Proposition 7

Let g be a Lie algebra, U its enveloping algebra, σ the canonical mapping of g into U and D a derivation of g. (a) *There exists one and only one derivation $D_U$ of U such that $\sigma \circ D = D_U \circ \sigma$ (that is such that $D_U$ extends D, when g can be identified with a submodule of U under σ). (b) $D_U$ leaves stable $U_n$ and the set $U^n$ of images in U of the homogeneous symmetric tensors of order n over g.

(c) $D_U$ commutes with the principal antiautomorphism of $U$.

(d) If $D$ is the inner derivation of $g$ defined by an element $x$ of $g$, $D_U$ is the inner derivation of $U$ defined by $\sigma(x)$.

Let $D_T$ be the derivation of the tensor algebra $T$ of $g$ which extends $D$ (Lemma 4). The two-sided ideal $J$ of $T$ generated by the

$(x, y$ in $g$) is stable under $D_T$. For:

On passing to the quotient, $D_T$ defines a derivation $D_U$ of $U$ such that $\sigma \circ D = D_U \circ \sigma$. The uniqueness of $D_U$ is immediate since 1 and $\sigma(g)$ generate the algebra $U$. Assertion (b) is obvious. Let $A$ be the principal antiautomorphism of $U$. We now prove (c). If $x_1, \ldots, x_n$ are in $g$, then

Finally, let $x \in g$. Let $\Delta$ be the inner derivation $y \mapsto \sigma(x)y - y\sigma(x)$ of $U$ (Algebra, Chapter IV, § 4, no. 3, Example 2). Then, for $x' \in g$,

whence $\Delta \circ \sigma = \sigma \circ \mathrm{ad}\, x$. This completes the proof.

Applying Proposition 7 to the case of a commutative Lie algebra, it is seen that every endomorphism $u$ of a $K$-module can be extended uniquely to a derivation of the symmetric algebra of this module; this derivation is derived on passing to the quotient from the derivation of the tensor algebra which extends $u$.

We again take a Lie algebra $g$ over $K$ and let $D$ be a derivation of $g$. We use the earlier notation $T, S, U, G$. Let $D_T, D_S$ be the derivations of $T, S$ which extend $D$ and let $D_U$ be the unique derivation of $U$ such that $\sigma \circ D = D_U \circ \sigma$. Since $D_U$ leaves the $U_n$ stable, $D_U$ defines on taking quotients a derivation $D_G$ of $G$. Since $D_U$ and $D_S$ are derived from $D_T$ when passing to quotients, the commutative diagram (3) proves that $D_G$ can also be derived from $D_S$ by the homomorphism $\omega$ defined in no. 6. If further $K$ is a field of characteristic 0, the isomorphisms of diagram (4) map one into another the restrictions of $D_T, D_S, D_U, D_G$ to ${S'}^n, S^n, U^n, G^n$. Hence the canonical isomorphism of $S$ onto $U$ maps $D_S$ to $D_U$.