Proposition 9

*Let $S_1$ and $S_2$ be two disjoint sets and $d$ a mapping of $S_1 \times S_2$ into $L(S_2)$. Let $g$ be the quotient Lie algebra of $L(S_1 \cup S_2)$ by the ideal generated by the elements $[s_1, s_2] - d(s_1, s_2)$ with $s_1 \in S_1, s_2 \in S_2$; let $\psi$ be the canonical mapping of $L(S_1 \cup S_2)$ onto $g$.

(a) *For $i = 1, 2$, the restriction $\phi_i$ of $\psi$ to $S_i$ can be extended to an isomorphism of $L(S_i)$ onto a subalgebra $a_i$ of $g$.

(b) $g = a_1 + a_2, a_1 \cap a_2 = \{0\}$ and $a_2$ is an ideal of $g$.

For $i = 1, 2$, let $\psi_i$ denote the homomorphism of $L(S_i)$ into $g$ which extends $\phi_i$ and $a_i$ its image. Clearly $\phi_i(S_i)$ generates $a_i$.

Let $s_1 \in S_1$; we write $D = \operatorname{ad} \phi_1(s_1)$. The derivation $D$ of $g$ maps $\phi_2(S_2)$ into $a_2$ by the relation

as the subalgebra $a_2$ of $g$ is generated by $\phi_2(S_2)$, therefore $D(a_2) \subset a_2$. The set of $x \in g$ such that $\operatorname{ad} x$ leaves $a_2$ invariant is a Lie subalgebra of $g$ which contains $\phi_1(S_1)$ by the above and hence also $a_1$. Hence

Therefore $a_1 + a_2$ is a Lie subalgebra of $g$ and, as it contains the generating set $\phi_1(S_1) \cup \phi_2(S_2)$,

For all $s_1 \in S_1$ there exists a derivation $D_{s_1}$ of $L(S_2)$ such that

for all $s_2$ in $S_2$ (no. 8, Corollary to Proposition 8). The mapping $s_1 \mapsto D_{s_1}$ can be extended to a homomorphism $D$ of $L(S_1)$ into the Lie algebra of derivations of $L(S_2)$. Let $\mathfrak{h}$ be the semi-direct product of $L(S_1)$ by $L(S_2)$ corresponding to $D$ (Chapter I, § 1, no. 8). As a module $\mathfrak{h}$ is equal to $L(S_1) \times L(S_2)$ and in particular

for $s_1 \in S_1$ and $s_2 \in S_2$.

From (25) we deduce the existence of a homomorphism $f$ of $g$ into $\mathfrak{h}$ such that $f(\phi_1(s_1)) = (s_1, 0)$ and $f(\phi_2(s_2)) = (0, s_2)$ for $s_1 \in S_1$ and $s_2 \in S_2$. We deduce immediately the relation

for $a_1 \in L(S_1)$ and $a_2 \in L(S_2)$.

Relation (26) shows that $\psi_1$ and $\psi_2$ are injective and that $a_1 \cap a_2 = \{0\}$. Formulae (23) and (24) then imply the proposition.