatorname*{gr}_n\left(U\left
%(\mathfrak{g}\right) \diagup\left(U\left(\mathfrak{g}\right)\cdot\mathfrak
%{h}\right)\right)
%}}}%
%BeginExpansion
\xymatrixcolsep{5pc}
\xymatrix{
\operatorname*{gr}_n\left(\otimes\mathfrak{g}\right) \ar@{->>}%
[r]^{\operatorname*{gr}_n\psi} \ar@{->>}[d]_{\operatorname*{gr}_n\zeta}
& \operatorname*{gr}_n\left(U\left(\mathfrak{g}\right)\right) \ar@
{->>}[d]^{\operatorname*{gr}_n\rho} \\
\operatorname*{gr}_n\left(\left(\otimes\mathfrak{g}\right) \diagup
\left(J + \left(\otimes\mathfrak{g}\right)\cdot\mathfrak{h}\right)\right
) \ar[r]_{\operatorname*{gr}_n\tau} & \operatorname*{gr}_n\left(U\left
(\mathfrak{g}\right) \diagup\left(U\left(\mathfrak{g}\right)\cdot\mathfrak
{h}\right)\right)
}%
%EndExpansion
. \label{thm.l42.diag.gr}%
\end{equation}
\newline\textbf{(d)} Assume that both $\mathfrak{h}$ and $N$ are free
$k$-modules. Let $n\in\mathbb{N}$. Then, $\left(  \operatorname*{grad}%
\nolimits_{\mathfrak{n},n}^{-1}\circ\omega_{n}\right)  \left(
\operatorname*{Ker}\left(  \operatorname*{gr}\nolimits_{n}\tau\right)
\right)  =K_{n}\left(  \mathfrak{n}\right)  $. (For the definition of
$K_{n}\left(  \mathfrak{n}\right)  $, see Definition \ref{defs.symm-alg},
applied to $V=\mathfrak{n}$. For the definition of $\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}$ and $\omega_{n}$, see Theorem \ref{thm.l25}
\textbf{(c)}.)
\end{theorem}

The condition that both $\mathfrak{h}$ and $N$ are free in \textbf{(d)} is
somewhat restrictive - we will partly lift it in Subsection
\ref{subsect.goodwillie}.

\textit{Proof of Theorem \ref{thm.l42}.} \textbf{(a)} Let $J_{0}$ denote the
$k$-submodule
\[
\left\langle v\otimes w-w\otimes v-\left[  v,w\right]  \ \mid\ \left(
v,w\right)  \in\mathfrak{g}\times\mathfrak{h}\right\rangle
\]
of $\otimes\mathfrak{g}$. Then, $J=\left(  \otimes\mathfrak{g}\right)  \cdot
J_{0}\cdot\left(  \otimes\mathfrak{g}\right)  $ (by Proposition
\ref{prop.l25a} \textbf{(b)}). On the other hand,%
\begin{align*}
J_{0}  &  =\left\langle v\otimes w-w\otimes v-\left[  v,w\right]
\ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}\right\rangle \\
&  \subseteq\left\langle v\otimes w-w\otimes v-\left[  v,w\right]
\ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{g}\right\rangle
\end{align*}
(since $\mathfrak{g}\times\mathfrak{h}\subseteq\mathfrak{g}\times\mathfrak{g}%
$). Now
\begin{align*}
J  &  =\left(  \otimes\mathfrak{g}\right)  \cdot\underbrace{J_{0}}%
_{\subseteq\left\langle v\otimes w-w\otimes v-\left[  v,w\right]
\ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{g}\right\rangle
}\cdot\left(  \otimes\mathfrak{g}\right) \\
&  \subseteq\left(  \otimes\mathfrak{g}\right)  \cdot\left\langle v\otimes
w-w\otimes v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)  \in
\mathfrak{g}\times\mathfrak{g}\right\rangle \cdot\left(  \otimes
\mathfrak{g}\right)  =I_{\mathfrak{g}}.
\end{align*}


With the help of%
\begin{align*}
\psi\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
&  =\psi\left(  \underbrace{J}_{\subseteq I_{\mathfrak{g}}}\right)
+\underbrace{\psi\left(  \left(  \otimes\mathfrak{g}\right)  \cdot
\mathfrak{h}\right)  }_{\substack{=\psi\left(  \otimes\mathfrak{g}\right)
\cdot\psi\left(  \mathfrak{h}\right)  \\\text{(since }\psi\text{ is a
}k\text{-algebra}\\\text{homomorphism)}}}\subseteq\underbrace{\psi\left(
I_{\mathfrak{g}}\right)  }_{\substack{=0\text{ (since }\psi\text{ is
the}\\\text{projection on }\left(  \otimes\mathfrak{g}\right)  \diagup
I_{\mathfrak{g}}\text{)}}}+\underbrace{\psi\left(  \otimes\mathfrak{g}\right)
}_{\subseteq U\left(  \mathfrak{g}\right)  }\cdot\psi\left(  \mathfrak{h}%
\right) \\
&  \subseteq0+U\left(  \mathfrak{g}\right)  \cdot\psi\left(  \mathfrak{h}%
\right)  =U\left(  \mathfrak{g}\right)  \cdot\psi\left(  \mathfrak{h}\right)
=U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h},
\end{align*}
we obtain%
\[
\left(  \rho\circ\psi\right)  \left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  =\rho\left(  \underbrace{\psi\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  }_{\subseteq U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}}\right)  \subseteq\rho\left(  U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  =0
\]
(since $\rho$ is the projection on $U\left(  \mathfrak{g}\right)
\diagup\left(  U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $). In
other words, $\left(  \rho\circ\psi\right)  \left(  J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  =0$.

Thus, by the homomorphism theorem, the map $\rho\circ\psi:\otimes
\mathfrak{g}\rightarrow U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ factors through the factor
map $\zeta:\otimes\mathfrak{g}\rightarrow\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
$. In other words, there exists one and only one map $\theta:\left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  \rightarrow U\left(  \mathfrak{g}\right)
\diagup\left(  U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $
satisfying $\theta\circ\zeta=\rho\circ\psi$, and this map $\theta$ is a
$k$-module homomorphism. Since $\theta\circ\zeta=\rho\circ\psi$ is equivalent
to the commutativity of the diagram (\ref{thm.l42.diag}), this result rewrites
as follows: There exists one and only one map $\theta:\left(  \otimes
\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  \rightarrow U\left(  \mathfrak{g}\right)
\diagup\left(  U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ for
which the diagram (\ref{thm.l42.diag}) commutes, and this map $\theta$ is a
$k$-module homomorphism. This proves Theorem \ref{thm.l42} \textbf{(a)}, and
even a part of Theorem \ref{thm.l42} \textbf{(b) }(namely, the part saying
that $\tau$ is a $k$-module homomorphism).

\textbf{(b)} We have already proven that $\tau$ is a $k$-module homomorphism.
Now, we need only show that $\tau$ is surjective and an $\mathfrak{h}$-module homomorphism.

By the definition of $\tau$, the diagram (\ref{thm.l42.diag}) commutes for
$\theta=\tau$. That is, $\tau\circ\zeta=\rho\circ\psi$. Since the maps $\rho$
and $\psi$ are surjective (because $\rho$ and $\psi$ are projections), their
composition $\rho\circ\psi$ is surjective. Thus, $\tau\circ\zeta=\rho\circ
\psi$ is surjective, so that $\tau$ must too be surjective.

We know that $\zeta$ is surjective, and that $\tau\circ\zeta$ is an
$\mathfrak{h}$-module homomorphism (since $\tau\circ\zeta=\rho\circ\psi$, and
since both $\rho$ and $\psi$ are $\mathfrak{h}$-module homomorphisms).
Applying Lemma \ref{lem.surj} to $\otimes\mathfrak{g}$, $\left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  $, $U\left(  \mathfrak{g}\right)
\diagup\left(  U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $,
$\zeta$ and $\tau$ instead of $A$, $B$, $C$, $f$ and $g$, we thus obtain that
$\tau$ is an $\mathfrak{h}$-module homomorphism.

This completes the proof of Theorem \ref{thm.l42} \textbf{(b)}.

\textbf{(c)} It is pretty much trivial that $\left(  W_{n}\right)  _{n\geq0}$
is indeed a filtration of $U\left(  \mathfrak{g}\right)  \diagup\left(
U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ (all we use here is
that $\tau\circ\zeta$ is surjective). Now, $W_{n}=\left(  \tau\circ
\zeta\right)  \left(  \mathfrak{g}^{\otimes\leq n}\right)  $ yields
$W_{n}=\left(  \rho\circ\psi\right)  \left(  \mathfrak{g}^{\otimes\leq
n}\right)  $ because of $\tau\circ\zeta=\rho\circ\psi$. Also, $W_{n}=\left(
\tau\circ\zeta\right)  \left(  \mathfrak{g}^{\otimes\leq n}\right)
=\tau\left(  \underbrace{\zeta\left(  \mathfrak{g}^{\otimes\leq n}\right)
}_{=F_{n}}\right)  =\tau\left(  F_{n}\right)  $ and $W_{n}=\left(  \rho
\circ\psi\right)  \left(  \mathfrak{g}^{\otimes\leq n}\right)  =\rho\left(
\underbrace{\psi\left(  \mathfrak{g}^{\otimes\leq n}\right)  }_{=U_{\leq
n}\left(  \mathfrak{g}\right)  }\right)  =\rho\left(  U_{\leq n}\left(
\mathfrak{g}\right)  \right)  $. It is now absolutely obvious that the maps
$\psi$, $\rho$, $\zeta$ and $\tau$ all respect the filtration. Theorem
\ref{thm.l42} \textbf{(c)} is now proven.

\textbf{(d)} Assume that both $\mathfrak{h}$ and $N$ are free $k$-modules.
This, of course, yields that $\mathfrak{g}=\mathfrak{h}\oplus N$ must also be
free, and thus $\mathfrak{g}$ satisfies the $n$-PBW condition (by Theorem
\ref{thm.pbw} \textbf{(a)}). Thus, Proposition \ref{prop.pbw} \textbf{(b)}
yields $\operatorname*{Ker}\left(  \operatorname*{gr}\nolimits_{n}\psi\right)
=\operatorname*{grad}\nolimits_{\mathfrak{g},n}\left(  K_{n}\left(
\mathfrak{g}\right)  \right)  $. In other words, $\operatorname*{grad}%
\nolimits_{\mathfrak{g},n}^{-1}\left(  \operatorname*{Ker}\left(
\operatorname*{gr}\nolimits_{n}\psi\right)  \right)  =K_{n}\left(
\mathfrak{g}\right)  $ (since $\operatorname*{grad}\nolimits_{\mathfrak{g},n}$
is an isomorphism).

On the other hand, the construction of $K_{n}\left(  V\right)  $ for a
$k$-module $V$ yields that whenever $f:A\rightarrow B$ is a surjective
$k$-linear map between two $k$-modules $A$ and $B$, then $f^{\otimes n}\left(
K_{n}\left(  A\right)  \right)  =K_{n}\left(  B\right)  $. Applying this to
$A=\mathfrak{g}$, $B=\mathfrak{n}$ and $f=\pi$, we obtain $\pi^{\otimes
n}\left(  K_{n}\left(  \mathfrak{g}\right)  \right)  =K_{n}\left(
\mathfrak{n}\right)  $.

Our goal is to show that $\left(  \operatorname*{grad}\nolimits_{\mathfrak{n}%
,n}^{-1}\circ\omega_{n}\right)  \left(  \operatorname*{Ker}\left(
\operatorname*{gr}\nolimits_{n}\tau\right)  \right)  =K_{n}\left(
\mathfrak{n}\right)  $. This will be done once we have proven that%
\begin{equation}
\operatorname*{Ker}\left(  \operatorname*{gr}\nolimits_{n}\tau\right)
=\left(  \operatorname*{gr}\nolimits_{n}\zeta\right)  \left(
\operatorname*{Ker}\left(  \operatorname*{gr}\nolimits_{n}\psi\right)
\right)  . \label{pf.l42.1}%
\end{equation}
In fact, once (\ref{pf.l42.1}) is shown, we have%
\begin{align*}
&  \left(  \operatorname*{grad}\nolimits_{\mathfrak{n},n}^{-1}\circ\omega
_{n}\right)  \left(  \underbrace{\operatorname*{Ker}\left(  \operatorname*{gr}%
\nolimits_{n}\tau\right)  }_{=\left(  \operatorname*{gr}\nolimits_{n}%
\zeta\right)  \left(  \operatorname*{Ker}\left(  \operatorname*{gr}%
\nolimits_{n}\psi\right)  \right)  }\right) \\
&  =\left(  \operatorname*{grad}\nolimits_{\mathfrak{n},n}^{-1}\circ\omega
_{n}\right)  \left(  \left(  \operatorname*{gr}\nolimits_{n}\zeta\right)
\left(  \operatorname*{Ker}\left(  \operatorname*{gr}\nolimits_{n}\psi\right)
\right)  \right)  =\left(  \operatorname*{grad}\nolimits_{\mathfrak{n},n}%
^{-1}\circ\underbrace{\omega_{n}\circ\operatorname*{gr}\nolimits_{n}\zeta
}_{\substack{=\operatorname*{gr}\nolimits_{n}\left(  \otimes\pi\right)
\\\text{(since the diagram}\\\text{(\ref{thm.l25.c.diag'}) commutes)}%
}}\right)  \left(  \operatorname*{Ker}\left(  \operatorname*{gr}%
\nolimits_{n}\psi\right)  \right) \\
&  =\underbrace{\left(  \operatorname*{grad}\nolimits_{\mathfrak{n},n}%
^{-1}\circ\operatorname*{gr}\nolimits_{n}\left(  \otimes\pi\right)  \right)
}_{\substack{=\pi^{\otimes n}\circ\operatorname*{grad}\nolimits_{\mathfrak{g}%
,n}^{-1}\\\text{(this follows easily from the}\\\text{definitions of
}\operatorname*{grad}\nolimits_{\mathfrak{g},n}\text{ and }%
\operatorname*{grad}\nolimits_{\mathfrak{n},n}\text{)}}}\left(
\operatorname*{Ker}\left(  \operatorname*{gr}\nolimits_{n}\psi\right)
\right)  =\left(  \pi^{\otimes n}\circ\operatorname*{grad}%
\nolimits_{\mathfrak{g},n}^{-1}\right)  \left(  \operatorname*{Ker}\left(
\operatorname*{gr}\nolimits_{n}\psi\right)  \right) \\
&  =\pi^{\otimes n}\left(  \underbrace{\operatorname*{grad}%
\nolimits_{\mathfrak{g},n}^{-1}\left(  \operatorname*{Ker}\left(
\operatorname*{gr}\nolimits_{n}\psi\right)  \right)  }_{=K_{n}\left(
\mathfrak{g}\right)  }\right)  =\pi^{\otimes n}\left(  K_{n}\left(
\mathfrak{g}\right)  \right)  =K_{n}\left(  \mathfrak{n}\right)  ,
\end{align*}
which proves Theorem \ref{thm.l42} \textbf{(d)}. So, in order to complete the
proof of Theorem \ref{thm.l42} \textbf{(d)}, the only thing we need to do is
verify (\ref{pf.l42.1}).

According to Proposition \ref{prop.9l} (applied to the diagram
(\ref{thm.l42.diag.gr}) instead of the diagram (\ref{prop.9l.diag})), the
equality (\ref{pf.l42.1}) will follow once we can show that
$\operatorname*{gr}\nolimits_{n}\zeta$ is surjective and that%
\begin{equation}
\operatorname*{Ker}\left(  \operatorname*{gr}\nolimits_{n}\rho\right)
\subseteq\left(  \operatorname*{gr}\nolimits_{n}\psi\right)  \left(
\operatorname*{Ker}\left(  \operatorname*{gr}\nolimits_{n}\zeta\right)
\right)  . \label{pf.l42.2}%
\end{equation}
But it is easy to see that $\operatorname*{gr}\nolimits_{n}\zeta$ is
surjective (and it was proven in detail during the proof of Proposition
\ref{prop.l25.Ker-alpha} \textbf{(e)}). Thus, we only have to prove
(\ref{pf.l42.2}) now.

Due to the commutative diagram (\ref{thm.l25.c.diag'}), and due to the fact
that $\omega_{n}$ is an isomorphism, we have $\operatorname*{Ker}\left(
\operatorname*{gr}\nolimits_{n}\zeta\right)  =\operatorname*{Ker}\left(
\operatorname*{gr}\nolimits_{n}\left(  \otimes\pi\right)  \right)  $.

Let $s\in\operatorname*{Ker}\left(  \operatorname*{gr}\nolimits_{n}%
\rho\right)  $ be arbitrary. Then, $s\in\operatorname*{gr}\nolimits_{n}\left(
U\left(  \mathfrak{g}\right)  \right)  =\left(  U_{\leq n}\left(
\mathfrak{g}\right)  \right)  \diagup\left(  U_{\leq\left(  n-1\right)
}\left(  \mathfrak{g}\right)  \right)  $, so that there exists some $S\in
U_{\leq n}\left(  \mathfrak{g}\right)  $ such that $s=\overline{S}%
$.\ \ \ \ {\footnotemark}\footnotetext{Henceforth until the end of this proof
of Theorem \ref{thm.l42}, the term $\overline{S}$ always denotes the residue
class of $S$ modulo $U_{\leq\left(  n-1\right)  }\left(  \mathfrak{g}\right)
$. Similarly, the terms $\overline{\rho\left(  S\right)  }$ and $\overline
{S-S^{\prime}}$ will have to be interpreted.} Consider this $S$. Since
$s\in\operatorname*{Ker}\left(  \operatorname*{gr}\nolimits_{n}\rho\right)  $,
we have $\left(  \operatorname*{gr}\nolimits_{n}\rho\right)  \left(  s\right)
=0$. But $s=\overline{S}$ shows that $\left(  \operatorname*{gr}%
\nolimits_{n}\rho\right)  \left(  s\right)  =\left(  \operatorname*{gr}%
\nolimits_{n}\rho\right)  \left(  \overline{S}\right)  =\overline{\rho\left(
S\right)  }$, so that $\left(  \operatorname*{gr}\nolimits_{n}\rho\right)
\left(  s\right)  =0$ becomes $\overline{\rho\left(  S\right)  }=0$. In other
words, $\rho\left(  S\right)  \in W_{n-1}=\rho\left(  U_{\leq\left(
n-1\right)  }\left(  \mathfrak{g}\right)  \right)  $. This means that there
exists some $S^{\prime}\in U_{\leq\left(  n-1\right)  }\left(  \mathfrak{g}%
\right)  $ such that $\rho\left(  S\right)  =\rho\left(  S^{\prime}\right)  $.
Consider this $S^{\prime}$. Then, $\rho\left(  S\right)  =\rho\left(
S^{\prime}\right)  $ yields $0=\rho\left(  S\right)  -\rho\left(  S^{\prime
}\right)  =\rho\left(  S-S^{\prime}\right)  $, so that $S-S^{\prime}%
\in\operatorname*{Ker}\rho=U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}$.
On the other hand, $S\in U_{\leq n}\left(  \mathfrak{g}\right)  $ and
$S^{\prime}\in U_{\leq\left(  n-1\right)  }\left(  \mathfrak{g}\right)
\subseteq U_{\leq n}\left(  \mathfrak{g}\right)  $ lead to $S-S^{\prime}\in
U_{\leq n}\left(  \mathfrak{g}\right)  $ (since $U_{\leq n}\left(
\mathfrak{g}\right)  $ is a $k$-module). Combining this with $S-S^{\prime}\in
U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}$, we obtain%
\begin{align*}
S-S^{\prime}  &  \in U_{\leq n}\left(  \mathfrak{g}\right)  \cap\left(
U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  =U_{\leq\left(
n-1\right)  }\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\\
&  \ \ \ \ \ \ \ \ \ \ \left(  \text{by Proposition \ref{prop.pbw.U(g)h},
applied to }n\text{ instead of }m\right) \\
&  =\psi\left(  \mathfrak{g}^{\otimes\leq\left(  n-1\right)  }\cdot
\mathfrak{h}\right)  .
\end{align*}
Thus,
\begin{align*}
\overline{S-S^{\prime}}  &  \in\overline{\psi\left(  \mathfrak{g}^{\otimes
\leq\left(  n-1\right)  }\cdot\mathfrak{h}\right)  }\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, }\overline{\psi\left(  \mathfrak{g}^{\otimes\leq\left(
n-1\right)  }\cdot\mathfrak{h}\right)  }\text{ means the image of }\psi\left(
\mathfrak{g}^{\otimes\leq\left(  n-1\right)  }\cdot\mathfrak{h}\right) \\
\text{under the canonical projection }U_{\leq n}\left(  \mathfrak{g}\right)
\rightarrow\operatorname{gr}_{n}\left(  U\left(  \mathfrak{g}\right)  \right)
\end{array}
\right) \\
&  =\left(  \operatorname*{gr}\nolimits_{n}\psi\right)  \left(
\underbrace{\overline{\mathfrak{g}^{\otimes\leq\left(  n-1\right)  }%
\cdot\mathfrak{h}}}_{\subseteq\operatorname*{Ker}\left(  \operatorname*{gr}%
\nolimits_{n}\left(  \otimes\pi\right)  \right)  =\operatorname*{Ker}\left(
\operatorname*{gr}\nolimits_{n}\zeta\right)  }\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{here, }\overline{\mathfrak{g}^{\otimes\leq\left(  n-1\right)  }%
\cdot\mathfrak{h}}\text{ means the image of }\mathfrak{g}^{\otimes\leq\left(
n-1\right)  }\cdot\mathfrak{h}\\
\text{under the canonical projection }\mathfrak{g}^{\otimes\leq n}%
\rightarrow\operatorname{gr}_{n}\left(  \otimes\mathfrak{g}\right)
\end{array}
\right) \\
&  \subseteq\left(  \operatorname*{gr}\nolimits_{n}\psi\right)  \left(
\operatorname*{Ker}\left(  \operatorname*{gr}\nolimits_{n}\zeta\right)
\right)  .
\end{align*}
Now,%
\begin{align*}
s  &  =\overline{S}=\overline{S-S^{\prime}}\ \ \ \ \ \ \ \ \ \ \left(
\text{since }S^{\prime}\in U_{\leq\left(  n-1\right)  }\left(  \mathfrak{g}%
\right)  \right) \\
&  \in\left(  \operatorname*{gr}\nolimits_{n}\psi\right)  \left(
\operatorname*{Ker}\left(  \operatorname*{gr}\nolimits_{n}\zeta\right)
\right)  .
\end{align*}
Since this is shown for every $s\in\operatorname*{Ker}\left(
\operatorname*{gr}\nolimits_{n}\rho\right)  $, we thus conclude that
$\operatorname*{Ker}\left(  \operatorname*{gr}\nolimits_{n}\rho\right)
\subseteq\left(  \operatorname*{gr}\nolimits_{n}\psi\right)  \left(
\operatorname*{Ker}\left(  \operatorname*{gr}\nolimits_{n}\zeta\right)
\right)  $. Thus, (\ref{pf.l42.2}) is shown. This completes the proof of
Theorem \ref{thm.l42} \textbf{(d)}.

\subsection{The associated graded object of $U\left(  \mathfrak{g}\right)
\diagup\left(  U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $}

Here an important consequence of Theorem \ref{thm.l42}:

\begin{corollary}
\label{cor.l42}Consider the situation of Theorem \ref{thm.l42} \textbf{(d)}.
Let $n\in\mathbb{N}$.\newline Consider the canonical projection
$\operatorname{sym}_{\mathfrak{n},n}:\mathfrak{n}^{\otimes n}\rightarrow
\operatorname{Sym}^{n}\mathfrak{n}$.\newline\textbf{(a)} The map
$\operatorname{gr}_{n}\tau:\operatorname{gr}_{n}\left(  \left(  \otimes
\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  \right)  \rightarrow\operatorname{gr}_{n}\left(
U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}\right)  \right)  $ is surjective.\newline\textbf{(b)} There
exists one and only one $k$-module homomorphism $\Xi:\operatorname{gr}%
_{n}\left(  U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \right)  \rightarrow
\operatorname{Sym}^{n}\mathfrak{n}$ for which the diagram%
\begin{equation}
\xymatrixcolsep{7pc}\xymatrix{ \operatorname{gr}_n\left(\left(\otimes \mathfrak g\right)\diagup \left(J+\left(\otimes \mathfrak g\right)\cdot \mathfrak h\right)\right) \ar[r]^{\operatorname*{gr}_n\tau} \ar[d]_{\omega_n}^{\cong} & \operatorname*{gr}_n\left( U\left(\mathfrak g\right) \diagup \left(U\left(\mathfrak g\right)\cdot\mathfrak h\right) \right) \ar[dd]^{\Xi} \\ \operatorname{gr}_n\left(\otimes\mathfrak n\right) \ar[d]_{\operatorname*{grad}_{\mathfrak n,n}^{-1}}^{\cong} & \\ \mathfrak n^{\otimes n} \ar[r]_{\operatorname*{sym}_{\mathfrak n,n}} & \operatorname*{Sym}^n \mathfrak n }
\label{cor.l42.diag}%
\end{equation}
commutes.\newline\textbf{(c)} Let us denote this map $\Xi$ by $\Theta_{n}$.
Then, $\Theta_{n}$ is an $\mathfrak{h}$-module isomorphism $\operatorname{gr}%
_{n}\left(  U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \right)  \rightarrow
\operatorname{Sym}^{n}\mathfrak{n}$ which satisfies $\operatorname{sym}%
_{\mathfrak{n},n}\circ\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega
_{n}=\Theta_{n}\circ\operatorname{gr}_{n}\tau$.\newline\textbf{(d)} Let $\pi$
be the canonical projection $\mathfrak{g}\rightarrow\mathfrak{g}%
\diagup\mathfrak{h}=\mathfrak{n}$. The diagram%
\begin{equation}
\xymatrixcolsep{7pc}\xymatrix{ \operatorname{gr}_n\left(\otimes \mathfrak g\right) \ar[r]^-{\operatorname*{gr}_n\left(\rho\circ\psi\right)} \ar[d]_{\operatorname*{gr}_n\left(\otimes \pi\right)} & \operatorname*{gr}_n\left( U\left(\mathfrak g\right) \diagup \left(U\left(\mathfrak g\right)\cdot\mathfrak h\right) \right) \ar[dd]^{\Theta_n} \\ \operatorname{gr}_n\left(\otimes\mathfrak n\right) \ar[d]_{\operatorname*{grad}_{\mathfrak n,n}^{-1}}^{\cong} & \\ \mathfrak n^{\otimes n} \ar[r]_{\operatorname*{sym}_{\mathfrak n,n}} & \operatorname*{Sym}^n \mathfrak n }
\label{cor.l42.diag2}%
\end{equation}
commutes.
\end{corollary}

\textit{Proof of Corollary \ref{cor.l42}.} \textbf{(a)} This follows from
Proposition \ref{prop.gr.surj}, because $\tau\left(  F_{n}\right)  =W_{n}$.

\textbf{(b)} We have $\operatorname{Ker}\operatorname{sym}_{\mathfrak{n}%
,n}=K_{n}\left(  \mathfrak{n}\right)  $ (since $\operatorname{sym}%
_{\mathfrak{n},n}$ is the projection from $\mathfrak{n}^{\otimes n}$ on
$\mathfrak{n}^{\otimes n}\diagup K_{n}\left(  \mathfrak{n}\right)  $).

Since $\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega_{n}$ is a
$k$-module isomorphism (which is because $\operatorname{grad}_{\mathfrak{n}%
,n}^{-1}$ and $\omega_{n}$ are $k$-module isomorphisms), we have
\begin{align*}
\operatorname{Ker}\left(  \operatorname{sym}_{\mathfrak{n},n}\circ
\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega_{n}\right)   &  =\left(
\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega_{n}\right)  ^{-1}\left(
\underbrace{\operatorname{Ker}\operatorname{sym}_{\mathfrak{n},n}}%
_{=K_{n}\left(  \mathfrak{n}\right)  }\right)  =\left(  \operatorname{grad}%
_{\mathfrak{n},n}^{-1}\circ\omega_{n}\right)  ^{-1}\left(  K_{n}\left(
\mathfrak{n}\right)  \right) \\
&  =\operatorname{Ker}\left(  \operatorname{gr}_{n}\tau\right)
\end{align*}
(because Theorem \ref{thm.l42} \textbf{(d)} says that $\left(
\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega_{n}\right)  \left(
\operatorname{Ker}\left(  \operatorname{gr}_{n}\tau\right)  \right)
=K_{n}\left(  \mathfrak{n}\right)  $, and because $\operatorname{grad}%
_{\mathfrak{n},n}^{-1}\circ\omega_{n}$ is an isomorphism). In particular, this
yields that $\operatorname{Ker}\left(  \operatorname{gr}_{n}\tau\right)
\subseteq\operatorname{Ker}\left(  \operatorname{sym}_{\mathfrak{n},n}%
\circ\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega_{n}\right)  $. Since
$\operatorname{gr}_{n}\tau$ is surjective, the homomorphism theorem thus
yields that there exists one and only one $k$-module homomorphism
$\Xi:\operatorname{gr}_{n}\left(  U\left(  \mathfrak{g}\right)  \diagup\left(
U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \right)
\rightarrow\operatorname{Sym}^{n}\mathfrak{n}$ which satisfies $\Xi
\circ\operatorname{gr}_{n}\tau=\operatorname{sym}_{\mathfrak{n},n}%
\circ\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega_{n}$. In other
words, there exists one and only one $k$-module homomorphism $\Xi
:\operatorname{gr}_{n}\left(  U\left(  \mathfrak{g}\right)  \diagup\left(
U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \right)
\rightarrow\operatorname{Sym}^{n}\mathfrak{n}$ for which the diagram
(\ref{cor.l42.diag}) commutes. This proves Corollary \ref{cor.l42}
\textbf{(b)}.

\textbf{(c)} The map $\operatorname{sym}_{\mathfrak{n},n}\circ
\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega_{n}:\operatorname{gr}%
_{n}\left(  \left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \right)  \rightarrow
\operatorname{Sym}^{n}\mathfrak{n}$ is surjective (since $\operatorname{sym}%
_{\mathfrak{n},n}$ is surjective while $\operatorname{grad}_{\mathfrak{n}%
,n}^{-1}$ and $\omega_{n}$ are isomorphisms). Since $\operatorname{sym}%
_{\mathfrak{n},n}\circ\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega
_{n}=\Theta_{n}\circ\operatorname{gr}_{n}\tau$ (because the map $\Theta_{n}$
is defined as the map $\Xi$ for which the diagram (\ref{cor.l42.diag})
commutes), this yields that the map $\Theta_{n}\circ\operatorname{gr}_{n}\tau$
is surjective. Hence, the map $\Theta_{n}$ is surjective.

On the other hand, let $i$ be an arbitrary element of $\operatorname{Ker}%
\Theta_{n}$. Then, we can write $i$ in the form $i=\left(  \operatorname{gr}%
_{n}\tau\right)  \left(  i^{\prime}\right)  $ for some $i^{\prime}%
\in\operatorname{gr}_{n}\left(  \left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\right)  $ (since $\operatorname{gr}_{n}\tau$ is surjective). Now,
$i\in\operatorname{Ker}\Theta_{n}$ yields $\Theta_{n}\left(  i\right)  =0$, so
that%
\[
\left(  \underbrace{\operatorname{sym}_{\mathfrak{n},n}\circ
\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega_{n}}_{=\Theta_{n}%
\circ\operatorname{gr}_{n}\tau}\right)  \left(  i^{\prime}\right)  =\left(
\Theta_{n}\circ\operatorname{gr}_{n}\tau\right)  \left(  i^{\prime}\right)
=\Theta_{n}\left(  \underbrace{\left(  \operatorname{gr}_{n}\tau\right)
\left(  i^{\prime}\right)  }_{=i}\right)  =\Theta_{n}\left(  i\right)  =0,
\]
thus $i^{\prime}\in\operatorname{Ker}\left(  \operatorname{sym}_{\mathfrak{n}%
,n}\circ\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\omega_{n}\right)
=\operatorname{Ker}\left(  \operatorname{gr}_{n}\tau\right)  $, so that
$0=\left(  \operatorname{gr}_{n}\tau\right)  \left(  i^{\prime}\right)  =i$.
We have thus shown that every $i\in\operatorname{Ker}\Theta_{n}$ satisfies
$i=0$. Thus, $\operatorname{Ker}\Theta_{n}=0$, so that $\Theta_{n}$ is injective.

Since $\Theta_{n}$ is surjective and injective and a $k$-module homomorphism,
we conclude that $\Theta_{n}$ is a $k$-module isomorphism.

We know that $\operatorname{gr}_{n}\tau$ is a surjective $\mathfrak{h}$-module
homomorphism (since $\operatorname{gr}_{n}\tau$ is surjective according to
part \textbf{(a)}, and is an $\mathfrak{h}$-module homomorphism since $\tau$
is an $\mathfrak{h}$-module homomorphism (by Theorem \ref{thm.l42}
\textbf{(b)})), and we know that $\Theta_{n}\circ\operatorname{gr}_{n}\tau$ is
an $\mathfrak{h}$-module homomorphism (since $\Theta_{n}\circ\operatorname{gr}%
_{n}\tau=\operatorname{sym}_{\mathfrak{n},n}\circ\operatorname{grad}%
_{\mathfrak{n},n}^{-1}\circ\omega_{n}$, and since all of the maps
$\operatorname{sym}_{\mathfrak{n},n}$, $\operatorname{grad}_{\mathfrak{n}%
,n}^{-1}$ and $\omega_{n}$ are $\mathfrak{h}$-module homomorphisms). Applying
Lemma \ref{lem.surj} to $\operatorname{gr}_{n}\left(  \otimes\mathfrak{g}%
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\right)  $, $\operatorname{gr}_{n}\left(  U\left(  \mathfrak{g}\right)
\diagup\left(  U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\right)  $, $\operatorname{Sym}^{n}\mathfrak{n}$, $\operatorname{gr}_{n}\tau$
and $\Theta_{n}$ instead of $A$, $B$, $C$, $f$ and $g$, we thus conclude that
$\Theta_{n}$ is an $\mathfrak{h}$-module homomorphism. Combining this with the
fact that $\Theta_{n}$ is a $k$-module isomorphism, we conclude that
$\Theta_{n}$ is an $\mathfrak{h}$-module isomorphism. This proves Corollary
\ref{cor.l42} \textbf{(c)}.

\textbf{(d)} By the functoriality of $\operatorname{gr}_{n}$, we have
$\operatorname{gr}_{n}\left(  \tau\circ\zeta\right)  =\operatorname{gr}%
_{n}\tau\circ\operatorname{gr}_{n}\zeta$. On the other hand, the commutative
diagram (\ref{thm.l25.c.diag'}) yields $\omega_{n}\circ\operatorname{gr}%
_{n}\zeta=\operatorname{gr}_{n}\left(  \otimes\pi\right)  $. Now,%
\begin{align*}
\Theta_{n}\circ\operatorname{gr}_{n}\left(  \underbrace{\rho\circ\psi
}_{\substack{=\tau\circ\zeta\\\text{(by Theorem \ref{thm.l42} \textbf{(b)})}%
}}\right)   &  =\Theta_{n}\circ\underbrace{\operatorname{gr}_{n}\left(
\tau\circ\zeta\right)  }_{=\operatorname{gr}_{n}\tau\circ\operatorname{gr}%
_{n}\zeta}=\underbrace{\Theta_{n}\circ\operatorname{gr}_{n}\tau}%
_{\substack{=\operatorname{sym}_{\mathfrak{n},n}\circ\operatorname{grad}%
_{\mathfrak{n},n}^{-1}\circ\omega_{n}\\\text{(by Corollary \ref{cor.l42}
\textbf{(c)})}}}\circ\operatorname{gr}_{n}\zeta\\
&  =\operatorname{sym}_{\mathfrak{n},n}\circ\operatorname{grad}_{\mathfrak{n}%
,n}^{-1}\circ\underbrace{\omega_{n}\circ\operatorname{gr}_{n}\zeta
}_{=\operatorname{gr}_{n}\left(  \otimes\pi\right)  }=\operatorname{sym}%
_{\mathfrak{n},n}\circ\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ
\operatorname{gr}_{n}\left(  \otimes\pi\right)  .
\end{align*}
In other words, the diagram (\ref{cor.l42.diag2}) commutes. This proves
Corollary \ref{cor.l42} \textbf{(d)}.

Note that Corollary \ref{cor.l42} can be used to prove Theorem \ref{thm.pbw}
\textbf{(a)}. This is not particularly surprising and not particularly useful,
as we have used Theorem \ref{thm.pbw} \textbf{(a)} in our proof of Corollary
\ref{cor.l42}. But let us give the proof for the sake of completeness:

\textit{Proof of Theorem \ref{thm.pbw} \textbf{(a)} using Corollary
\ref{cor.l42}:} Let $n\in\mathbb{N}$. Let $\mathfrak{g}$ be a $k$-Lie algebra
which is a free $k$-module. Let $\mathfrak{h}=0$ and $N=\mathfrak{g}$. Then,
$\mathfrak{h}$ is a Lie subalgebra of $\mathfrak{g}$ such that $\mathfrak{g}%
=\mathfrak{h}\oplus N$. Moreover, both $\mathfrak{h}$ and $N$ are free
$k$-modules. Moreover, the inclusion $\mathfrak{h}\hookrightarrow\mathfrak{g}$
splits \textit{as a }$k$\textit{-module inclusion}. Thus, we can apply
Corollary \ref{cor.l42} \textbf{(d)} and obtain that the diagram
(\ref{cor.l42.diag2}) commutes (where we are using the notations of Corollary
\ref{cor.l42}, of course). But since we are in a situation where
$\mathfrak{h}=0$ and $\mathfrak{n}=\mathfrak{g}\diagup\mathfrak{h}%
=\mathfrak{g}$, this diagram simplifies to%
\[
\xymatrixcolsep{7pc}\xymatrix{
\operatorname{gr}_n\left(\otimes \mathfrak g\right) \ar[r]^-{\operatorname*{gr}_n\psi} \ar[d]_{\operatorname*{id}} & \operatorname*{gr}_n\left( U\left(\mathfrak g\right) \right) \ar[dd]^{\Theta_n} \\
\operatorname{gr}_n\left(\otimes\mathfrak g\right) \ar[d]_{\operatorname*{grad}_{\mathfrak g,n}^{-1}}^{\cong} &
\\ \mathfrak g^{\otimes n} \ar[r]_{\operatorname*{sym}_{\mathfrak g,n}} & \operatorname*{Sym}^n \mathfrak g }
\]
(since $\pi=\operatorname{id}$, $\otimes\pi=\operatorname{id}$,
$\operatorname{gr}_{n}\left(  \otimes\pi\right)  =\operatorname{id}$ and
$\rho=\operatorname{id}$). Thus, $\Theta_{n}\circ\operatorname{gr}_{n}%
\psi=\operatorname{sym}_{\mathfrak{g},n}\circ\operatorname{grad}%
_{\mathfrak{g},n}^{-1}$. Thus, $\Theta_{n}\circ\operatorname{gr}_{n}\psi
\circ\operatorname{grad}_{\mathfrak{g},n}=\operatorname{sym}_{\mathfrak{g},n}%
$, so that%
\[
\operatorname{sym}_{\mathfrak{g},n}=\Theta_{n}\circ
\underbrace{\operatorname{gr}_{n}\psi\circ\operatorname{grad}_{\mathfrak{g}%
,n}}_{\substack{=\operatorname{PBW}_{\mathfrak{g},n}\circ\operatorname{sym}%
_{\mathfrak{g},n}\\\text{(since this is how we defined }\operatorname{PBW}%
_{\mathfrak{g},n}\text{)}}}=\Theta_{n}\circ\operatorname{PBW}_{\mathfrak{g}%
,n}\circ\operatorname{sym}_{\mathfrak{g},n}.
\]
Since $\operatorname{sym}_{\mathfrak{g},n}$ is surjective, this yields
$\operatorname{id}=\Theta_{n}\circ\operatorname{PBW}_{\mathfrak{g},n}$. Thus,
$\operatorname{PBW}_{\mathfrak{g},n}$ is the inverse of the $\mathfrak{h}%
$-module isomorphism $\Theta_{n}$ (here, we are using the fact that
$\Theta_{n}$ is an $\mathfrak{h}$-module isomorphism; this follows from
Corollary \ref{cor.l42} \textbf{(c)}). This yields that $\operatorname{PBW}%
_{\mathfrak{g},n}$ itself is an $\mathfrak{h}$-module isomorphism. In other
words, $\mathfrak{g}$ satisfies the $n$-PBW condition, and our proof of
Theorem \ref{thm.pbw} \textbf{(a)} is complete. As already explained, this
proof does not take us far, as the proof of Corollary \ref{cor.l42} given
above made substantial use of Theorem \ref{thm.pbw} \textbf{(a)}; but at least
it shows that Corollary \ref{cor.l42} is indeed a generalization of Theorem
\ref{thm.pbw} \textbf{(a)}.

\subsection{The splitting of the filtration}

The next theorem encompasses a part of \cite[Theorem 4.5]{calaque1}, and
(together with the theorems we have already proven) will (almost) complete the
proof of the main result of \cite{calaque1} ("almost" because the converse
direction will still be missing, but it is rather easy and straightforward).

\begin{theorem}
\label{thm.l45}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a $k$-Lie
algebra. Let $\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$. Assume that
the inclusion $\mathfrak{h}\hookrightarrow\mathfrak{g}$ splits \textit{as a
}$k$\textit{-module inclusion} (but not necessarily as an $\mathfrak{h}%
$-module inclusion). This means that there exists a $k$-submodule $N$ of
$\mathfrak{g}$ such that $\mathfrak{g}=\mathfrak{h}\oplus N$.\newline Assume
that both $\mathfrak{h}$ and $N$ are free $k$-modules.\newline Let us consider
the $\mathfrak{g}$-module $U\left(  \mathfrak{g}\right)  \ \ \ \ $%
.{\footnotemark}\newline Let us also use the notations introduced in Theorem
\ref{thm.l25}, in particularly the $\mathfrak{h}$-module $\mathfrak{n}%
$.\newline We also consider the $\mathfrak{h}$-module $\operatorname{Sym}%
^{n}\mathfrak{n}$ (see Definition \ref{defs.symm-alg.g} for its definition)
for every $n\in\mathbb{N}$.\newline\textbf{Assume that this }$\mathfrak{h}%
$\textbf{-module }$\mathfrak{n}$ \textbf{is actually the restriction of some
}$\left(  \mathfrak{g},\mathfrak{h}\right)  $\textbf{-semimodule to
}$\mathfrak{h}$\textbf{.}\newline Now let us assume the following statement,
which we call the \textbf{symmetric splitting assumption: The canonical
projection }$\operatorname{sym}_{\mathfrak{n},n}:\mathfrak{n}^{\otimes
n}\rightarrow\operatorname{Sym}^{n}\mathfrak{n}$\textbf{ (defined in
Definition \ref{defs.symm-alg}) splits as an }$\mathfrak{h}$\textbf{-module
projection for every }$n\in\mathbb{N}$\textbf{.\newline}(Note that the
symmetric splitting assumption is automatically satisfied in the case when
every positive integer is invertible in the ring $k$, because in this case we
can split the projection $\operatorname{sym}_{\mathfrak{n},n}:\mathfrak{n}%
^{\otimes n}\rightarrow\operatorname{Sym}^{n}\mathfrak{n}$ by the map
$\operatorname{Sym}^{n}\mathfrak{n}\rightarrow\mathfrak{n}^{\otimes n}$ which
sends $\overline{v_{1}\otimes v_{2}\otimes...\otimes v_{n}}$ to $\dfrac{1}%
{n!}\sum\limits_{\sigma\in S_{n}}v_{\sigma\left(  1\right)  }\otimes
v_{\sigma\left(  2\right)  }\otimes...\otimes v_{\sigma\left(  n\right)  }$
for every $\left(  v_{1},v_{2},...,v_{n}\right)  \in\mathfrak{n}^{n}$. But
this is not the only case in which the symmetric splitting assumption
holds.)\newline Define a filtration $\left(  W_{n}\right)  _{n\geq0}$ on
$U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}\right)  $ as in Theorem \ref{thm.l42} \textbf{(c)}.\newline
Then, the filtration $\left(  W_{n}\right)  _{n\geq0}$ is an $\mathfrak{h}%
$-split $\mathfrak{h}$-module filtration.\newline More precisely, we can
construct a splitting for the $\mathfrak{h}$-module projection $W_{n}%
\rightarrow W_{n}\diagup W_{n-1}$ for every $n\geq1$ as follows:\newline
Consider the map $\mathfrak{h}$-module isomorphism $\Theta_{n}%
:\operatorname{gr}_{n}\left(  U\left(  \mathfrak{g}\right)  \diagup\left(
U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \right)
\rightarrow\operatorname{Sym}^{n}\mathfrak{n}$ from Corollary \ref{cor.l42}
\textbf{(c)}. Since $\operatorname{gr}_{n}\left(  U\left(  \mathfrak{g}%
\right)  \diagup\left(  U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}%
\right)  \right)  =W_{n}\diagup W_{n-1}$, this $\Theta_{n}$ is thus an
$\mathfrak{h}$-module isomorphism $W_{n}\diagup W_{n-1}\rightarrow
\operatorname{Sym}^{n}\mathfrak{n}$.\newline Let $\mathbf{I}_{n}%
:\operatorname{Sym}^{n}\mathfrak{n}\rightarrow\mathfrak{n}^{\otimes n}$ be an
$\mathfrak{h}$-module homomorphism such that $\operatorname{sym}%
_{\mathfrak{n},n}\circ\mathbf{I}_{n}=\operatorname*{id}$. (In other words, let
$\mathbf{I}_{n}:\operatorname{Sym}^{n}\mathfrak{n}\rightarrow\mathfrak{n}%
^{\otimes n}$ be an $\mathfrak{h}$-module homomorphism which splits the
projection $\operatorname{sym}_{\mathfrak{n},n}:\mathfrak{n}^{\otimes
n}\rightarrow\operatorname{Sym}^{n}\mathfrak{n}$. Such an $\mathbf{I}_{n}$
exists due to the symmetric splitting assumption. If every positive integer is
invertible in the ring $k$, then we can even find a \textit{canonical}
$\mathbf{I}_{n}$.)\newline Since the filtration $\left(  F_{n}\right)
_{n\geq0}$ is $\mathfrak{h}$-split (by Theorem \ref{thm.l37ba}), the exact
sequence $%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{5pc}
%\xymatrix{
%0 \ar[r] & F_{n-1} \ar[r]^-{\text{inclusion}} & F_n \ar[r]^-{\text
%{projection}} & F_n \diagup F_{n-1} \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrixcolsep{5pc}
\xymatrix{
0 \ar[r] & F_{n-1} \ar[r]^-{\text{inclusion}} & F_n \ar[r]^-{\text
{projection}} & F_n \diagup F_{n-1} \ar[r] & 0
}%
%EndExpansion
$ is $\mathfrak{h}$-split. So there exists some $\mathfrak{h}$-module
homomorphism $\vartheta_{n}:F_{n}\diagup F_{n-1}\rightarrow F_{n}$ which
splits the projection $F_{n}\rightarrow F_{n}\diagup F_{n-1}$. Consider this
$\vartheta_{n}$.\newline Let $\tau\mid_{F_{n}}^{W_{n}}$ denote the map
$F_{n}\rightarrow W_{n}$ obtained by restricting $\tau$ to $F_{n}$ (since we
know that $\tau\left(  F_{n}\right)  =W_{n}$).\newline Then, the map $\left(
\tau\mid_{F_{n}}^{W_{n}}\right)  \circ\vartheta_{n}\circ\omega_{n}^{-1}%
\circ\operatorname*{grad}\nolimits_{\mathfrak{n},n}\circ\mathbf{I}_{n}%
\circ\Theta_{n}^{-1}:W_{n}\diagup W_{n-1}\rightarrow W_{n}$ is an
$\mathfrak{h}$-module homomorphism which splits the projection $W_{n}%
\rightarrow W_{n}\diagup W_{n-1}$.
\end{theorem}

\begin{vershort}
\footnotetext{Again, let us remind ourselves that "the $\mathfrak{g}$-module
$U\left(  \mathfrak{g}\right)  $" is to be understood in accordance to Remark
\ref{rmk.g-mod-U(g).short} here.}
\end{vershort}

\begin{verlong}
\footnotetext{Again, let us remind ourselves that "the $\mathfrak{g}$-module
$U\left(  \mathfrak{g}\right)  $" is to be understood in accordance to Remark
\ref{rmk.g-mod-U(g).long} here.}
\end{verlong}

\textit{Proof of Theorem \ref{thm.l45}.} First, let us make sure that the map
$\left(  \tau\mid_{F_{n}}^{W_{n}}\right)  \circ\vartheta_{n}\circ\omega
_{n}^{-1}\circ\operatorname*{grad}\nolimits_{\mathfrak{n},n}\circ
\mathbf{I}_{n}\circ\Theta_{n}:W_{n}\diagup W_{n-1}\rightarrow W_{n}$ is
well-defined at all. This is since $\Theta_{n}:W_{n}\diagup W_{n-1}%
\rightarrow\operatorname{Sym}^{n}\mathfrak{n}$ (in fact, we know that
$\Theta_{n}:\operatorname{gr}_{n}\left(  U\left(  \mathfrak{g}\right)
\diagup\left(  U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\right)  \rightarrow\operatorname{Sym}^{n}\mathfrak{n}$, but by definition of
$\operatorname*{gr}\nolimits_{n}$ we have $\operatorname{gr}_{n}\left(
U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}\right)  \right)  =W_{n}\diagup W_{n-1}$), since
$\mathbf{I}_{n}:\operatorname{Sym}^{n}\mathfrak{n}\rightarrow\mathfrak{n}%
^{\otimes n}$, since $\operatorname*{grad}\nolimits_{\mathfrak{n}%
,n}:\mathfrak{n}^{\otimes n}\rightarrow\operatorname*{gr}\nolimits_{n}\left(
\otimes\mathfrak{n}\right)  $, since $\omega_{n}^{-1}:\operatorname*{gr}%
\nolimits_{n}\left(  \otimes\mathfrak{n}\right)  \rightarrow F_{n}\diagup
F_{n-1}$ (because $\omega_{n}:F_{n}\diagup F_{n-1}\rightarrow
\operatorname*{gr}\nolimits_{n}\left(  \otimes\mathfrak{n}\right)  $), since
$\vartheta_{n}:F_{n}\diagup F_{n-1}\rightarrow F_{n}$ and since $\tau
\mid_{F_{n}}^{W_{n}}:F_{n}\rightarrow W_{n}$.

Second, this map $\left(  \tau\mid_{F_{n}}^{W_{n}}\right)  \circ\vartheta
_{n}\circ\omega_{n}^{-1}\circ\operatorname*{grad}\nolimits_{\mathfrak{n}%
,n}\circ\mathbf{I}_{n}\circ\Theta_{n}$ is indeed an $\mathfrak{h}$-module
homomorphism, since it is a composition of six $\mathfrak{h}$-module homomorphisms.

We must now prove that this map $\left(  \tau\mid_{F_{n}}^{W_{n}}\right)
\circ\vartheta_{n}\circ\omega_{n}^{-1}\circ\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}\circ\mathbf{I}_{n}\circ\Theta_{n}$ splits the
projection $W_{n}\rightarrow W_{n}\diagup W_{n-1}$. In fact, let us denote the
projection $W_{n}\rightarrow W_{n}\diagup W_{n-1}$ by $\Gamma_{n}$. Let us
denote the projection $F_{n}\rightarrow F_{n}\diagup F_{n-1}$ by $\Gamma
_{n}^{\prime}$. Then, $\Gamma_{n}\circ\left(  \tau\mid_{F_{n}}^{W_{n}}\right)
=\operatorname*{gr}\nolimits_{n}\tau\circ\Gamma_{n}^{\prime}$ follows easily
from the definition of $\operatorname*{gr}\nolimits_{n}$. On the other hand,
$\Gamma_{n}^{\prime}\circ\vartheta_{n}=\operatorname*{id}$ (since
$\vartheta_{n}$ is defined as a splitting of the projection $F_{n}\rightarrow
F_{n}\diagup F_{n-1}$, but that projection $F_{n}\rightarrow F_{n}\diagup
F_{n-1}$ is $\Gamma_{n}^{\prime}$). Thus,
\begin{align*}
&  \Gamma_{n}\circ\left(  \left(  \tau\mid_{F_{n}}^{W_{n}}\right)
\circ\vartheta_{n}\circ\omega_{n}^{-1}\circ\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}\circ\mathbf{I}_{n}\circ\Theta_{n}\right) \\
&  =\underbrace{\Gamma_{n}\circ\left(  \tau\mid_{F_{n}}^{W_{n}}\right)
}_{=\operatorname*{gr}\nolimits_{n}\tau\circ\Gamma_{n}^{\prime}}\circ
\vartheta_{n}\circ\omega_{n}^{-1}\circ\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}\circ\mathbf{I}_{n}\circ\Theta_{n}\\
&  =\operatorname*{gr}\nolimits_{n}\tau\circ\underbrace{\Gamma_{n}^{\prime
}\circ\vartheta_{n}}_{=\operatorname*{id}}\circ\omega_{n}^{-1}\circ
\operatorname*{grad}\nolimits_{\mathfrak{n},n}\circ\mathbf{I}_{n}\circ
\Theta_{n}=\operatorname*{gr}\nolimits_{n}\tau\circ\omega_{n}^{-1}%
\circ\operatorname*{grad}\nolimits_{\mathfrak{n},n}\circ\mathbf{I}_{n}%
\circ\Theta_{n}\\
&  =\Theta_{n}^{-1}\circ\operatorname{sym}_{\mathfrak{n},n}\circ
\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\underbrace{\omega_{n}%
\circ\omega_{n}^{-1}}_{=\operatorname*{id}}\circ\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}\circ\mathbf{I}_{n}\circ\Theta_{n}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }\operatorname{sym}_{\mathfrak{n},n}\circ\operatorname{grad}%
_{\mathfrak{n},n}^{-1}\circ\omega_{n}=\Theta_{n}\circ\operatorname{gr}_{n}%
\tau\text{ (by Corollary \ref{cor.l42} \textbf{(c)}),}\\
\text{and thus }\operatorname{gr}_{n}\tau=\Theta_{n}^{-1}\circ
\operatorname{sym}_{\mathfrak{n},n}\circ\operatorname{grad}_{\mathfrak{n}%
,n}^{-1}\circ\omega_{n}\text{ (since }\Theta_{n}\text{ is an isomorphism)}%
\end{array}
\right) \\
&  =\Theta_{n}^{-1}\circ\operatorname{sym}_{\mathfrak{n},n}\circ
\underbrace{\operatorname{grad}_{\mathfrak{n},n}^{-1}\circ\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}}_{=\operatorname*{id}}\circ\mathbf{I}_{n}%
\circ\Theta_{n}=\Theta_{n}^{-1}\circ\underbrace{\operatorname{sym}%
_{\mathfrak{n},n}\circ\mathbf{I}_{n}}_{=\operatorname*{id}}\circ\Theta
_{n}=\Theta_{n}^{-1}\circ\Theta_{n}=\operatorname*{id}.
\end{align*}
In other words, the map $\left(  \tau\mid_{F_{n}}^{W_{n}}\right)
\circ\vartheta_{n}\circ\omega_{n}^{-1}\circ\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}\circ\mathbf{I}_{n}\circ\Theta_{n}$ splits the
projection $W_{n}\rightarrow W_{n}\diagup W_{n-1}$ (because $\Gamma_{n}$ is
the projection $W_{n}\rightarrow W_{n}\diagup W_{n-1}$). Therefore, the exact
sequence $%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{5pc}
%\xymatrix{
%0 \ar[r] & W_{n-1} \ar[r]^-{\text{inclusion}} & W_n \ar[r]^-{\text
%{projection}} & W_n \diagup W_{n-1} \ar[r] & 0
%}}}%
%BeginExpansion
\xymatrixcolsep{5pc}
\xymatrix{
0 \ar[r] & W_{n-1} \ar[r]^-{\text{inclusion}} & W_n \ar[r]^-{\text
{projection}} & W_n \diagup W_{n-1} \ar[r] & 0
}%
%EndExpansion
$ is $\mathfrak{h}$-split (since the map $\left(  \tau\mid_{F_{n}}^{W_{n}%
}\right)  \circ\vartheta_{n}\circ\omega_{n}^{-1}\circ\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}\circ\mathbf{I}_{n}\circ\Theta_{n}$ is an
$\mathfrak{h}$-module homomorphism). Since we have proven this for each
$n\in\mathbb{N}$, we thus conclude that the filtration $\left(  W_{n}\right)
_{n\geq0}$ is $\mathfrak{h}$-split.

This proves Theorem \ref{thm.l45}.

\subsection{Non-canonical isomorphisms}

Corollary \ref{cor.l42} \textbf{(c)} gave us a canonical $\mathfrak{h}$-module
isomorphism $\operatorname{gr}_{n}\left(  U\left(  \mathfrak{g}\right)
\diagup\left(  U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\right)  \rightarrow\operatorname{Sym}^{n}\mathfrak{n}$ for every
$n\in\mathbb{N}$. Therefore,
\[
\bigoplus\limits_{n\in\mathbb{N}}\operatorname{gr}_{n}\left(  U\left(
\mathfrak{g}\right)  \diagup\left(  U\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}\right)  \right)  \cong\bigoplus\limits_{n\in\mathbb{N}%
}\operatorname{Sym}^{n}\mathfrak{n}=\operatorname*{Sym}\mathfrak{n}%
\]
by a canonical isomorphism. Now, the standard intuition for the direct sum
$\bigoplus\limits_{n\in\mathbb{N}}\operatorname{gr}_{n}V$ (where $V$ is some
filtered $k$-module) is that this sum is a kind of "approximation" for $V$,
which is usually simpler than $V$ itself (for example, $U\left(
\mathfrak{g}\right)  $ is generally a non-commutative algebra, while
$\bigoplus\limits_{n\in\mathbb{N}}\operatorname{gr}_{n}\left(  U\left(
\mathfrak{g}\right)  \right)  $ is a commutative one). So now, knowing that
the "approximation" $\bigoplus\limits_{n\in\mathbb{N}}\operatorname{gr}%
_{n}\left(  U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \right)  $ of $U\left(
\mathfrak{g}\right)  \diagup\left(  U\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}\right)  $ is isomorphic to $\operatorname*{Sym}\mathfrak{n}%
$, we can ask ourselves what $U\left(  \mathfrak{g}\right)  \diagup\left(
U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ itself is isomorphic
to. It turns out that in the situation of Theorem \ref{thm.l45}, we have the
same answer, but the isomorphism is not canonical anymore:

\begin{proposition}
\label{prop.l49}Consider the situation of Theorem \ref{thm.l45}.\newline We
have $U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(  \mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  \cong\operatorname*{Sym}\mathfrak{n}$ as
$\mathfrak{h}$-modules.\newline More precisely, there is an $\mathfrak{h}%
$-module isomorphism $U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \rightarrow\operatorname*{Sym}%
\mathfrak{n}$ which respects the filtration such that the inverse of this
isomorphism also respects the filtration.
\end{proposition}

\textit{Proof of Proposition \ref{prop.l49}.} We know from Theorem
\ref{thm.l45} that the filtration $\left(  W_{n}\right)  _{n\geq0}$ is
$\mathfrak{h}$-split. Thus, Proposition \ref{prop.split.filtr-g-2} (applied to
$U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}\right)  $, $\left(  W_{n}\right)  _{n\geq0}$ and
$\mathfrak{h}$ instead of $V$, $\left(  V_{n}\right)  _{n\geq0}$ and
$\mathfrak{g}$) yields that there exists a bifiltered $\mathfrak{h}$-module
isomorphism $U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \rightarrow\bigoplus
\limits_{p\in\mathbb{N}}\operatorname*{gr}\nolimits_{p}\left(  U\left(
\mathfrak{g}\right)  \diagup\left(  U\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}\right)  \right)  $.

On the other hand, every $p\in\mathbb{N}$ satisfies $\operatorname*{gr}%
\nolimits_{p}\left(  U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \right)  \cong%
\operatorname*{Sym}\nolimits^{p}\mathfrak{n}$ as $\mathfrak{h}$-modules
(because Corollary \ref{cor.l42} \textbf{(c)} (applied to $p$ instead of $n$)
shows that $\Theta_{p}$ is an $\mathfrak{h}$-module isomorphism
$\operatorname{gr}_{p}\left(  U\left(  \mathfrak{g}\right)  \diagup\left(
U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \right)
\rightarrow\operatorname{Sym}^{p}\mathfrak{n}$). Therefore, there exists a
bifiltered $\mathfrak{h}$-module isomorphism $\bigoplus\limits_{p\in
\mathbb{N}}\operatorname*{gr}\nolimits_{p}\left(  U\left(  \mathfrak{g}%
\right)  \diagup\left(  U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}%
\right)  \right)  \rightarrow\bigoplus\limits_{p\in\mathbb{N}}%
\operatorname*{Sym}\nolimits^{p}\mathfrak{n}=\operatorname*{Sym}\mathfrak{n}$.
Composing this isomorphism with our bifiltered $\mathfrak{h}$-module
isomorphism $U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \rightarrow\bigoplus
\limits_{p\in\mathbb{N}}\operatorname*{gr}\nolimits_{p}\left(  U\left(
\mathfrak{g}\right)  \diagup\left(  U\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}\right)  \right)  $, we obtain a bifiltered $\mathfrak{h}%
$-module isomorphism $U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \rightarrow\operatorname*{Sym}%
\mathfrak{n}$. By the definition of "bifiltered", this is an isomorphism which
respects the filtration such that the inverse of this isomorphism also
respects the filtration. This proves Proposition \ref{prop.l49}.

Even if the (somewhat restrictive) conditions of Theorem \ref{thm.l45} are not
satisfied, we can still obtain a $k$\textit{-module} isomorphy $U\left(
\mathfrak{g}\right)  \diagup\left(  U\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}\right)  \cong\operatorname*{Sym}\mathfrak{n}$ under somewhat
weaker conditions:

\begin{proposition}
\label{prop.l50}Consider the situation of Theorem \ref{thm.l42}. Assume that
$\mathfrak{h}$ and $N$ are free $k$-modules.\newline We have $U\left(
\mathfrak{g}\right)  \diagup\left(  U\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}\right)  \cong\operatorname*{Sym}\mathfrak{n}$ as
$k$-modules.\newline More precisely, there is a $k$-module isomorphism
$U\left(  \mathfrak{g}\right)  \diagup\left(  U\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}\right)  \rightarrow\operatorname*{Sym}\mathfrak{n}$ which
respects the filtration such that the inverse of this isomorphism also
respects the filtration.
\end{proposition}

\textit{Proof of Proposition \ref{prop.l50}.} The idea of the proof of
Proposition \ref{prop.l50} is to proceed as in the proof of Proposition
\ref{prop.l49}, but to read "$k$-split", "$k$-module" and "$k$-module
homomorphism" instead of each "$\mathfrak{h}$-split", "$\mathfrak{h}$-module"
and "$\mathfrak{h}$-module homomorphism", respectively. Do the same
modifications in the proof of Theorem \ref{thm.l45}.

Of course, it is not really that easy, because Theorem \ref{thm.l45} has some
more conditions than we have assumed in Proposition \ref{prop.l50}. Here is
how to deal with them:

\begin{itemize}
\item Since we read "$k$-module" instead of "$\mathfrak{h}$-module", the
symmetric splitting assumption (which is needed for Theorem \ref{thm.l45} to
hold) now takes the following form: "The canonical projection
$\operatorname{sym}_{\mathfrak{n},n}:\mathfrak{n}^{\otimes n}\rightarrow
\operatorname{Sym}^{n}\mathfrak{n}$ (defined in Definition \ref{defs.symm-alg}%
) splits as a $k$-module projection for every $n\in\mathbb{N}$." But this is
obviously satisfied, because $\mathfrak{n}\cong N$ (as $k$-modules) is a free
$k$-module.

\item The assumption that the $\mathfrak{h}$-module $\mathfrak{n}$ be the
restriction of some $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule to
$\mathfrak{h}$ is not granted anymore. Fortunately, we only use it to make
sure that the filtration $\left(  F_{n}\right)  _{n\geq0}$ is $\mathfrak{h}%
$-split. Since we read "$k$-split" instead of "$\mathfrak{h}$-split", we
therefore just need a new argument for why the filtration $\left(
F_{n}\right)  _{n\geq0}$ is $k$-split. We have more or less done this in
Section \ref{sect.L25} already: From Proposition \ref{prop.l25.Ker-alpha}
\textbf{(b)} and \textbf{(d)}, the map $\overline{\varphi}$ (constructed in
Proposition \ref{prop.l25.Ker-alpha} \textbf{(a)}) is a bifiltered isomorphism
$\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \rightarrow\otimes N$. Since
the filtration $\left(  N^{\otimes\leq n}\right)  _{n\geq0}$ of $\otimes N$ is
$k$-split, we can now easily conclude that so is the filtration $\left(
F_{n}\right)  _{n\geq0}$ of $\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
$.
\end{itemize}

This proves Proposition \ref{prop.l50}.

\section{\label{sect.gen}Generalizations, improvements and analogues}

While the results we gave above were already somewhat more general than those
of \cite{calaque1}, some of them can be extended even further, and/or shown to
have analogues. These extensions and analogues have never been studied in
detail, and neither am I going to do so in the present paper, but I will
discuss them in brief in this Section \ref{sect.gen}.

\subsection{\label{subsect.general}When $\mathfrak{g}$ is not a Lie algebra}

The results of Sections \ref{sect.L25} and \ref{sect.split} can be
substantially generalized once we notice the following: All results of
Sections \ref{sect.L25} and \ref{sect.split} were formulated in the following
setup:%
\begin{align*}
&  \mathfrak{g}\text{ is a }k\text{-Lie algebra, and }\mathfrak{h}\text{ is a
Lie subalgebra of }\mathfrak{g}\text{.}\\
&  \text{We consider }\mathfrak{g}\text{ as an }\mathfrak{h}\text{-module (by
restricting the }\mathfrak{g}\text{-module }\mathfrak{g}\text{).}%
\end{align*}
However, we have never used this setup in its full glory in Sections
\ref{sect.L25} and \ref{sect.split}, and everything done in these Sections can
be extended to the case when this setup is replaced by the following one:%
\begin{align*}
&  \mathfrak{h}\text{ is a }k\text{-Lie algebra, and }\mathfrak{g}\text{ is an
}\mathfrak{h}\text{-module which}\\
&  \text{happens to contain the }\mathfrak{h}\text{-module }\mathfrak{h}\text{
as an }\mathfrak{h}\text{-submodule.}%
\end{align*}
For example, Theorem \ref{thm.l25} takes the following form in this case:

\begin{theorem}
\label{thm.l25.gen}Let $k$ be a commutative ring. Let $\mathfrak{h}$ be a
$k$-Lie algebra, and let $\mathfrak{g}$ be an $\mathfrak{h}$-module (not
necessarily a Lie algebra itself!). Assume that the $\mathfrak{h}$-module
$\mathfrak{h}$ itself (this $\mathfrak{h}$-module $\mathfrak{h}$ is defined
according to Definition \ref{defs.g-mod-g}, applied to $\mathfrak{h}$ instead
of $\mathfrak{g}$) is an $\mathfrak{h}$-submodule of $\mathfrak{g}$.\newline
Assume that the inclusion $\mathfrak{h}\hookrightarrow\mathfrak{g}$ splits
\textit{as a }$k$\textit{-module inclusion} (but not necessarily as an
$\mathfrak{h}$-module inclusion).\newline Let $J$ be the two-sided ideal%
\[
\left(  \otimes\mathfrak{g}\right)  \cdot\left\langle v\otimes w-w\otimes
v+w\rightharpoonup v\ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times
\mathfrak{h}\right\rangle \cdot\left(  \otimes\mathfrak{g}\right)
\]
of the $k$-algebra $\otimes\mathfrak{g}$.\newline Let $\mathfrak{n}%
=\mathfrak{g}\diagup\mathfrak{h}$. This $\mathfrak{n}$ is an $\mathfrak{h}%
$-module (because both $\mathfrak{g}$ and $\mathfrak{h}$ are $\mathfrak{h}%
$-modules).\newline Let $\pi:\mathfrak{g}\rightarrow\mathfrak{n}$ be the
canonical projection with kernel $\mathfrak{h}$. Obviously, $\pi$ is an
$\mathfrak{h}$-module homomorphism. Thus, $\otimes\pi:\otimes\mathfrak{g}%
\rightarrow\otimes\mathfrak{n}$ is also an $\mathfrak{h}$-module homomorphism
(according to Proposition \ref{prop.(X)-hom-is-hom}).\newline We consider
$\mathfrak{h}$ as an $\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$ by
means of the embedding $\mathfrak{h}\hookrightarrow\mathfrak{g}\hookrightarrow
\otimes\mathfrak{g}$.\newline\textbf{(a)} Both $J$ and $\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}$ are $\mathfrak{h}$-submodules of
$\otimes\mathfrak{g}$. Thus, $\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
$ is an $\mathfrak{h}$-module. Let $\zeta:\otimes\mathfrak{g}\rightarrow
\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ be the canonical projection.
Then, $\zeta$ is an $\mathfrak{h}$-module homomorphism.\newline\textbf{(b)}
For every $n\in\mathbb{N}$, let $F_{n}$ be the $\mathfrak{h}$-submodule
$\zeta\left(  \mathfrak{g}^{\otimes\leq n}\right)  $ of $\left(
\otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  $.\ \ \ \ {\footnotemark} Also define an
$\mathfrak{h}$-submodule $F_{-1}$ of $\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
$ by $F_{-1}=0$. Then, $\left(  F_{n}\right)  _{n\geq0}$ is an $\mathfrak{h}%
$-module filtration of $\left(  \otimes\mathfrak{g}\right)  \diagup\left(
J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ and
satisfies $F_{n}\diagup F_{n-1}\cong\mathfrak{n}^{\otimes n}$ as
$\mathfrak{h}$-modules for every $n\in\mathbb{N}$.\newline\textbf{(c)} Let
$n\in\mathbb{N}$. There exists one and only one $k$-module homomorphism
$\Omega_{n}:F_{n}\diagup F_{n-1}\rightarrow\operatorname*{gr}\nolimits_{n}%
\left(  \otimes\mathfrak{n}\right)  $ for which the diagram%
\[%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{9pc}
%\xymatrix{
%\operatorname*{gr}_n\left(\otimes\mathfrak{g}\right) \ar[d]_{\operatorname
%*{gr}_n\zeta} \ar[dr]^{\operatorname*{gr}_n\left(\otimes\pi\right)} & \\
%\operatorname*{gr}_n\left(\left(\otimes\mathfrak{g}\right)\diagup
%\left(J + \left(\otimes\mathfrak{g}\right)\cdot\mathfrak{h}\right
%) \right) = F_n \diagup F_{n-1} \ar[r]_-{\Omega_n} & \operatorname*{gr}%
%_n\left(\otimes\mathfrak{n}\right)
%}}}%
%BeginExpansion
\xymatrixcolsep{9pc}
\xymatrix{
\operatorname*{gr}_n\left(\otimes\mathfrak{g}\right) \ar[d]_{\operatorname
*{gr}_n\zeta} \ar[dr]^{\operatorname*{gr}_n\left(\otimes\pi\right)} & \\
\operatorname*{gr}_n\left(\left(\otimes\mathfrak{g}\right)\diagup
\left(J + \left(\otimes\mathfrak{g}\right)\cdot\mathfrak{h}\right
) \right) = F_n \diagup F_{n-1} \ar[r]_-{\Omega_n} & \operatorname*{gr}%
_n\left(\otimes\mathfrak{n}\right)
}%
%EndExpansion
\]
commutes. Denote this homomorphism $\Omega_{n}$ by $\omega_{n}$. Then,
$\omega_{n}$ is an $\mathfrak{h}$-module isomorphism, and the diagram%
\[%
%TCIMACRO{\TeXButton{xymatrix}{\xymatrixcolsep{9pc}
%\xymatrix{
%\operatorname*{gr}_n\left(\otimes\mathfrak{g}\right) \ar[d]_{\operatorname
%*{gr}_n\zeta} \ar[dr]^{\operatorname*{gr}_n\left(\otimes\pi\right)} & \\
%\operatorname*{gr}_n\left(\left(\otimes\mathfrak{g}\right)\diagup
%\left(J + \left(\otimes\mathfrak{g}\right)\cdot\mathfrak{h}\right
%) \right) = F_n \diagup F_{n-1} \ar[r]_-{\omega_n} & \operatorname*{gr}%
%_n\left(\otimes\mathfrak{n}\right)
%}}}%
%BeginExpansion
\xymatrixcolsep{9pc}
\xymatrix{
\operatorname*{gr}_n\left(\otimes\mathfrak{g}\right) \ar[d]_{\operatorname
*{gr}_n\zeta} \ar[dr]^{\operatorname*{gr}_n\left(\otimes\pi\right)} & \\
\operatorname*{gr}_n\left(\left(\otimes\mathfrak{g}\right)\diagup
\left(J + \left(\otimes\mathfrak{g}\right)\cdot\mathfrak{h}\right
) \right) = F_n \diagup F_{n-1} \ar[r]_-{\omega_n} & \operatorname*{gr}%
_n\left(\otimes\mathfrak{n}\right)
}%
%EndExpansion
\]
commutes.\newline Applying Definition \ref{defs.filtr.grad} to $\mathfrak{n}$
and $n$ instead of $V$ and $p$, we obtain a map $\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}:\mathfrak{n}^{\otimes n}\rightarrow
\operatorname*{gr}\nolimits_{n}\left(  \otimes\mathfrak{n}\right)  $.
According to Proposition \ref{prop.filtr.grad.g} (applied to $\mathfrak{h}$,
$n$ and $\mathfrak{n}$ instead of $\mathfrak{g}$, $p$ and $V$), this map
$\operatorname*{grad}\nolimits_{\mathfrak{n},n}$ is a canonical $\mathfrak{h}%
$-module isomorphism. Thus, its inverse $\operatorname*{grad}%
\nolimits_{\mathfrak{n},n}^{-1}$ is an $\mathfrak{h}$-module isomorphism as
well. The composition $\operatorname*{grad}\nolimits_{\mathfrak{n},n}%
^{-1}\circ\omega_{n}:F_{n}\diagup F_{n-1}\rightarrow\mathfrak{n}^{\otimes n}$
is an $\mathfrak{h}$-module isomorphism (because $\omega_{n}$ and
$\operatorname*{grad}\nolimits_{\mathfrak{n},n}^{-1}$ are $\mathfrak{h}%
$-module isomorphisms).
\end{theorem}

\footnotetext{In fact, $\zeta\left(  \mathfrak{g}^{\otimes\leq n}\right)  $ is
indeed an $\mathfrak{h}$-submodule (because $\zeta$ is an $\mathfrak{h}%
$-module homomorphism and because $\mathfrak{g}^{\otimes\leq n}$ is an
$\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$).}

This is indeed a generalization of Theorem \ref{thm.l25}, because in the
situation of Theorem \ref{thm.l25}, we have:%
\begin{equation}
\left(
\begin{array}
[c]{c}%
\text{the ideal }J\text{ defined in Theorem \ref{thm.l25} is identical}\\
\text{with the ideal }J\text{ defined in Theorem \ref{thm.l25.gen}}%
\end{array}
\right)  . \label{thm.l25.gen.deriv.1}%
\end{equation}


\begin{vershort}
This is easy to see (and left to the reader).
\end{vershort}

\begin{verlong}
\textit{Proof of (\ref{thm.l25.gen.deriv.1}).} Every $\left(  v,w\right)
\in\mathfrak{g}\times\mathfrak{h}$ satisfies $\left[  v,w\right]  =-\left[
w,v\right]  $ (due to (\ref{[]-3})) and $w\rightharpoonup v=\left[
w,v\right]  $ (due to (\ref{prop.g-is-g-module.form}), applied to $w$ and $v$
instead of $v$ and $w$). Thus, every $\left(  v,w\right)  \in\mathfrak{g}%
\times\mathfrak{h}$ satisfies
\begin{align}
v\otimes w-w\otimes v-\underbrace{\left[  v,w\right]  }_{=-\left[  w,v\right]
}  &  =v\otimes w-w\otimes v+\underbrace{\left[  w,v\right]  }%
_{=w\rightharpoonup v}\nonumber\\
&  =v\otimes w-w\otimes v+w\rightharpoonup v. \label{pf.l25.deriv.J}%
\end{align}


Now,%
\begin{align*}
&  \left(  \text{the ideal }J\text{ defined in Theorem \ref{thm.l25.gen}%
}\right) \\
&  =\left(  \otimes\mathfrak{g}\right)  \cdot\left\langle \underbrace{v\otimes
w-w\otimes v+w\rightharpoonup v}_{=v\otimes w-w\otimes v-\left[  v,w\right]
}\ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times\mathfrak{h}\right\rangle
\cdot\left(  \otimes\mathfrak{g}\right) \\
&  =\left(  \otimes\mathfrak{g}\right)  \cdot\left\langle v\otimes w-w\otimes
v-\left[  v,w\right]  \ \mid\ \left(  v,w\right)  \in\mathfrak{g}%
\times\mathfrak{h}\right\rangle \cdot\left(  \otimes\mathfrak{g}\right) \\
&  =\left(  \text{the ideal }J\text{ defined in Theorem \ref{thm.l25}}\right)
.
\end{align*}
This proves (\ref{thm.l25.gen.deriv.1}).
\end{verlong}

To give a proof of Theorem \ref{thm.l25.gen}, we just have to repeat the proof
of Theorem \ref{thm.l25} that we gave in Section \ref{sect.L25} (including all
the auxiliary facts we showed in Section \ref{sect.L25}) up to the following changes:

\begin{itemize}
\item Replace the words "$\mathfrak{g}$-module" by "$\mathfrak{h}$-module".

\item Replace the words "$\mathfrak{g}$-algebra" by "$\mathfrak{h}$-algebra".

\item Replace the words "$\mathfrak{g}$-submodule" by "$\mathfrak{h}$-submodule".

\item Whenever a term of the form $\left[  x,y\right]  $ for some
$x\in\mathfrak{g}$ and $y\in\mathfrak{g}$ appears in Section \ref{sect.L25},
proceed by the following rules:\newline- If $x$ is known to lie in
$\mathfrak{h}$, replace this term by $x\rightharpoonup y$.\newline- If $y$ is
known to lie in $\mathfrak{h}$, replace this term by $-y\rightharpoonup
x$.\newline(If both $x$ and $y$ are known to lie in $\mathfrak{h}$, then it
does not matter which of these two rules is being followed, because
$x\rightharpoonup y=\left[  x,y\right]  =-\left[  y,x\right]
=-y\rightharpoonup x$ for any $x\in\mathfrak{h}$ and $y\in\mathfrak{h}%
$.)\newline Fortunately, all terms of the form $\left[  x,y\right]  $ that
appear in Section \ref{sect.L25} have either $x$ or $y$ lying in
$\mathfrak{h}$, so that after these replacements, no terms of the form
$\left[  x,y\right]  $ remain anymore.
\end{itemize}

Most results in Section \ref{sect.(g,h)} can be generalized as soon as we
extend Definition \ref{defs.semimod} (the definition of a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule) as follows:

\begin{definition}
\label{defs.semimod.gen}Let $k$ be a commutative ring. Let $\mathfrak{h}$ be a
$k$-Lie algebra, and let $\mathfrak{g}$ be an $\mathfrak{h}$-module (not
necessarily a Lie algebra itself!). Assume that the $\mathfrak{h}$-module
$\mathfrak{h}$ itself (this $\mathfrak{h}$-module $\mathfrak{h}$ is defined
according to Definition \ref{defs.g-mod-g}, applied to $\mathfrak{h}$ instead
of $\mathfrak{g}$) is an $\mathfrak{h}$-submodule of $\mathfrak{g}$.\newline
Let $V$ be a $k$-module. Let $\mu:\mathfrak{g}\times V\rightarrow V$ be a
$k$-bilinear map. We say that $\left(  V,\mu\right)  $ is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $\textit{-semimodule} if and only if
\begin{equation}
\left(  \mu\left(  a\rightharpoonup b,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{h}\text{, }b\in\mathfrak{g}\text{ and }v\in V\right)  .
\label{defs.semimod.gen.mu}%
\end{equation}
If $\left(  V,\mu\right)  $ is a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule, then the $k$-bilinear map $\mu:\mathfrak{g}\times V\rightarrow
V$ is called the \textit{Lie action} of the $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $-semimodule $V$.\newline 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{defs.semimod.gen.mu}) rewrites as%
\[
\left(  \left(  a\rightharpoonup b\right)  \rightharpoonup v=a\rightharpoonup
\left(  b\rightharpoonup v\right)  -b\rightharpoonup\left(  a\rightharpoonup
v\right)  \text{ for every }a\in\mathfrak{h}\text{, }b\in\mathfrak{g}\text{
and }v\in V\right)  .
\]
\newline Also, an abuse of notation allows us to write "$V$ is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule" instead of "$\left(
V,\mu\right)  $ is a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule"
if the map $\mu$ is clear from the context or has not been introduced
yet.\newline Besides, when $\left(  V,\mu\right)  $ is a $\left(
\mathfrak{g},\mathfrak{h}\right)  $-semimodule, we will say that $\mu$ is a
$\left(  \mathfrak{g},\mathfrak{h}\right)  $-\textit{semimodule structure on}
$V$. In other words, if $V$ is a $k$-module, then a $\left(  \mathfrak{g}%
,\mathfrak{h}\right)  $\textit{-semimodule structure on }$V$ means a map
$\mu:\mathfrak{g}\times V\rightarrow V$ such that $\left(  V,\mu\right)  $ is
a $\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule. (Thus, in order to
make a $k$-module into a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule, we must define a $\left(  \mathfrak{g},\mathfrak{h}\right)
$-semimodule structure on it.)
\end{definition}

Theorem \ref{thm.l37ba} generalizes as follows:

\begin{theorem}
\label{thm.l37ba.gen}Let $k$ be a commutative ring. Let $\mathfrak{h}$ be a
$k$-Lie algebra, and let $\mathfrak{g}$ be an $\mathfrak{h}$-module (not
necessarily a Lie algebra itself!). Assume that the $\mathfrak{h}$-module
$\mathfrak{h}$ itself (this $\mathfrak{h}$-module $\mathfrak{h}$ is defined
according to Definition \ref{defs.g-mod-g}, applied to $\mathfrak{h}$ instead
of $\mathfrak{g}$) is an $\mathfrak{h}$-submodule of $\mathfrak{g}$.\newline
Assume that the inclusion $\mathfrak{h}\hookrightarrow\mathfrak{g}$ splits
\textit{as a }$k$\textit{-module inclusion} (but not necessarily as an
$\mathfrak{h}$-module inclusion).\newline Let $J$ be the two-sided ideal%
\[
\left(  \otimes\mathfrak{g}\right)  \cdot\left\langle v\otimes w-w\otimes
v+w\rightharpoonup v\ \mid\ \left(  v,w\right)  \in\mathfrak{g}\times
\mathfrak{h}\right\rangle \cdot\left(  \otimes\mathfrak{g}\right)
\]
of the $k$-algebra $\otimes\mathfrak{g}$.\newline Let $\mathfrak{n}%
=\mathfrak{g}\diagup\mathfrak{h}$. This $\mathfrak{n}$ is an $\mathfrak{h}%
$-module (because both $\mathfrak{g}$ and $\mathfrak{h}$ are $\mathfrak{h}%
$-modules). \textbf{Assume that this }$\mathfrak{h}$\textbf{-module
}$\mathfrak{n}$ \textbf{is actually the restriction of some }$\left(
\mathfrak{g},\mathfrak{h}\right)  $\textbf{-semimodule to }$\mathfrak{h}$
(where "$\left(  \mathfrak{g},\mathfrak{h}\right)  $-semimodule" is to be
understood according to Definition \ref{defs.semimod.gen}).\newline Let
$\pi:\mathfrak{g}\rightarrow\mathfrak{n}$ be the canonical projection with
kernel $\mathfrak{h}$. Obviously, $\pi$ is an $\mathfrak{h}$-module
homomorphism. Thus, $\otimes\pi:\otimes\mathfrak{g}\rightarrow\otimes
\mathfrak{n}$ is also an $\mathfrak{h}$-module homomorphism (according to
Proposition \ref{prop.(X)-hom-is-hom}).\newline We consider $\mathfrak{h}$ as
an $\mathfrak{h}$-submodule of $\otimes\mathfrak{g}$ by means of the embedding
$\mathfrak{h}\hookrightarrow\mathfrak{g}\hookrightarrow\otimes\mathfrak{g}%
$.\newline\textbf{(a)} Both $J$ and $\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}$ are $\mathfrak{h}$-submodules of $\otimes\mathfrak{g}$.
Thus, $\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(
\otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ is an $\mathfrak{h}%
$-module. Let $\zeta:\otimes\mathfrak{g}\rightarrow\left(  \otimes
\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes\mathfrak{g}\right)
\cdot\mathfrak{h}\right)  $ be the canonical projection. Then, $\zeta$ is an
$\mathfrak{h}$-module homomorphism.\newline\textbf{(b)} For every
$n\in\mathbb{N}$, let $F_{n}$ be the $\mathfrak{h}$-submodule $\zeta\left(
\mathfrak{g}^{\otimes\leq n}\right)  $ of $\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
$. (That $F_{n}$ indeed is an $\mathfrak{h}$-submodule was proven in Theorem
\ref{thm.l25} already.) Then, $\left(  F_{n}\right)  _{n\geq0}$ is an
$\mathfrak{h}$-module filtration of $\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
$ and satisfies $F_{n}\cong\mathfrak{n}^{\otimes\leq n}$ as $\mathfrak{h}%
$-modules for every $n\in\mathbb{N}$.\newline\textbf{(c)} There exists an
$\mathfrak{h}$-module isomorphism $\left(  \otimes\mathfrak{g}\right)
\diagup\left(  J+\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\rightarrow\otimes\mathfrak{n}$ such that for every $n\in\mathbb{N}$, the
image of $F_{n}$ under this isomorphism is $\mathfrak{n}^{\otimes\leq n}%
$.\newline\textbf{(d)} The filtration $\left(  F_{n}\right)  _{n\geq0}$ of
$\left(  \otimes\mathfrak{g}\right)  \diagup\left(  J+\left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ is $\mathfrak{h}$-split.
\end{theorem}

Again, the proof of this theorem is a repetition of the proof of Theorem
\ref{thm.l37ba} with the same replacements as we had to do to obtain a proof
of Theorem \ref{thm.l25.gen}.

The results of Section \ref{sect.pbw} probably cannot be generalized in a
similar fashion.

A further surprise seems to be that the proofs of Theorems \ref{thm.l25.gen}
and \ref{thm.l37ba.gen} apparently never use the axioms (\ref{[]-1}) and
(\ref{[]-2}) for the $k$-Lie algebra $\mathfrak{h}$, instead only using%
\[
\left[  u,\left[  v,w\right]  \right]  =\left[  \left[  u,v\right]  ,w\right]
+\left[  v,\left[  u,w\right]  \right]  \ \ \ \ \ \ \ \ \ \ \text{for all
}u\in\mathfrak{h}\text{, }v\in\mathfrak{h}\text{ and }w\in\mathfrak{h}.
\]
"Apparently" because I have not had enough time to check that this indeed is
the case. If it is, this means that Theorems \ref{thm.l25.gen} and
\ref{thm.l37ba.gen} extend to Leibniz algebras in lieu of Lie algebras. I am
not aware of a similar extension of the Poincar\'{e}-Birkhoff-Witt theorem.

\subsection{\label{subsect.super}The case of Lie superalgebras}

The notion of a \textit{Lie superalgebra} (also known under the name
\textit{super Lie algebra} and studied in \cite{scheunert}, \cite{deligne}) is
one of the most well-understood generalizations of that of a Lie algebra.
While classification results for Lie superalgebras are significantly harder
than their non-super counterparts, most "purely algebraic" properties of Lie
algebras tend to have their analogues for Lie superalgebras, which usually are
even proven in more or less the same manner. This has to do with the fact that
Lie superalgebras are just Lie algebras in the category of super-$k$-modules;
however there is also a much more pedestrian approach to proving properties of
Lie superalgebras by re-reading the proofs of the corresponding facts about
Lie algebras and adding signs via the Koszul rule.

Different sources disagree about the correct way to define the notion of a Lie
superalgebra. This might have to do with the fact that the primary interest
lies in Lie superalgebras over a field of characteristic $0$ (rather than an
arbitrary field, let alone a commutative ring), and all the definitions of a
Lie superalgebra are equivalent to each other if we are over a field of
characteristic $0$. As I am interested in the general case, let me give the
following definition of a Lie superalgebra (which is one of the most
restrictive ones, but not as restrictive as \cite[Definition 8.1.1]%
{neisendorfer}):

\begin{definition}
\label{defs.super.liealg}Let $k$ be a commutative ring. A $k$\textit{-Lie
superalgebra} will mean a super-$k$-module $\mathfrak{g}$ (see Definition
\ref{defs.super.mod} below) together with a $k$-bilinear map $\beta
:\mathfrak{g}\times\mathfrak{g}\rightarrow\mathfrak{g}$ satisfying the
conditions%
\begin{align}
&  \left(  \beta\left(  v,v\right)  =0\text{ for every }v\in\mathfrak{g}%
_{0}\right)  ;\label{defs.super.liealg.1}\\
&  \left(
\begin{array}
[c]{c}%
\left(  -1\right)  ^{i\ell}\beta\left(  u,\beta\left(  v,w\right)  \right)
+\left(  -1\right)  ^{ji}\beta\left(  v,\beta\left(  w,u\right)  \right)
+\left(  -1\right)  ^{\ell j}\beta\left(  w,\beta\left(  u,v\right)  \right)
=0\\
\text{ for every }i\in\mathbb{Z}\diagup2\mathbb{Z}\text{, }j\in\mathbb{Z}%
\diagup2\mathbb{Z}\text{ and }\ell\in\mathbb{Z}\diagup2\mathbb{Z}\\
\text{and every }u\in\mathfrak{g}_{i}\text{, }v\in\mathfrak{g}_{j}\text{ and
}w\in\mathfrak{g}_{\ell}%
\end{array}
\right)  ;\label{defs.super.liealg.2}\\
&  \left(
\begin{array}
[c]{c}%
\beta\left(  v,w\right)  =-\left(  -1\right)  ^{ij}\beta\left(  w,v\right) \\
\text{for every }i\in\mathbb{Z}\diagup2\mathbb{Z}\text{ and }j\in
\mathbb{Z}\diagup2\mathbb{Z}\text{ and every }v\in\mathfrak{g}_{i}\text{ and
}w\in\mathfrak{g}_{j}%
\end{array}
\right)  ;\label{defs.super.liealg.3}\\
&  \left(  \beta\left(  v,\beta\left(  v,v\right)  \right)  =0\text{ for every
}v\in\mathfrak{g}_{1}\right)  ;\label{defs.super.liealg.4}\\
&  \left(  \beta\left(  \mathfrak{g}_{i}\times\mathfrak{g}_{j}\right)
\subseteq\mathfrak{g}_{i+j}\text{ for every }i\in\mathbb{Z}\diagup
2\mathbb{Z}\text{ and }j\in\mathbb{Z}\diagup2\mathbb{Z}\right)  .
\label{defs.super.liealg.5}%
\end{align}
\newline This $k$-bilinear map $\beta:\mathfrak{g}\times\mathfrak{g}%
\rightarrow\mathfrak{g}$ will be called the \textit{Lie bracket} of the
$k$-Lie superalgebra $\mathfrak{g}$. We will often use the square brackets
notation for $\beta$, which means that we are going to abbreviate
$\beta\left(  v,w\right)  $ by $\left[  v,w\right]  $ for any $v\in
\mathfrak{g}$ and $w\in\mathfrak{g}$. Using this notation, the equations
(\ref{defs.super.liealg.1}), (\ref{defs.super.liealg.2}),
(\ref{defs.super.liealg.3}), (\ref{defs.super.liealg.4}) and
(\ref{defs.super.liealg.5}) rewrite as%
\begin{align}
&  \left(  \left[  v,v\right]  =0\text{ for every }v\in\mathfrak{g}%
_{0}\right)  ;\label{defs.super.liealg.[1]}\\
&  \left(
\begin{array}
[c]{c}%
\left(  -1\right)  ^{i\ell}\left[  u,\left[  v,w\right]  \right]  +\left(
-1\right)  ^{ji}\left[  v,\left[  w,u\right]  \right]  +\left(  -1\right)
^{\ell j}\left[  w,\left[  u,v\right]  \right]  =0\\
\text{ for every }i\in\mathbb{Z}\diagup2\mathbb{Z}\text{, }j\in\mathbb{Z}%
\diagup2\mathbb{Z}\text{ and }\ell\in\mathbb{Z}\diagup2\mathbb{Z}\\
\text{and every }u\in\mathfrak{g}_{i}\text{, }v\in\mathfrak{g}_{j}\text{ and
}w\in\mathfrak{g}_{\ell}%
\end{array}
\right)  ;\label{defs.super.liealg.[2]}\\
&  \left(
\begin{array}
[c]{c}%
\left[  v,w\right]  =-\left(  -1\right)  ^{ij}\left[  w,v\right] \\
\text{for every }i\in\mathbb{Z}\diagup2\mathbb{Z}\text{ and }j\in
\mathbb{Z}\diagup2\mathbb{Z}\text{ and every }v\in\mathfrak{g}_{i}\text{ and
}w\in\mathfrak{g}_{j}%
\end{array}
\right)  ;\label{defs.super.liealg.[3]}\\
&  \left(  \left[  v,\left[  v,v\right]  \right]  =0\text{ for every }%
v\in\mathfrak{g}_{1}\right)  ;\label{defs.super.liealg.[4]}\\
&  \left(  \left[  \mathfrak{g}_{i},\mathfrak{g}_{j}\right]  \subseteq
\mathfrak{g}_{i+j}\text{ for every }i\in\mathbb{Z}\diagup2\mathbb{Z}\text{ and
}j\in\mathbb{Z}\diagup2\mathbb{Z}\right)  \label{defs.super.liealg.[5]}%
\end{align}
(where $\left[  \mathfrak{g}_{i},\mathfrak{g}_{j}\right]  $ means the
$k$-linear span $\left\langle \left[  v,w\right]  \ \mid\ \left(  v,w\right)
\in\mathfrak{g}_{i}\times\mathfrak{g}_{j}\right\rangle $).\newline The
equation (\ref{defs.super.liealg.2}) (or its equivalent version
(\ref{defs.super.liealg.[2]})) is called the \textit{super-Jacobi identity}.
\end{definition}

Here we have used the following definition:

\begin{definition}
\label{defs.super.mod}Let $k$ be a commutative ring. A \textit{super-}%
$k$\textit{-module} will mean a $k$-module $V$ together with a pair $\left(
V_{0},V_{1}\right)  $ of $k$-submodules of $V$ such that $V=V_{0}\oplus V_{1}%
$.\newline Here, $0$ and $1$ are considered not as integers, but as elements
of $\mathbb{Z}\diagup2\mathbb{Z}$ (so that $1+1=0$). This sounds like a
useless requirement, but it helps us in handling super-$k$-modules
notationally; for example, the equation (\ref{defs.super.liealg.5}) would not
make sense if $0$ and $1$ would be considered as integers (because in the case
$i=1$ and $j=1$, we would have $i+j=2$, but there is no such thing as
$\mathfrak{g}_{2}$ unless $2$ is treated as an element of $\mathbb{Z}%
\diagup2\mathbb{Z}$).\newline The $k$-submodule $V_{0}$ of $V$ is called the
\textit{even part of }$V$. The $k$-submodule $V_{1}$ of $V$ is called the
\textit{odd part of }$V$.\newline An element of $V$ is said to be
\textit{homogeneous} if it lies in $V_{0}$ or in $V_{1}$.
\end{definition}

\begin{condition}
\label{conv.super.conv}We are going to use the notation $V_{0}$ as a universal
notation for the even part of a super-$k$-module $V$. This means that whenever
we have some super-$k$-module $V$ (it needs not be actually called $V$; I only
refer to it by $V$ here in this Convention), the even part of $V$ will be
called $V_{0}$.\newline Similarly, we are going to use the notation $V_{1}$ as
a universal notation for the odd part of a super-$k$-module $V$.
\end{condition}

\begin{remark}
Our Definition \ref{defs.super.liealg} differs from some definitions of a Lie
superalgebra given in literature by having the axioms
(\ref{defs.super.liealg.1}) and (\ref{defs.super.liealg.4}). These axioms are
indeed dispensable when one is only interested in the case of $k$ being a
field of characteristic $0$ (or, more generally, of $k$ being a commutative
ring in which $2$ and $3$ are invertible){\footnotemark}. However, for the
sake of generality, we keep these axioms in.
\end{remark}

\footnotetext{In fact,
\par
\begin{itemize}
\item axiom (\ref{defs.super.liealg.1}) follows from axiom
(\ref{defs.super.liealg.3}) if $2$ is invertible in $k$;
\par
\item axiom (\ref{defs.super.liealg.4}) follows from axiom
(\ref{defs.super.liealg.2}) if $3$ is invertible in $k$.
\end{itemize}
}

Just as the notion of Lie algebras gives birth to that of $\mathfrak{g}%
$-modules, we can define the notion of a $\mathfrak{g}$\textit{-supermodule}
(or just $\mathfrak{g}$\textit{-module}) over a Lie superalgebra
$\mathfrak{g}$:

\begin{definition}
\label{defs.super.module}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be
a Lie superalgebra. (According to Convention \ref{conv.super.conv}, this
automatically entails that we denote by $\mathfrak{g}_{0}$ the even part of
$\mathfrak{g}$, and denote by $\mathfrak{g}_{1}$ the odd part of
$\mathfrak{g}$.)\newline Let $V$ be a $k$-supermodule. (According to
Convention \ref{conv.super.conv}, this automatically entails that we denote by
$V_{0}$ the even part of $V$, and denote by $V_{1}$ the odd part of
$V$.)\newline Let $\mu:\mathfrak{g}\times V\rightarrow V$ be a $k$-bilinear
map. We say that $\left(  V,\mu\right)  $ is a $\mathfrak{g}$%
\textit{-supermodule} if and only if
\begin{equation}
\left(
\begin{array}
[c]{c}%
\mu\left(  \left[  a,b\right]  ,v\right)  =\mu\left(  a,\mu\left(  b,v\right)
\right)  -\left(  -1\right)  ^{ij}\mu\left(  b,\mu\left(  a,v\right)  \right)
\\
\text{for every }i\in\mathbb{Z}\diagup2\mathbb{Z}\text{, }j\in\mathbb{Z}%
\diagup2\mathbb{Z}\text{ and every }a\in\mathfrak{g}_{i}\text{, }%
b\in\mathfrak{g}_{j}\text{ and }v\in V
\end{array}
\right)  \label{defs.super.module.1}%
\end{equation}
and%
\[
\left(  \mu\left(  \mathfrak{g}_{i}\times V_{j}\right)  \subseteq
V_{i+j}\text{ for every }i\in\mathbb{Z}\diagup2\mathbb{Z}\text{ and }%
j\in\mathbb{Z}\diagup2\mathbb{Z}\right)  .
\]
If $\left(  V,\mu\right)  $ is a $\mathfrak{g}$-supermodule, then the
$k$-bilinear map $\mu:\mathfrak{g}\times V\rightarrow V$ is called the
\textit{Lie action} of the $\mathfrak{g}$-supermodule $V$.
\end{definition}

(This definition seems to be agreed on in most references. I have not seen any
conflicting definitions as in the case of Definition \ref{defs.super.liealg}.)

While I have not checked the details, I am convinced that all results of
Sections \ref{sect.L25}, \ref{sect.(g,h)} and \ref{sect.split} (and Subsection
\ref{subsect.general}) carry over to Lie superalgebras (and Lie supermodules)
as long as $2$ is invertible in the ground ring $k$. Even the invertibility of
$2$ might actually be redundant for most of these results (and it seems that
the reason for its redundancy is the fact that most of the results still hold
for Leibniz algebras - but, as I already said, this is not thoroughly
checked). As for Section \ref{sect.pbw}, trouble might come from the
Poincar\'{e}-Birkhoff-Witt theorem (Theorem \ref{thm.pbw}), whose validity in
the Lie superalgebra case has not been studied to the extent it has been
studied in the original, Lie algebraic case. However, there \textit{are} two
known results:

\begin{theorem}
\label{thm.pbw.super}Let $k$ be a commutative ring. Let $\mathfrak{g}$ be a
$k$-Lie superalgebra. Let $n\in\mathbb{N}$.\newline\textbf{(a)} If
$\mathfrak{g}_{0}$ and $\mathfrak{g}_{1}$ are free $k$-modules, and if $2$ is
invertible in the ring $k$, then $\mathfrak{g}$ satisfies the $n$-PBW
condition.\newline\textbf{(b)} If $k$ is a $\mathbb{Q}$-algebra, then
$\mathfrak{g}$ satisfies the $n$-PBW condition.
\end{theorem}

A proof of Theorem \ref{thm.pbw.super} \textbf{(b)} was given in \cite[Part 1,
Chapter 1, \S 1.3.7]{deligne} and \cite[\S 2.5]{petracci}; a proof of Theorem
\ref{thm.pbw.super} \textbf{(a)} can be found in \cite[\S 2.3, Theorem
1]{scheunert}. Note that whoever claims that $3$ must be invertible in the
ring $k$ in order for Theorem \ref{thm.pbw.super} \textbf{(a)} to hold is
probably using a definition of Lie superalgebra which does not contain the
axiom (\ref{defs.super.liealg.4}). \textit{However,} even having the axiom
(\ref{defs.super.liealg.1}) does not prevent us from having to require the
invertibility of $2$, unless we replace our definition of a Lie superalgebra
by a significantly more complicated one (\cite[Definition 8.1.1]%
{neisendorfer}), in which case we can indeed drop the invertibility of $2$
(\cite[Theorem 8.2.2]{neisendorfer}). Having said this, we are \textit{not}
going to use \cite[Definition 8.1.1]{neisendorfer} as the definition of a Lie
superalgebra in this paper; instead we will keep understanding a Lie
superalgebra according to Definition \ref{defs.super.liealg}. As a
consequence, we will not be able to get rid of the condition that $2$ be
invertible in $k$ in the Poincar\'{e}-Birkhoff-Witt theorem and its consequences.

The correct analogue of Theorem \ref{thm.pbw.basis} now says:

\begin{theorem}
\label{thm.pbw.basis.super}Let $k$ be a commutative ring. Let $\mathfrak{g}$
be a $k$-Lie superalgebra. Assume that $2$ is invertible in the ring $k$. Also
assume that the $k$-module $\mathfrak{g}$ has a basis $\left(  e_{i}\right)
_{i\in I}$, where $I$ is a totally ordered set, and where $e_{i}$ is
homogeneous for every $i\in I$. Then,
\[
\left(  \overline{e_{i_{1}}\otimes e_{i_{2}}\otimes...\otimes e_{i_{n}}%
}\right)  _{\substack{n\in\mathbb{N};\ \left(  i_{1},i_{2},...,i_{n}\right)
\in I^{n};\\i_{1}\leq i_{2}\leq...\leq i_{n};\\\text{every }p\text{ which
satisfies }\left(  e_{i_{p}}\in\mathfrak{g}_{1}\text{ and }e_{i_{p+1}}%
\in\mathfrak{g}_{1}\right)  \text{ satisfies }i_{p}<i_{p+1}}}
\]
is a basis of the $k$-module $U\left(  \mathfrak{g}\right)  $.
\end{theorem}

A proof of Theorem \ref{thm.pbw.basis.super} in the case when $k$ is a field
of characteristic $\neq2$ and $\neq3$ can be found in \cite[Theorem
6.1.1]{musson}.

These changes in the formulation of the Poincar\'{e}-Birkhoff-Witt theorem(s)
don't seem to keep Proposition \ref{prop.pbw.U(g)h} from retaining its
validity in the case of $\mathfrak{g}$ being a Lie superalgebra, at least as
long as $2$ is assumed to be invertible in $k$ and we assume the even and the
odd parts of $\mathfrak{h}$ and $N$ to be free $k$-modules (and not just
$\mathfrak{h}$ and $N$ themselves). As a consequence, nothing speaks against
the other results of Section \ref{sect.pbw} holding in this case, although
this has yet to be verified more accurately.

\begin{noncompile}
\label{subsect.restr}Restricted universal enveloping algebras and the
characteristic $p$ case
\end{noncompile}

\begin{noncompile}
Over fields of prime characteristic, there exists a notion of Lie algebras
with a particular additional structure: the so-called restricted Lie algebras.

[...]
\end{noncompile}

\subsection{\label{subsect.cliff}Poincar\'{e}-Birkhoff-Witt type theorems for
Clifford algebras}

\subsubsection{Clifford algebras}

There is an analogy between Lie algebras and quadratic spaces, with universal
enveloping algebras of Lie algebras on the one side corresponding to Clifford
algebras of quadratic spaces on the other. This analogy, however, is marred by
an imbalance: Numerous results which hold in high generality on the quadratic
spaces side require additional assumptions or weakenings on the Lie algebras
side. As a basic example, let me show the quadratic-spaces counterpart of the
Poincar\'{e}-Birkhoff-Witt theorem. First, the relevant definitions (I am not
really working with quadratic spaces, but rather with spaces with bilinear forms):

\begin{definition}
\label{defs.cliff}Let $k$ be a commutative ring. Let $L$ be a $k$%
-module.\newline\textbf{(a)} Let $f:L\times L\rightarrow k$ be a $k$-bilinear
form on $L$. We define the \textit{Clifford algebra} $\operatorname*{Cl}%
\left(  L,f\right)  $ to be the $k$-algebra $\left(  \otimes L\right)  \diagup
I_{f}$, where $I_{f}$ is the two-sided ideal%
\[
\left(  \otimes L\right)  \cdot\left\langle v\otimes v-f\left(  v,v\right)
\ \mid\ v\in L\right\rangle \cdot\left(  \otimes L\right)
\]
of the $k$-algebra $\otimes L$.\newline\textbf{(b)} We denote by $\wedge L$
the exterior algebra of the $k$-module $L$. Clearly, $\wedge
L=\operatorname*{Cl}\left(  L,\mathbf{0}\right)  $, where $\mathbf{0}$ denotes
the bilinear form $L\times L\rightarrow k$ which sends every pair $\left(
x,y\right)  \in L\times L$ to $0$.
\end{definition}

\begin{remark}
You are reading right: In this Definition \ref{defs.cliff}, the form $f$ is
not required to be symmetric, but only the values of $f\left(  v,v\right)  $
for $v\in L$ are actually used. Over a field of characteristic $\neq2$ (and
more generally, over a ring where $2$ is invertible), every bilinear form
$f:L\times L\rightarrow k$ has a "symmetrization", which means a symmetric
bilinear form $\widetilde{f}$ satisfying $\left(  \widetilde{f}\left(
v,v\right)  =f\left(  v,v\right)  \text{ for every }v\in L\right)  $. But, in
my experience \cite{dg1}, restricting one's attention to symmetric bilinear
forms is not really of much use in the theory of Clifford algebras; most
important facts don't require this.\newline Most texts define Clifford
algebras for quadratic forms rather than bilinear forms. Unfortunately, I was
not able to spot an undisputed definition of what a quadratic form on an
arbitrary $k$-module is. If one uses the definition of a quadratic form given
in \cite[\S 3, n$%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
$4, D\'{e}finition 2]{bourbaki-1}, then the main result I want to state
(Theorem \ref{thm.cliff.pbw}) is not valid for Clifford algebras of quadratic
forms (that is, the Clifford algebra of a quadratic form on a $k$-module $L$
is not necessarily isomorphic to $\wedge L$; see \cite{grinberg-cliff-MO} for
a counterexample). It is still valid when the $k$-module is free, but this
should not come as a surprise: For free $k$-modules, every quadratic form can
be written in the form $v\mapsto f\left(  v,v\right)  $ for some bilinear form
$f$ (but not necessarily a symmetric bilinear form $f$; this is yet another
reason not to require $f$ to be symmetric in Definition \ref{defs.cliff});
this shows that Definition \ref{defs.cliff} encompasses the notion of the
Clifford algebra of a quadratic form at least for free $k$-modules.
\end{remark}

\subsubsection{Poincar\'{e}-Birkhoff-Witt for Clifford algebras}

Now, how would an analogue of the Poincar\'{e}-Birkhoff-Witt theorem for
Clifford algebras look like?

We can consider a commutative ring $k$, some $k$-module $L$ with a
$k$-bilinear form $f:L\times L\rightarrow k$, and some $n\in\mathbb{N}$. Let
$\psi:\otimes L\rightarrow\operatorname*{Cl}\left(  L,f\right)  $ be the
canonical projection. Consider the canonical filtration on $\operatorname*{Cl}%
\left(  L,f\right)  $ (that is, the filtration obtained from the degree
filtration on $\otimes L$ via the projection $\psi$). Let $\wedge^{n}L$ denote
the $n$-th exterior power of the $k$-module $L$, and let $\wedge_{L,n}$ be the
canonical projection $L^{\otimes n}\rightarrow\wedge^{n}L$. Then, it can be
easily seen that there exists a unique $k$-module homomorphism $\mathbf{P}%
:\wedge^{n}L\rightarrow\operatorname*{gr}\nolimits_{n}\left(
\operatorname*{Cl}\left(  L,f\right)  \right)  $ such that $\operatorname*{gr}%
\nolimits_{n}\psi\circ\operatorname{grad}_{L,n}=\mathbf{P}\circ\wedge_{L,n}$.
In analogy to Definition \ref{defs.PBWmap} \textbf{(a)}, we can call this map
$\mathbf{P}$ the $n$\textit{-PBW homomorphism} of the pair $\left(
L,f\right)  $. In analogy to Definition \ref{defs.PBWmap} \textbf{(b)}, we
could now say that the pair $\left(  L,f\right)  $ satisfies \textit{the }%
$n$\textit{-PBW condition} if the $n$-PBW homomorphism of $\left(  L,f\right)
$ is a $k$-module isomorphism.

We could now expect that an analogue of Theorem \ref{thm.pbw} would state
several conditions under which the pair $\left(  L,f\right)  $ satisfies the
$n$-PBW condition. But things turn out simpler this time: The $n$-PBW
condition is a tautological condition, since the $n$-PBW homomorphism of
$\left(  L,f\right)  $ always is an isomorphism! But actually things are even
simpler: Not only do we have an isomorphism $\wedge^{n}L\cong%
\operatorname*{gr}\nolimits_{n}\left(  \operatorname*{Cl}\left(  L,f\right)
\right)  $ (the $n$-PBW homomorphism), but also this isomorphism has the form
$\operatorname*{gr}\nolimits_{n}\left(  \underline{\alpha}_{\mathbf{0}}%
^{f}\right)  $ for an isomorphism $\underline{\alpha}_{\mathbf{0}}^{f}:\wedge
L\rightarrow\operatorname*{Cl}\left(  L,f\right)  $ (a $k$-module isomorphism,
not a $k$-algebra isomorphism, of course). This means that not only the
associated graded modules of $\wedge L$ and $\operatorname*{Cl}\left(
L,f\right)  $ are isomorphic, but also the $k$-modules $\wedge L$ and
$\operatorname*{Cl}\left(  L,f\right)  $ are isomorphic themselves. We record
this as a theorem:

\begin{theorem}
\label{thm.cliff.pbw}Let $k$ be a commutative ring. Let $L$ be a $k$-module.
Let $f:L\times L\rightarrow k$ be a $k$-bilinear form on $L$. Then, there
exists an isomorphism $\operatorname*{Cl}\left(  L,f\right)  \cong\wedge L$ of
filtered $k$-modules.
\end{theorem}

This theorem is not new. It is an obvious consequence of combining \cite[\S 9,
n$%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
$3, Proposition 3]{bourbaki-algIX} and \cite[\S 9, n$%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
$3, Lemme 4]{bourbaki-algIX}. It is also stated in \cite[Theorem
(2.16)]{baeza} for the case when $L$ is a finitely-generated projective
$k$-module, but the proof uses neither the finite generation nor the
projectivity assumption.

The particular case of Theorem \ref{thm.cliff.pbw} when $k$ is a field is a
rather well-known fact, which is unfortunately usually proven in ways which
don't extend to the general case. In 2010, I rediscovered Theorem
\ref{thm.cliff.pbw} as a generalization of this fact, and wrote down a proof
in \cite[Theorem 1]{dg1}, unaware of the result already been known.

\subsubsection{A very rough outline of the proof}

The proofs of Theorem \ref{thm.cliff.pbw} given in \cite{bourbaki-algIX},
\cite{baeza} and \cite{dg1} are essentially one and the same argument (but
vary in notation and in the level of detail). I will sketch this
argument\footnote{The notations I will be using in the following are those of
\cite[Theorem 1]{dg1}.}, because it was the archetype for my construction of
the map $\varphi$ in Definition \ref{defs.l25.core} and for my construction of
the map $\gamma$ in Definition \ref{defs.l37.core}.

The proof of Theorem \ref{thm.cliff.pbw} proceeds in a purely computational
way by recursively constructing both an isomorphism $\underline{\alpha
}_{\mathbf{0}}^{f}:\wedge L\rightarrow\operatorname*{Cl}\left(  L,f\right)  $
(which actually turns out to be induced by an automorphism $\alpha^{f}:\otimes
L\rightarrow\otimes L$\ \ \ \ \footnote{This automorphism $\alpha^{f}$ would
be called $\lambda_{-f}$ in the notations of \cite[\S 9, n$%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
$2]{bourbaki-algIX}, and would be called $\widehat{-f}$ in the notations of
\cite[Chapter 2, \S 2]{baeza}.}) and its inverse (which turns out to be
induced by an automorphism $\alpha^{-f}:\otimes L\rightarrow\otimes L$ which
is constructed in the same way as $\alpha^{f}:\otimes L\rightarrow\otimes L$
except that it is based on the form $-f$ rather than $f$). Note that the
isomorphism $\underline{\alpha}_{\mathbf{0}}^{f}:\wedge L\rightarrow
\operatorname*{Cl}\left(  L,f\right)  $ is called the \textit{quantization
map} in \cite[\S 2.5]{meinrenken}, while its inverse is called the
\textit{symbol map}. Here is how these isomorphisms are constructed:

The construction starts off by defining a tensor $v\overset{f}{\llcorner}%
U\in\otimes L$ for every $v\in L$ and $U\in\otimes L$.\ \ \ \ \footnote{What I
call $v\overset{f}{\llcorner}U$ would be called $i_{v}^{f}\left(  U\right)  $
in the notations of \cite[\S 9, n$%
%TCIMACRO{\U{b0}}%
%BeginExpansion
{{}^\circ}%
%EndExpansion
$2]{bourbaki-algIX}, and would be called $\overline{f_{v}}\left(  U\right)  $
in the notations of \cite[Chapter 2, \S 2]{baeza}.} This tensor
$v\overset{f}{\llcorner}U$ is defined in such a way that it bilinearly depends
on $\left(  v,U\right)  $, and satisfies%
\begin{align*}
v\overset{f}{\llcorner}\left(  u_{1}\otimes u_{2}\otimes...\otimes
u_{p}\right)   &  =\sum_{i=1}^{p}\left(  -1\right)  ^{i-1}f\left(
v,u_{i}\right)  \cdot u_{1}\otimes u_{2}\otimes...\otimes\widehat{u_{i}%
}\otimes...\otimes u_{p}\\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{for every }v\in L\text{ and }u_{1}%
,u_{2},...,u_{p}\in L\right.
\end{align*}
(where the hat over $u_{i}$ means "omit the tensorand $u_{i}$ from this tensor
product"). (This is easily seen to be well-defined. The definition given in
\cite{dg1} is slightly different, but easily shown equivalent to the one given here.)

Now, we define a $k$-linear map $\alpha^{f}:\otimes L\rightarrow\otimes L$ by%
\begin{align*}
\alpha^{f}\left(  \lambda\right)   &  =\lambda\ \ \ \ \ \ \ \ \ \ \text{for
every }\lambda\in k=L^{\otimes0};\\
\alpha^{f}\left(  u\cdot U\right)   &  =u\cdot\alpha^{f}\left(  U\right)
-u\overset{f}{\llcorner}\alpha^{f}\left(  U\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}\text{, }u\in L\text{ and
}U\in L^{\otimes n}\text{.}%
\end{align*}
After a bit of work, we see that this map $\alpha^{f}$ is well-defined and
respects the degree filtration of $\otimes L$.

We can write down explicit formulae for $\alpha^{f}$ in low degrees:%
\begin{align*}
\alpha^{f}\left(  \lambda\right)   &  =\lambda\ \ \ \ \ \ \ \ \ \ \text{for
every }\lambda\in k=L^{\otimes0};\\
\alpha^{f}\left(  u\right)   &  =u\ \ \ \ \ \ \ \ \ \ \text{for any }u\in L;\\
\alpha^{f}\left(  u\otimes v\right)   &  =u\otimes v-f\left(  u,v\right)
\ \ \ \ \ \ \ \ \ \ \text{for any }u,v\in L;\\
\alpha^{f}\left(  u\otimes v\otimes w\right)   &  =u\otimes v\otimes
w-f\left(  v,w\right)  u+f\left(  u,w\right)  v-f\left(  u,v\right)
w\ \ \ \ \ \ \ \ \ \ \text{for any }u,v,w\in L;\\
\alpha^{f}\left(  u\otimes v\otimes w\otimes t\right)   &  =u\otimes v\otimes
w\otimes t-f\left(  w,t\right)  u\otimes v+f\left(  v,t\right)  u\otimes
w-f\left(  v,w\right)  u\otimes t\\
&  \ \ \ \ \ \ \ \ \ \ -f\left(  u,v\right)  w\otimes t+f\left(  u,w\right)
v\otimes t-f\left(  u,t\right)  v\otimes w\\
&  \ \ \ \ \ \ \ \ \ \ +f\left(  w,t\right)  f\left(  u,v\right)  -f\left(
v,t\right)  f\left(  u,w\right)  +f\left(  v,w\right)  f\left(  u,t\right) \\
&  \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ for any }u,v,w,t\in L.
\end{align*}
See also \cite[\S 5]{dg1} for a general (but rather unwieldy) combinatorial
expression for $\alpha^{f}\left(  u_{1}\otimes u_{2}\otimes...\otimes
u_{n}\right)  $ for arbitrary $n$.

The maps $\alpha^{f}$ for various bilinear forms $f$ satisfy some surprising
properties: First, $\alpha^{\mathbf{0}}=\operatorname*{id}$ and $\alpha
^{f}\circ\alpha^{g}=\alpha^{f+g}$ for any two bilinear forms $f$ and $g$ (see
\cite[Theorem 32]{dg1}). (This means that $f\mapsto\alpha^{f}$ defines a
representation of the additive group $\left\{  f:L\times L\rightarrow
k\ \mid\ f\text{ is }k\text{-bilinear}\right\}  $ on $\otimes L$. In how far
this can be related to the general representation theory of Lie/algebraic
groups is unclear to me at the moment.) As a consequence, $\alpha^{f}$ is an
automorphism of $\otimes L$, and $\alpha^{-f}$ is its inverse.

Another property that can be showed by computation (\cite[Theorem 31]{dg1}) is
that $\alpha^{f}\left(  I_{g}\right)  =I_{f+g}$ for any two $k$-bilinear forms
$f$ and $g$. In particular, this yields $\alpha^{f}\left(  I_{\mathbf{0}%
}\right)  =I_{f}$ and $\alpha^{-f}\left(  I_{f}\right)  =I_{\mathbf{0}}$.
Since $\alpha^{f}\left(  I_{\mathbf{0}}\right)  =I_{f}$, the map $\alpha^{f}$
induces a $k$-module homomorphism from $\left(  \otimes L\right)  \diagup
I_{\mathbf{0}}=\wedge L$ to $\left(  \otimes L\right)  \diagup I_{f}%
=\operatorname*{Cl}\left(  L,f\right)  $. Similarly, the map $\alpha^{-f}$
induces the inverse of this $k$-module homomorphism. So we have constructed
our isomorphism between $\wedge L$ and $\operatorname*{Cl}\left(  L,f\right)
$.

All steps of this argument can be found in the detailed version of \cite{dg1}.
However, the reader will probably be able to reconstruct them on her own using
from the sketch given above, since most of what has been omitted is
straightforward computation and induction arguments.

When $k$ is a field of characteristic $\neq2$, much shorter proofs of Theorem
\ref{thm.cliff.pbw} abound (for example, a standard proof proceeds by
symmetrization of the bilinear form $f$ and subsequent construction of a
Gram-Schmidt orthogonal basis of $L$, which gives a "canonical" form for the
Clifford algebra). Probably because most users of Clifford algebras come from
a geometrical or physical background and have little use for the luxury of
allowing $k$ to be an arbitrary commutative ring (or a field of characteristic
$2$), the general case of Theorem \ref{thm.cliff.pbw} appears to be little
known to the mathematical community. However, it is the general, computational
proof of Theorem \ref{thm.cliff.pbw} which, by its inductive construction of
the map $\alpha^{f}$, motivated my arguments in Sections \ref{sect.L25} and
\ref{sect.split} of the present paper. In fact, compare the above inductive
definition%
\begin{align*}
\alpha^{f}\left(  \lambda\right)   &  =\lambda\ \ \ \ \ \ \ \ \ \ \text{for
every }\lambda\in k=L^{\otimes0};\\
\alpha^{f}\left(  u\cdot U\right)   &  =u\cdot\alpha^{f}\left(  U\right)
-u\overset{f}{\llcorner}\alpha^{f}\left(  U\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}\text{, }u\in L\text{ and
}U\in L^{\otimes n}%
\end{align*}
of the map $\alpha^{f}$ in \cite{dg1} with the inductive definition%
\begin{align*}
\varphi\left(  \lambda\right)   &  =\lambda\ \ \ \ \ \ \ \ \ \ \text{for every
}\lambda\in k=\mathfrak{g}^{\otimes0};\\
\varphi\left(  u\cdot U\right)   &  =t\left(  u\right)  \cdot\varphi\left(
U\right)  +\varphi\left(  s\left(  u\right)  \rightharpoonup U\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}\text{, }u\in
\mathfrak{g}\text{ and }U\in\mathfrak{g}^{\otimes n}%
\end{align*}
of the map $\varphi$ in Section \ref{sect.L25} of the present paper (this is
not exactly the way we defined $\varphi$ in Definition \ref{defs.l25.core},
but it is easily seen to be equivalent) and with the inductive definition%
\begin{align*}
\gamma\left(  \lambda\right)   &  =\lambda\ \ \ \ \ \ \ \ \ \ \text{for every
}\lambda\in k=\mathfrak{g}^{\otimes0};\\
\gamma\left(  u\cdot U\right)   &  =\pi\left(  u\right)  \cdot\gamma\left(
U\right)  +u\rightharpoonup\left(  \gamma\left(  U\right)  \right)
\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}\text{, }u\in
\mathfrak{g}\text{ and }U\in\mathfrak{g}^{\otimes n}%
\end{align*}
of the map $\gamma$ in Section \ref{sect.split} of the present paper (this is
not exactly the way we defined $\gamma$ in Definition \ref{defs.l37.core}, but
it is easily seen to be equivalent). The similarity between the terms
$u\overset{f}{\llcorner}\alpha^{f}\left(  U\right)  $ and $u\rightharpoonup
\left(  \gamma\left(  U\right)  \right)  $ is particularly obvious, since
$\overset{f}{\llcorner}$ is a quadratic-space analogue of the $\mathfrak{g}%
$-action $\rightharpoonup$.

\subsubsection{The heuristics of the proof}

What idea was behind the recursive definition of the map $\alpha^{f}$ in
\cite{dg1}? It was inspired by the standard construction of the quantization
map in characteristic $0$, which, I think, goes back to Chevalley. This
construction gives the following formula for the quantization map $q:\wedge
L\rightarrow\operatorname*{Cl}\left(  L,f\right)  $ (this $q$ is my
$\underline{\alpha}_{\mathbf{0}}^{f}$) when $k$ is a field of characteristic
$0$ and $f$ is a symmetric bilinear form:%
\[
q\left(  u_{1}\wedge u_{2}\wedge...\wedge u_{n}\right)  =\dfrac{1}{n!}%
\sum_{s\in S_{n}}\left(  -1\right)  ^{s}u_{s\left(  1\right)  }u_{s\left(
2\right)  }...u_{s\left(  n\right)  }\ \ \ \ \ \ \ \ \ \ \text{for every
}u_{1},u_{2},...,u_{n}\in L
\]
(the right hand side is to be understood as a product in $\operatorname*{Cl}%
\left(  L,f\right)  $). This appears, e. g., in \cite[Chapter 2, Proposition
2.9]{meinrenken}. I tried to transform this formula for $q$ into an equivalent
form which did not require $k$ to have characteristic $0$ anymore (i. e.,
which did not contain the $\dfrac{1}{n!}$ in front of the sum). By trial and
error, I came up with the following:%
\begin{align*}
q\left(  u_{1}\right)   &  =u_{1}\ \ \ \ \ \ \ \ \ \ \left(  \text{this is
already okay}\right)  ;\\
q\left(  u_{1}\wedge u_{2}\right)   &  =\dfrac{1}{2}\left(  u_{1}u_{2}%
-u_{2}u_{1}\right)  =\dfrac{1}{2}\left(  u_{1}u_{2}+u_{1}u_{2}-2f\left(
u_{1},u_{2}\right)  \right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{since }u_{1}u_{2}+u_{2}u_{1}=\left(  u_{1}+u_{2}\right)  ^{2}-u_{1}%
^{2}-u_{2}^{2}\\
=f\left(  u_{1}+u_{2},u_{1}+u_{2}\right)  -f\left(  u_{1},u_{1}\right)
-f\left(  u_{2},u_{2}\right)  =2f\left(  u_{1},u_{2}\right) \\
\text{ in }\operatorname*{Cl}\left(  L,f\right)  \text{ (because }f\text{ is
symmetric)}%
\end{array}
\right) \\
&  =u_{1}u_{2}-f\left(  u_{1},u_{2}\right)  ;\\
q\left(  u_{1}\wedge u_{2}\wedge u_{3}\right)   &  =\dfrac{1}{6}\left(
u_{1}u_{2}u_{3}-u_{1}u_{3}u_{2}-u_{2}u_{1}u_{3}-u_{3}u_{2}u_{1}+u_{2}%
u_{3}u_{1}+u_{3}u_{1}u_{2}\right) \\
&  =...\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{another computation using the symmetry of }f\text{ and}\\
\text{the identity }v^{2}=f\left(  v,v\right)  \text{ in }\operatorname*{Cl}%
\left(  L,f\right)  \text{ omitted}%
\end{array}
\right) \\
&  =u_{1}u_{2}u_{3}-f\left(  u_{2},u_{3}\right)  u_{1}+f\left(  u_{1}%
,u_{3}\right)  u_{2}-f\left(  u_{1},u_{2}\right)  u_{3}.
\end{align*}
These results suggested me the recursive equation $q\left(  u\wedge U\right)
=u\cdot q\left(  U\right)  -u\overset{f}{\llcorner}q\left(  U\right)  $ (where
$u\overset{f}{\llcorner}q\left(  U\right)  $ makes sense because
$u\overset{f}{\llcorner}I_{f}\subseteq I_{f}$). Once this equation was found,
the next obvious step was to lift the map $q:\wedge L\rightarrow
\operatorname*{Cl}\left(  L,f\right)  $ to a map $\alpha^{f}:\otimes
L\rightarrow\otimes L$ because tensors are easier to deal with than elements
of $\wedge L$. The most straightforward approach to construct such a lifting
is by lifting the recursive equation $q\left(  u\wedge U\right)  =u\cdot
q\left(  U\right)  -u\overset{f}{\llcorner}q\left(  U\right)  $ to $\otimes
L$; so, I defined a map $\alpha^{f}:\otimes L\rightarrow\otimes L$ by
\begin{align*}
\alpha^{f}\left(  \lambda\right)   &  =\lambda\ \ \ \ \ \ \ \ \ \ \text{for
every }\lambda\in k=L^{\otimes0};\\
\alpha^{f}\left(  u\cdot U\right)   &  =u\cdot\alpha^{f}\left(  U\right)
-u\overset{f}{\llcorner}\alpha^{f}\left(  U\right)
\ \ \ \ \ \ \ \ \ \ \text{for every }n\in\mathbb{N}\text{, }u\in L\text{ and
}U\in L^{\otimes n}.
\end{align*}


It turned out that this map $\alpha^{f}$ is an isomorphism (due to Proposition
\ref{prop.filtr.iso}) and that $\alpha^{f}\left(  I_{\mathbf{0}}\right)
\subseteq I_{f}$ (by computation). Yet, this did not yet prove that $q$ is an
isomorphism; in fact, the latter would require showing that $\alpha^{f}\left(
I_{\mathbf{0}}\right)  =I_{f}$, and not only $\alpha^{f}\left(  I_{\mathbf{0}%
}\right)  \subseteq I_{f}$. Again, the most straightforward (to a
constructivist) approach to this problem was to construct the inverse of
$\alpha^{f}$ by recursion. Some experimentation showed that its inverse
$\left(  \alpha^{f}\right)  ^{-1}$ satisfies exactly the same recursive
equation as $\alpha^{f}$, up to a sign change:%
\[
\left(  \alpha^{f}\right)  ^{-1}\left(  u\cdot U\right)  =u\cdot\left(
\alpha^{f}\right)  ^{-1}\left(  U\right)  +u\overset{f}{\llcorner}\left(
\alpha^{f}\right)  ^{-1}\left(  U\right)  .
\]
This means this inverse is $\alpha^{-f}$. Searching for a reason why
$\alpha^{-f}$ is the inverse of $\alpha^{f}$, I began to suspect the
$\alpha^{f}\circ\alpha^{g}=\alpha^{f+g}$ identity, and it did not take long
for this identity to be proven (as everything is defined recursively, making
induction easy). The rest was automatic. At the end of the journey, I was met
by the surprising realization that $f$ was nowhere required to be symmetric.

Note that \cite[Theorem 38]{dg1} shows that my map $\underline{\alpha
}_{\mathbf{0}}^{f}$ is indeed the same as $q$ as long as the form $f$ is symmetric.

\subsubsection{A relative Poincar\'{e}-Birkhoff-Witt for Clifford algebras}

Just like the standard Poincar\'{e}-Birkhoff-Witt theorem, the relative
Poincar\'{e}-Birkhoff-Witt theorem (for example, in the avatar of Lemma
\ref{intro.lem4.2}), too, has a quadratic-space analogue with weaker
conditions and a stronger assertion:

\begin{theorem}
\label{thm.cliff.pbw.rel}Let $k$ be a commutative ring. Let $L$ be a
$k$-module. Let $f:L\times L\rightarrow k$ be a $k$-bilinear form on $L$. Let
$M$ be a $k$-submodule of $L$ such that $f\left(  M\times M\right)  =0$ and
such that the $k$-module inclusion $M\hookrightarrow L$ splits. Then, there
exists a $k$-module isomorphism $\operatorname*{Cl}\left(  L,f\right)
\cong\wedge L$ which maps $\operatorname*{Cl}\left(  L,f\right)  \cdot M$ to
$\left(  \wedge L\right)  \cdot M$. Therefore, $\left(  \operatorname*{Cl}%
\left(  L,f\right)  \right)  \diagup\left(  \operatorname*{Cl}\left(
L,f\right)  \cdot M\right)  \cong\left(  \wedge L\right)  \diagup\left(
\left(  \wedge L\right)  \cdot M\right)  \cong\wedge\left(  L\diagup M\right)
$.
\end{theorem}

A\ proof of this theorem can be found in \cite[Theorem 61 \textbf{(b)}]{dg1}.
The condition $f\left(  M\times M\right)  =0$ is a quadratic-space analogue of
the condition that $\mathfrak{h}$ is a Lie subalgebra of $\mathfrak{g}$; it
cannot be improved. The condition that the $k$-module inclusion $M\rightarrow
L$ splits cannot be dropped either, but possibly can be weakened.

\subsubsection{Remark on Weyl algebras}

It is a known fact that if we extend the notion of Clifford algebras to
$k$-supermodules rather than $k$-modules only, then we obtain the tensor
product of the Clifford algebra of the even part and the Weyl algebra of the
odd part - however, at the price of requiring that $2$ is invertible in $k$.
As long as we are ready to pay this price, all of our results on Clifford
algebras carry over to Weyl algebras.

\subsubsection{A relative version of Theorem \ref{thm.l25}}

In the spirit of the above quadratic-space versions, here is an analogue of
Theorem \ref{thm.l25}:

\begin{theorem}
\label{thm.cliff.l25}Let $k$ be a commutative ring. Let $L$ be a $k$-module,
and let $M$ be a $k$-submodule of $L$. Assume that the inclusion
$M\hookrightarrow L$ splits. Let $N$ be the $k$-module $L\diagup M$.\newline
Let $f:L\times M\rightarrow k$ be a $k$-bilinear form such that $f\left(
M\times M\right)  =0$.\newline Let $J$ be the two-sided ideal%
\[
\left(  \otimes L\right)  \cdot\left\langle v\otimes w-w\otimes v-f\left(
v,w\right)  \ \mid\ \left(  v,w\right)  \in L\times M\right\rangle
\cdot\left(  \otimes L\right)
\]
of the $k$-algebra $\otimes L$.\newline The $k$-module $\left(  \otimes
L\right)  \diagup\left(  J+\left(  \otimes L\right)  \cdot M\right)  $ is
isomorphic to the $k$-module $\otimes N$. (More detailed assertions are left
to the reader.)
\end{theorem}

\subsection{\label{subsect.goodwillie}Flat modules}

Our formulation of Theorem \ref{thm.l45} (the main result of this paper)
easily provokes the question whether some of its many conditions can be lifted
or, at least, weakened. The latter is indeed the case:

\begin{theorem}
\label{thm.goodwillie.l45}Theorem \ref{thm.l45} still holds if we replace the
sentence "Assume that both $\mathfrak{h}$ and $N$ are free $k$-modules" by
"Assume that $\mathfrak{g}$ and $N$ are flat $k$-modules".
\end{theorem}

In order to prove this Theorem \ref{thm.goodwillie.l45}, we notice that most
steps of our proof of Theorem \ref{thm.l45} did not use the condition that
both $\mathfrak{h}$ and $N$ are free $k$-modules. The only steps that did were
the ones that used the $n$-PBW condition, and the one that used Proposition
\ref{prop.pbw.U(g)h}. As for the $n$-PBW condition, it still holds under the
weakened assumption ("Assume that $\mathfrak{g}$ and $N$ are flat
$k$-modules") due to Theorem \ref{thm.pbw} \textbf{(g)}, so there is no
trouble to expect from its direction. As for Proposition \ref{prop.pbw.U(g)h},
we have to generalize it as follows:

\begin{proposition}
\label{prop.pbw.goodwillie.U(g)h}Let $k$ be a commutative ring. Let
$\mathfrak{g}$ be a $k$-Lie algebra. Let $m\in\mathbb{N}$.\newline Let
$\mathfrak{h}$ be a Lie subalgebra of $\mathfrak{g}$ such that there exists a
flat $k$-submodule $N$ of $\mathfrak{g}$ such that $\mathfrak{g}%
=\mathfrak{h}\oplus N$. Assume further that the $k$-Lie algebra $\mathfrak{g}$
itself satisfies the $n$-PBW condition for every $n\in\mathbb{N}$.\newline
Then, $U_{\leq m}\left(  \mathfrak{g}\right)  \cap\left(  U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  =U_{\leq\left(  m-1\right)
}\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}$. (Here, we are using the
notation of Definition \ref{conv.Un}, and we are abbreviating the
$k$-submodule $U\left(  \mathfrak{g}\right)  \cdot\psi\left(  \mathfrak{h}%
\right)  $ of $U\left(  \mathfrak{g}\right)  $ by $U\left(  \mathfrak{g}%
\right)  \cdot\mathfrak{h}$.)
\end{proposition}

For the proof of this proposition, we need a lemma that was proven by Thomas
Goodwillie \cite{goodwillie-MO}:

\begin{lemma}
[Goodwillie]\label{lem.goodwillie}Let $k$ be a commutative ring. Let $A$ be
some $k$-module, and let $B$ be a $k$-submodule of $A$ such that the
$k$-module $A\diagup B$ is flat.\newline Let $i\in\mathbb{N}$ be such that
$i\geq1$.\newline Let $\mathbf{m}_{1}$ denote the canonical map $K_{i}\left(
A\right)  \otimes B\rightarrow A^{\otimes i}\otimes B$.\newline Let
$\mathbf{m}_{2}$ denote the canonical map $A^{\otimes i}\otimes B\rightarrow
A^{\otimes i}\otimes A\overset{\cong}{\rightarrow}A^{\otimes\left(
i+1\right)  }$.\newline Let $\mathbf{m}_{3}$ denote the canonical map
$A^{\otimes\left(  i-1\right)  }\otimes K_{2}\left(  B\right)  \rightarrow
A^{\otimes\left(  i-1\right)  }\otimes B^{\otimes2}\rightarrow A^{\otimes
\left(  i-1\right)  }\otimes A^{\otimes2}\overset{\cong}{\rightarrow
}A^{\otimes\left(  i+1\right)  }$.\newline Then,%
\begin{equation}
K_{i+1}\left(  A\right)  \cap\mathbf{m}_{2}\left(  A^{\otimes i}\otimes
B\right)  =\mathbf{m}_{2}\left(  \mathbf{m}_{1}\left(  K_{i}\left(  A\right)
\otimes B\right)  \right)  +\mathbf{m}_{3}\left(  A^{\otimes\left(
i-1\right)  }\otimes K_{2}\left(  B\right)  \right)  .
\label{lem.goodwillie.1}%
\end{equation}

\end{lemma}

\begin{remark}
All three maps $\mathbf{m}_{1}$, $\mathbf{m}_{2}$, $\mathbf{m}_{3}$ in Lemma
\ref{lem.goodwillie} are obtained by tensoring some inclusions with identity
maps and composing. (For example, $\mathbf{m}_{1}$ is obtained by tensoring
the inclusion $K_{i}\left(  A\right)  \rightarrow A^{\otimes i}$ with the
identity map $B\rightarrow B$.) This yields that these maps are injective
whenever $k$ is a field (or at least some flatness conditions are satisfied).
Therefore, when $k$ is a field, these three maps are often regarded as
inclusions and thus suppressed from the equality (\ref{lem.goodwillie.1}) (so
that this equality takes the simple-looking form $K_{i+1}\left(  A\right)
\cap\left(  A^{\otimes i}\otimes B\right)  =K_{i}\left(  A\right)  \otimes
B+A^{\otimes\left(  i-1\right)  }\otimes K_{2}\left(  B\right)  $). However,
we are considering a more general case here, and I do not believe that these
maps $\mathbf{m}_{1}$, $\mathbf{m}_{2}$, $\mathbf{m}_{3}$ are always injective
in our case; thus, suppressing these maps from (\ref{lem.goodwillie.1}) is not
justified for us.
\end{remark}

Note that Lemma \ref{lem.goodwillie} does not involve any Lie algebras; it is
a purely module-theoretical lemma and probably has its right place in
homological algebra. We refer to \cite{goodwillie-MO} for a proof of this
lemma (where $i$ was called $m-1$).

Let us rewrite Lemma \ref{lem.goodwillie} with the help of the tensor algebra first:

\begin{lemma}
[Goodwillie]\label{lem.goodwillie.2}Let $k$ be a commutative ring. Let $A$ be
some $k$-module, and let $B$ be a $k$-submodule of $A$ such that the
$k$-module $A\diagup B$ is flat.\newline Let $i\in\mathbb{N}$ be such that
$i\geq1$.\newline Then, the following equality of subsets of the tensor
algebra $\otimes A$ holds:%
\[
K_{i+1}\left(  A\right)  \cap\left(  A^{\otimes i}\cdot B\right)
=K_{i}\left(  A\right)  \cdot B+A^{\otimes\left(  i-1\right)  }\cdot
K_{2,A}\left(  B\right)  .
\]
\newline Here, we are identifying $B$ with a submodule of $\otimes A$ (due to
$B\subseteq A\subseteq\otimes A$), and denoting by $K_{2,A}\left(  B\right)  $
the image of $K_{2}\left(  B\right)  $ under the canonical map $\otimes
B\rightarrow\otimes A$.
\end{lemma}

\textit{Proof of Lemma \ref{lem.goodwillie.2}.} With the notations of Lemma
\ref{lem.goodwillie}, we have $\mathbf{m}_{2}\left(  A^{\otimes i}\otimes
B\right)  =A^{\otimes i}\cdot B$, $\mathbf{m}_{2}\left(  \mathbf{m}_{1}\left(
K_{i}\left(  A\right)  \otimes B\right)  \right)  =K_{i}\left(  A\right)
\cdot B$ and $\mathbf{m}_{3}\left(  A^{\otimes\left(  i-1\right)  }\otimes
K_{2}\left(  B\right)  \right)  =A^{\otimes\left(  i-1\right)  }\cdot
K_{2,A}\left(  B\right)  $. Therefore, Lemma \ref{lem.goodwillie.2} follows
from Lemma \ref{lem.goodwillie}.

\textit{Proof of Proposition \ref{prop.pbw.goodwillie.U(g)h}.} Since it is
trivial that $U_{\leq\left(  m-1\right)  }\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}\subseteq U_{\leq m}\left(  \mathfrak{g}\right)  \cap\left(
U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ (just as in the
proof of Proposition \ref{prop.pbw.U(g)h}), we only have to prove that
$U_{\leq m}\left(  \mathfrak{g}\right)  \cap\left(  U\left(  \mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  \subseteq U_{\leq\left(  m-1\right)
}\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}$.

Let us prove that%
\begin{equation}
\left(  \text{every integer }i\geq m\text{ satisfies }U_{\leq m}\left(
\mathfrak{g}\right)  \cap\left(  U_{\leq i}\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}\right)  \subseteq U_{\leq m}\left(  \mathfrak{g}\right)
\cap\left(  U_{\leq\left(  i-1\right)  }\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}\right)  \right)  . \label{prop.pbw.goodwillie.1}%
\end{equation}


Why prove this? Because once it is proven, Proposition
\ref{prop.pbw.goodwillie.U(g)h} follows by a simple induction argument (which
we are going to show in more details after we have proven
(\ref{prop.pbw.goodwillie.1})).

\textit{Proof of (\ref{prop.pbw.goodwillie.1}).} We assume WLOG that $i\geq1$
(because otherwise, $i=0$ and $i\geq m$ lead to $m=0$, and the whole statement
of (\ref{prop.pbw.goodwillie.1}) boils down to a triviality).

Let $x\in U_{\leq m}\left(  \mathfrak{g}\right)  \cap\left(  U_{\leq i}\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ be arbitrary. Then, $x\in
U_{\leq m}\left(  \mathfrak{g}\right)  $ and $x\in U_{\leq i}\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}$.

The projection $\psi:\otimes\mathfrak{g}\rightarrow U\left(  \mathfrak{g}%
\right)  $ clearly satisfies $U_{\leq i}\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}=\psi\left(  \mathfrak{g}^{\otimes\leq i}\cdot\mathfrak{h}%
\right)  $. Thus, $x\in U_{\leq i}\left(  \mathfrak{g}\right)  \cdot
\mathfrak{h}=\psi\left(  \mathfrak{g}^{\otimes\leq i}\cdot\mathfrak{h}\right)
$, so that there exists some $y\in\mathfrak{g}^{\otimes\leq i}\cdot
\mathfrak{h}$ such that $x=\psi\left(  y\right)  $. Consider this $y$.

Let $n=i+1$. Then, $i=n-1$.

Now, $y\in\mathfrak{g}^{\otimes\leq i}\cdot\underbrace{\mathfrak{h}%
}_{\subseteq\mathfrak{g}}\subseteq\mathfrak{g}^{\otimes\leq i}\cdot
\mathfrak{g}\subseteq\mathfrak{g}^{\otimes\leq\left(  i+1\right)
}=\mathfrak{g}^{\otimes\leq n}$ (since $i+1=n$), so that we can speak of the
element $\overline{y}\in\operatorname*{gr}\nolimits_{n}\left(  \otimes
\mathfrak{g}\right)  $. This element satisfies $\left(  \operatorname*{gr}%
\nolimits_{n}\psi\right)  \left(  \overline{y}\right)  =\overline{\psi\left(
y\right)  }=0$ (since $\psi\left(  y\right)  =x\in U_{\leq m}\left(
\mathfrak{g}\right)  \subseteq U_{\leq\left(  n-1\right)  }\left(
\mathfrak{g}\right)  $ (because $m\leq i=n-1$)). Thus, $\overline{y}%
\in\operatorname{Ker}\left(  \operatorname*{gr}\nolimits_{n}\psi\right)
=\operatorname*{grad}\nolimits_{\mathfrak{g},n}\left(  K_{n}\left(
\mathfrak{g}\right)  \right)  $ (by Proposition \ref{prop.pbw} \textbf{(b)}).
Let $z=\operatorname*{grad}\nolimits_{\mathfrak{g},n}^{-1}\left(  \overline
{y}\right)  $. Then, $\overline{y}\in\operatorname*{grad}%
\nolimits_{\mathfrak{g},n}\left(  K_{n}\left(  \mathfrak{g}\right)  \right)  $
leads to $z\in K_{n}\left(  \mathfrak{g}\right)  $.

On the other hand, $\operatorname*{grad}\nolimits_{\mathfrak{g},n}^{-1}\left(
\overline{y}\right)  $ is the $n$-th graded component of the tensor
$y\in\otimes\mathfrak{g}$ (in fact, for \textit{every} tensor $T\in
\mathfrak{g}^{\otimes\leq n}$, it is clear that $\operatorname*{grad}%
\nolimits_{\mathfrak{g},n}^{-1}\left(  \overline{T}\right)  $ is the $n$-th
graded component of $T$). Since $z=\operatorname*{grad}\nolimits_{\mathfrak{g}%
,n}^{-1}\left(  \overline{y}\right)  $, this means that $z$ is the $n$-th
graded component of the tensor $y\in\otimes\mathfrak{g}$. Since $n=i+1$, this
yields that $z$ is the $\left(  i+1\right)  $-th graded component of the
tensor $y\in\otimes\mathfrak{g}$. Thus, $z\in\mathfrak{g}^{\otimes i}%
\cdot\mathfrak{h}$ because of $y\in\mathfrak{g}^{\otimes\leq i}\cdot
\mathfrak{h}$ (since the $\left(  i+1\right)  $-th graded component of a
tensor in $\mathfrak{g}^{\otimes\leq i}\cdot\mathfrak{h}$ must always lie in
$\mathfrak{g}^{\otimes i}\cdot\mathfrak{h}$). Combined with $z\in K_{n}\left(
\mathfrak{g}\right)  =K_{i+1}\left(  \mathfrak{g}\right)  $ (since $n=i+1$),
this yields $z\in K_{i+1}\left(  \mathfrak{g}\right)  \cap\left(
\mathfrak{g}^{\otimes i}\cdot\mathfrak{h}\right)  $.

Now, let us recall that $y\in\mathfrak{g}^{\otimes\leq n}$, and that $z$ is
the $n$-th graded component of the tensor $y\in\otimes\mathfrak{g}$. Thus,
$y-z\in\mathfrak{g}^{\otimes\leq\left(  n-1\right)  }=\mathfrak{g}%
^{\otimes\leq i}$ (since $n-1=i$). Combined with $y-z\in\mathfrak{g}%
^{\otimes\leq i}\cdot\mathfrak{h}$ (since $y\in\mathfrak{g}^{\otimes\leq
i}\cdot\mathfrak{h}$ and $z\in\mathfrak{g}^{\otimes i}\cdot\mathfrak{h}%
\subseteq\mathfrak{g}^{\otimes\leq i}\cdot\mathfrak{h}$), this yields%
\begin{align*}
y-z  &  \in\mathfrak{g}^{\otimes\leq i}\cap\left(  \underbrace{\mathfrak{g}%
^{\otimes\leq i}}_{\subseteq\otimes\mathfrak{g}}\cdot\mathfrak{h}\right)
\subseteq\mathfrak{g}^{\otimes\leq i}\cap\left(  \left(  \otimes
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  =\mathfrak{g}^{\otimes
\leq\left(  i-1\right)  }\cdot\mathfrak{h}\\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because }\left(  \otimes\mathfrak{g}\right)  \cdot\mathfrak{h}\text{ is
a \textit{homogeneous} right ideal of }\otimes\mathfrak{g}\text{,}\\
\text{whose }p\text{-th graded component is }\mathfrak{g}^{\otimes\left(
p-1\right)  }\cdot\mathfrak{h}\text{ for every }p\in\mathbb{N}%
\end{array}
\right)  .
\end{align*}
Thus, $\psi\left(  y-z\right)  \in\psi\left(  \mathfrak{g}^{\otimes\leq\left(
i-1\right)  }\cdot\mathfrak{h}\right)  =U_{\leq\left(  i-1\right)  }\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}$.

Since the $k$-module $\mathfrak{g}\diagup\mathfrak{h}$ is flat (since
$\mathfrak{g}=\mathfrak{h}\oplus N$ yields $\mathfrak{g}\diagup\mathfrak{h}%
\cong N$, and we know that $N$ is flat), we can apply Lemma
\ref{lem.goodwillie.2} to $\mathfrak{g}$ and $\mathfrak{h}$ instead of $A$ and
$B$ (it is here that we use $i\geq1$), and obtain%
\begin{equation}
K_{i+1}\left(  \mathfrak{g}\right)  \cap\left(  \mathfrak{g}^{\otimes i}%
\cdot\mathfrak{h}\right)  =K_{i}\left(  \mathfrak{g}\right)  \cdot
\mathfrak{h}+\mathfrak{g}^{\otimes\left(  i-1\right)  }\cdot K_{2,\mathfrak{g}%
}\left(  \mathfrak{h}\right)  . \label{pf.pbw.goodwillie.5}%
\end{equation}


Now, let us look at $\psi\left(  K_{i}\left(  \mathfrak{g}\right)  \right)  $
and $\psi\left(  K_{2,\mathfrak{g}}\left(  \mathfrak{h}\right)  \right)  $
more closely.

Proposition \ref{prop.pbw} \textbf{(b) }(applied to $i$ instead of $n$) yields
$\operatorname{Ker}\left(  \operatorname*{gr}\nolimits_{i}\psi\right)
=\operatorname*{grad}\nolimits_{\mathfrak{g},i}\left(  K_{i}\left(
\mathfrak{g}\right)  \right)  $. This yields $\left(  \operatorname*{gr}%
\nolimits_{i}\psi\right)  \left(  \operatorname*{grad}\nolimits_{\mathfrak{g}%
,i}\left(  K_{i}\left(  \mathfrak{g}\right)  \right)  \right)  =0$ (although
this is clear from much simpler reasons). Thus,%
\[
0=\left(  \operatorname*{gr}\nolimits_{i}\psi\right)  \left(
\operatorname*{grad}\nolimits_{\mathfrak{g},i}\left(  K_{i}\left(
\mathfrak{g}\right)  \right)  \right)  =\left(  \operatorname{gr}_{i}\psi
\circ\operatorname*{grad}\nolimits_{\mathfrak{g},i}\right)  \left(
K_{i}\left(  \mathfrak{g}\right)  \right)  .
\]


Now let $\operatorname{inc}_{i}$ be the canonical inclusion $\mathfrak{g}%
^{\otimes i}\rightarrow\mathfrak{g}^{\otimes\leq i}$. Furthermore, let
$\psi_{i}:\mathfrak{g}^{\otimes\leq i}\rightarrow U_{\leq i}\left(
\mathfrak{g}\right)  $ be the homomorphism obtained from $\psi$ by restricting
the domain to $\mathfrak{g}^{\otimes\leq i}$ and the codomain to $U_{\leq
i}\left(  \mathfrak{g}\right)  $ (this is well-defined since $\psi\left(
\mathfrak{g}^{\otimes\leq i}\right)  \subseteq U_{\leq i}\left(
\mathfrak{g}\right)  $). Further, let $\operatorname{proj}_{i}$ be the
canonical projection $U_{\leq i}\left(  \mathfrak{g}\right)  \rightarrow
\left(  U_{\leq i}\left(  \mathfrak{g}\right)  \right)  \diagup\left(
U_{\leq\left(  i-1\right)  }\left(  \mathfrak{g}\right)  \right)
=\operatorname{gr}_{i}\left(  U\left(  \mathfrak{g}\right)  \right)  $. Then,
the following diagram commutes:%
\[
\xymatrixcolsep{4pc}\xymatrix{
\mathfrak g^{\otimes i} \ar[r]^-{\operatorname*{grad}_{\mathfrak g,i}} \ar[d]_{\psi_i\circ \operatorname*{inc}_i} & \operatorname*{gr}_i \left(\otimes \mathfrak g\right) \ar[d]^{\operatorname*{gr}_i \psi} \\
U_{\leq i}\left(\mathfrak g\right) \ar[r]_{\operatorname*{proj}_i} & \operatorname*{gr}_i\left(U\left(\mathfrak g\right)\right)
}
\]
(this is clear from the definitions of the arrows involved). In other words,
$\operatorname{gr}_{i}\psi\circ\operatorname*{grad}\nolimits_{\mathfrak{g}%
,i}=\operatorname{proj}_{i}\circ\psi_{i}\circ\operatorname{inc}_{i}$. Thus,
every $p\in K_{i}\left(  \mathfrak{g}\right)  $ satisfies
\begin{align*}
0  &  =\left(  \underbrace{\operatorname{gr}_{i}\psi\circ\operatorname*{grad}%
\nolimits_{\mathfrak{g},i}}_{=\operatorname{proj}_{i}\circ\psi_{i}%
\circ\operatorname{inc}_{i}}\right)  \left(  p\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{since }0=\left(  \operatorname{gr}_{i}%
\psi\circ\operatorname*{grad}\nolimits_{\mathfrak{g},i}\right)  \left(
K_{i}\left(  \mathfrak{g}\right)  \right)  \right) \\
&  =\left(  \operatorname{proj}_{i}\circ\psi_{i}\circ\operatorname{inc}%
_{i}\right)  \left(  p\right)  =\left(  \operatorname{proj}_{i}\circ\psi
_{i}\right)  \underbrace{\left(  \operatorname{inc}_{i}\left(  p\right)
\right)  }_{\substack{=p\text{ (since }\operatorname{inc}_{i}\\\text{is an
inclusion map)}}}\\
&  =\left(  \operatorname{proj}_{i}\circ\psi_{i}\right)  \left(  p\right)
=\operatorname{proj}_{i}\left(  \underbrace{\psi_{i}\left(  p\right)
}_{\substack{=\psi\left(  p\right)  \\\text{(by the definition of }\psi
_{i}\text{)}}}\right)  =\operatorname{proj}_{i}\left(  \psi\left(  p\right)
\right)  ,
\end{align*}
so that $\psi\left(  p\right)  \in\operatorname{Ker}\left(
\operatorname{proj}_{i}\right)  =U_{\leq\left(  i-1\right)  }\left(
\mathfrak{g}\right)  $. In other words,%
\begin{equation}
\psi\left(  K_{i}\left(  \mathfrak{g}\right)  \right)  \subseteq
U_{\leq\left(  i-1\right)  }\left(  \mathfrak{g}\right)  .
\label{pf.pbw.goodwillie.6}%
\end{equation}


On the other hand, Proposition \ref{rmk.symm-alg} (applied to $\mathfrak{h}$
and $2$ instead of $V$ and $n$) yields%
\[
K_{2}\left(  \mathfrak{h}\right)  =\sum_{i=1}^{1}\left\langle v_{1}\otimes
v_{2}-v_{\tau_{i}\left(  1\right)  }\otimes v_{\tau_{i}\left(  2\right)
}\text{ (seen as a tensor in }\otimes\mathfrak{h}\text{)}\ \mid\ \left(
v_{1},v_{2}\right)  \in\mathfrak{h}^{2}\right\rangle ,
\]
where $\tau_{1}$ is the transposition $\left(  1,2\right)  \in S_{2}$. This
immediately simplifies to%
\begin{align*}
K_{2}\left(  \mathfrak{h}\right)   &  =\left\langle v_{1}\otimes
v_{2}-\underbrace{v_{\tau_{1}\left(  1\right)  }}_{=v_{2}}\otimes
\underbrace{v_{\tau_{1}\left(  2\right)  }}_{=v_{1}}\text{ (seen as a tensor
in }\otimes\mathfrak{h}\text{)}\ \mid\ \left(  v_{1},v_{2}\right)
\in\mathfrak{h}^{2}\right\rangle \\
&  =\left\langle v_{1}\otimes v_{2}-v_{2}\otimes v_{1}\text{ (seen as a tensor
in }\otimes\mathfrak{h}\text{)}\ \mid\ \left(  v_{1},v_{2}\right)
\in\mathfrak{h}^{2}\right\rangle .
\end{align*}
Thus,
\begin{align*}
&  K_{2,\mathfrak{g}}\left(  \mathfrak{h}\right) \\
&  =\left(  \text{the image of }K_{2}\left(  \mathfrak{h}\right)  \text{ under
the canonical map }\otimes\mathfrak{h}\rightarrow\otimes\mathfrak{g}\right) \\
&  =\left(  \text{the image of }\left\langle v_{1}\otimes v_{2}-v_{2}\otimes
v_{1}\text{ (seen as a tensor in }\otimes\mathfrak{h}\text{)}\ \mid\ \left(
v_{1},v_{2}\right)  \in\mathfrak{h}^{2}\right\rangle \right. \\
&  \ \ \ \ \ \ \ \ \ \ \left.  \text{under the canonical map }\otimes
\mathfrak{h}\rightarrow\otimes\mathfrak{g}\right) \\
&  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{because}\\
K_{2}\left(  \mathfrak{h}\right)  =\left\langle v_{1}\otimes v_{2}%
-v_{2}\otimes v_{1}\text{ (seen as a tensor in }\otimes\mathfrak{h}%
\text{)}\ \mid\ \left(  v_{1},v_{2}\right)  \in\mathfrak{h}^{2}\right\rangle
\end{array}
\right) \\
&  =\left\langle v_{1}\otimes v_{2}-v_{2}\otimes v_{1}\text{ (seen as a tensor
in }\otimes\mathfrak{g}\text{)}\ \mid\ \left(  v_{1},v_{2}\right)
\in\mathfrak{h}^{2}\right\rangle ,
\end{align*}
so that%
\begin{align}
\psi\left(  K_{2,\mathfrak{g}}\left(  \mathfrak{h}\right)  \right)   &
=\psi\left(  \left\langle v_{1}\otimes v_{2}-v_{2}\otimes v_{1}\text{ (seen as
a tensor in }\otimes\mathfrak{g}\text{)}\ \mid\ \left(  v_{1},v_{2}\right)
\in\mathfrak{h}^{2}\right\rangle \right) \nonumber\\
&  =\left\langle \underbrace{\psi\left(  v_{1}\otimes v_{2}-v_{2}\otimes
v_{1}\text{ (seen as a tensor in }\otimes\mathfrak{g}\text{)}\right)
}_{\substack{=\psi\left(  \left[  v_{1},v_{2}\right]  \right)  \\\text{(since
}v_{1}\otimes v_{2}-v_{2}\otimes v_{1}-\left[  v_{1},v_{2}\right]  \in
I_{\mathfrak{g}}=\operatorname{Ker}\psi\text{)}}}\ \mid\ \left(  v_{1}%
,v_{2}\right)  \in\mathfrak{h}^{2}\right\rangle \nonumber\\
&  =\left\langle \psi\left(  \left[  v_{1},v_{2}\right]  \right)
\ \mid\ \left(  v_{1},v_{2}\right)  \in\mathfrak{h}^{2}\right\rangle
\nonumber\\
&  =\psi\left(  \left\langle \underbrace{\left[  v_{1},v_{2}\right]
}_{\substack{\in\mathfrak{h}\text{ (since }\left(  v_{1},v_{2}\right)
\in\mathfrak{h}^{2}\\\text{and since }\mathfrak{h}\text{ is a Lie subalgebra
of }\mathfrak{g}\text{)}}}\ \mid\ \left(  v_{1},v_{2}\right)  \in
\mathfrak{h}^{2}\right\rangle \right) \nonumber\\
&  \subseteq\psi\left(  \underbrace{\left\langle \mathfrak{h}\right\rangle
}_{=\mathfrak{h}}\right)  =\psi\left(  \mathfrak{h}\right)  .
\label{pf.pbw.goodwillie.7}%
\end{align}


Since $z\in K_{i+1}\left(  \mathfrak{g}\right)  \cap\left(  \mathfrak{g}%
^{\otimes i}\cdot\mathfrak{h}\right)  =K_{i}\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}+\mathfrak{g}^{\otimes\left(  i-1\right)  }\cdot
K_{2,\mathfrak{g}}\left(  \mathfrak{h}\right)  $ (by
(\ref{pf.pbw.goodwillie.5})), we have%
\begin{align*}
\psi\left(  z\right)   &  \in\psi\left(  K_{i}\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}+\mathfrak{g}^{\otimes\left(  i-1\right)  }\cdot
K_{2,\mathfrak{g}}\left(  \mathfrak{h}\right)  \right) \\
&  \subseteq\underbrace{\psi\left(  K_{i}\left(  \mathfrak{g}\right)  \right)
}_{\substack{\subseteq U_{\leq\left(  i-1\right)  }\left(  \mathfrak{g}%
\right)  \\\text{(by (\ref{pf.pbw.goodwillie.6}))}}}\cdot\psi\left(
\mathfrak{h}\right)  +\underbrace{\psi\left(  \mathfrak{g}^{\otimes\left(
i-1\right)  }\right)  }_{\subseteq U_{\leq\left(  i-1\right)  }\left(
\mathfrak{g}\right)  }\cdot\underbrace{\psi\left(  K_{2,\mathfrak{g}}\left(
\mathfrak{h}\right)  \right)  }_{\substack{\subseteq\psi\left(  \mathfrak{h}%
\right)  \\\text{(by (\ref{pf.pbw.goodwillie.7}))}}}\\
&  \subseteq\underbrace{U_{\leq\left(  i-1\right)  }\left(  \mathfrak{g}%
\right)  \cdot\psi\left(  \mathfrak{h}\right)  }_{=U_{\leq\left(  i-1\right)
}\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}}+\underbrace{U_{\leq\left(
i-1\right)  }\left(  \mathfrak{g}\right)  \cdot\psi\left(  \mathfrak{h}%
\right)  }_{=U_{\leq\left(  i-1\right)  }\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}}=U_{\leq\left(  i-1\right)  }\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}+U_{\leq\left(  i-1\right)  }\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}\\
&  =U_{\leq\left(  i-1\right)  }\left(  \mathfrak{g}\right)  \cdot
\mathfrak{h}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }U_{\leq\left(
i-1\right)  }\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\text{ is a
}k\text{-module}\right)  .
\end{align*}
But let us recall that we have shown that $\psi\left(  y-z\right)  \in
U_{\leq\left(  i-1\right)  }\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}$.
Now,%
\begin{align*}
x  &  =\psi\left(  y\right)  =\psi\left(  \left(  y-z\right)  +z\right)
=\underbrace{\psi\left(  y-z\right)  }_{\in U_{\leq\left(  i-1\right)
}\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}}+\underbrace{\psi\left(
z\right)  }_{\in U_{\leq\left(  i-1\right)  }\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }\psi\text{ is
}k\text{-linear}\right) \\
&  \in U_{\leq\left(  i-1\right)  }\left(  \mathfrak{g}\right)  \cdot
\mathfrak{h}+U_{\leq\left(  i-1\right)  }\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}=U_{\leq\left(  i-1\right)  }\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}\ \ \ \ \ \ \ \ \ \ \left(  \text{since }U_{\leq\left(
i-1\right)  }\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\text{ is a
}k\text{-module}\right)  .
\end{align*}
Combined with $x\in U_{\leq m}\left(  \mathfrak{g}\right)  $, this yields
$x\in U_{\leq m}\left(  \mathfrak{g}\right)  \cap\left(  U_{\leq\left(
i-1\right)  }\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $. Since
we have shown this for every $x\in U_{\leq m}\left(  \mathfrak{g}\right)
\cap\left(  U_{\leq i}\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)
$, we have therefore proven (\ref{prop.pbw.goodwillie.1}).

Now let us finish verifying $U_{\leq m}\left(  \mathfrak{g}\right)
\cap\left(  U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \subseteq
U_{\leq\left(  m-1\right)  }\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}$:

Let $x\in U_{\leq m}\left(  \mathfrak{g}\right)  \cap\left(  U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $ be arbitrary. We want to
prove that $x\in U_{\leq\left(  m-1\right)  }\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}$.

We have
\begin{align*}
x  &  \in U_{\leq m}\left(  \mathfrak{g}\right)  \cap\left(  U\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right)  \subseteq\underbrace{U\left(
\mathfrak{g}\right)  }_{=\bigcup\limits_{i\in\mathbb{N}}U_{\leq i}\left(
\mathfrak{g}\right)  }\cdot\mathfrak{h}=\left(  \bigcup\limits_{i\in
\mathbb{N}}U_{\leq i}\left(  \mathfrak{g}\right)  \right)  \cdot\mathfrak{h}\\
&  =\bigcup\limits_{i\in\mathbb{N}}\left(  U_{\leq i}\left(  \mathfrak{g}%
\right)  \cdot\mathfrak{h}\right)  \ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{this is because }\bigcup\limits_{i\in\mathbb{N}}U_{\leq i}\left(
\mathfrak{g}\right)  \text{ is an increasing union,}\\
\text{and increasing unions commute with multiplication}%
\end{array}
\right)  .
\end{align*}
Thus, there exists some $j\in\mathbb{N}$ such that $x\in U_{\leq j}\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}$. Consider this $j$.

If $j\leq m-1$, then $x\in U_{\leq j}\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}$ leads to $x\in U_{\leq\left(  m-1\right)  }\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}$, which is exactly what we want to
have. So let us assume that $j>m-1$. Then, $j\geq m$. Now, combining $x\in
U_{\leq j}\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}$ with $x\in U_{\leq
m}\left(  \mathfrak{g}\right)  \cap\left(  U\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}\right)  \subseteq U_{\leq m}\left(  \mathfrak{g}\right)  $,
we obtain $x\in U_{\leq m}\left(  \mathfrak{g}\right)  \cap\left(  U_{\leq
j}\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)  $. Thus,%
\begin{align*}
x  &  \in U_{\leq m}\left(  \mathfrak{g}\right)  \cap\left(  U_{\leq j}\left(
\mathfrak{g}\right)  \cdot\mathfrak{h}\right) \\
&  \subseteq U_{\leq m}\left(  \mathfrak{g}\right)  \cap\left(  U_{\leq\left(
j-1\right)  }\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{due to (\ref{prop.pbw.goodwillie.1}),
applied to }i=j\right) \\
&  \subseteq U_{\leq m}\left(  \mathfrak{g}\right)  \cap\left(  U_{\leq\left(
j-2\right)  }\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{due to (\ref{prop.pbw.goodwillie.1}),
applied to }i=j-1\right) \\
&  \subseteq U_{\leq m}\left(  \mathfrak{g}\right)  \cap\left(  U_{\leq\left(
j-3\right)  }\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\ \ \ \ \ \ \ \ \ \ \left(  \text{due to (\ref{prop.pbw.goodwillie.1}),
applied to }i=j-2\right) \\
&  \subseteq...\\
&  \subseteq U_{\leq m}\left(  \mathfrak{g}\right)  \cap\left(  U_{\leq\left(
m-1\right)  }\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\ \ \ \ \ \ \ \ \ \ \left(
\begin{array}
[c]{c}%
\text{we must stop this chain of inclusions at }m-1\text{, }\\
\text{because (\ref{prop.pbw.goodwillie.1}) holds only for }i\geq m
\end{array}
\right) \\
&  \subseteq U_{\leq\left(  m-1\right)  }\left(  \mathfrak{g}\right)
\cdot\mathfrak{h},
\end{align*}
which is exactly what we wanted to prove.

Thus, $x\in U_{\leq\left(  m-1\right)  }\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}$ has been shown to hold for every $x\in U_{\leq m}\left(
\mathfrak{g}\right)  \cap\left(  U\left(  \mathfrak{g}\right)  \cdot
\mathfrak{h}\right)  $. This means that $U_{\leq m}\left(  \mathfrak{g}%
\right)  \cap\left(  U\left(  \mathfrak{g}\right)  \cdot\mathfrak{h}\right)
\subseteq U_{\leq\left(  m-1\right)  }\left(  \mathfrak{g}\right)
\cdot\mathfrak{h}$. Proposition \ref{prop.pbw.goodwillie.U(g)h} is now proven.

\begin{noncompile}
\label{subsect.character}[Deformation by a character]
\end{noncompile}

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\end{document}