Proposition 1

(a) The restriction $\pi_0$ of $\pi$ to $L(X)$ is a derivation of $L(X)$. (b) For every integer $n \geq 1$ and all $u$ in $L^n(X)$, $\pi(u) = n.u$.

(a) Let $E$ be the endomorphism algebra of the module $L(X)$ and $\theta$ the homomorphism of $A(X)$ into $E$ such that $\theta(x) = \operatorname{ad} x$ for all $x \in X$. The restriction of $\theta$ to $L(X)$ is a Lie algebra homomorphism of $L(X)$ into $E$, which coincides on $X$ with the adjoint representation of $L(X)$, whence

Let $a$ be in $A(X)$ and $b$ in $A^+(X)$; then

It suffices to consider the case $a = x_1 \ldots x_p, \ b = x_{p+1} \ldots x_{p+q}$ with $p \geq 0, q \geq 1$ and $x_1, \ldots, x_{p+q}$ in $X$; but then (5) follows immediately from (3) since $\theta(x) = \operatorname{ad} x$ for $x \in X$.

Let $u$ and $v$ be in $L(X)$; by (4) and (5),

hence $\pi_0$ is a derivation of $L(X)$.

(b) Let $\pi_1$ be the endomorphism of the module $L(X)$ which coincides on $L^n(X)$ with multiplication by the integer $n \geq 1$. The formula $[L^n(X), L^m(X)] \subset L^{n+m}(X)$ shows that $\pi_1$ is a derivation (Algebra, Chapter III, § 10, no. 3, Example 6). The derivation $\pi_1 - \pi_0$ of $L(X)$ is zero on $X$ and, as $X$ generates $L(X)$, $\pi_0 = \pi_1$, whence (b).