Proposition 8

Let $X$ be a set, let $M$ be an $L(X)$-module and let $d$ be a mapping of $X$ into $M$. There exists one and only one linear mapping $D$ of $L(X)$ into $M$ extending $d$ and satisfying the relation:

We define a Lie algebra $g$ with underlying module $M \times L(X)$ by means of the bracket

for $a, a'$ in $L(X)$ and $m, m'$ in $M$ (Chapter I, § 1, no. 8). Let $f$ be the homomorphism of $L(X)$ into $g$ such that $f(x) = (d(x), x)$ for all $x$ in $X$; let $f(a) = (D(a), u(a))$ for all $a$ in $L(X)$. By formula (21), $u$ is a homomorphism of $L(X)$ into itself; as $u(x) = x$ for $x$ in $X$, $u(a) = a$ for all $a$ in $L(X)$, whence

By (21) and (22), relation (20) then follows from $f([a, a']) = [f(a), f(a')]$.

Conversely, let $D'$ be a mapping of $L(X)$ into $M$ satisfying relation (20') analogous to (20) and extending $d$. Let $f'(a) = (D'(a), a)$ for $a \in L(X)$; by (20') and (21), $f'$ is a homomorphism of $L(X)$ into $g$, coinciding with $f$ on $X$, whence $f' = f$ and $D' = D$.