Proposition 10 (elimination theorem)

Let X be a set, S a subset of X and T the set of sequences $(s_1, \ldots, s_n, x)$ with $n \geq 0, s_1, \ldots, s_n$ in S and x in X $- S.$†

(a) The module $L(X)$ is the direct sum of the subalgebra $L(S)$ of $L(X)$ and the ideal $a$ of $L(X)$ generated by $X - S$.

(b) There exists a Lie algebra isomorphism $\phi$ of $L(T)$ onto $a$ which maps $(s_1, \ldots, s_n, x)$ to $(\operatorname{ad} s_1 \circ \cdots \circ \operatorname{ad} s_n)(x)$.

Let $g$ be the Lie algebra constructed as in Proposition 9 given

for $t = (s_1, \ldots, s_n, x)$ in T and $s \in S_1$. We identify $L(S)$ and $L(T)$ with their canonical images in $g$ (Proposition 9 (a)).

Let $\psi$ be the mapping $(s_1, \ldots, s_n, x) \mapsto (\operatorname{ad} s_1 \circ \cdots \circ \operatorname{ad} s_n)(x)$ of T into $L(X)$. Obviously $\psi(d(s, t)) = [s, \psi(t)]$ for $s \in S$ and $t \in T$ and hence there exists a homomorphism $\alpha : g \to L(X)$ whose restriction to S is the identity and whose restriction to T is $\psi$. Now $X - S \subset T$, whence there is a homomorphism $\beta : L(X) \to g$ whose restriction to $X = S \cup (X - S)$ is the identity.

We show that $\alpha$ is an isomorphism and $\beta$ the inverse isomorphism. As $\psi(x) = x$ for $x$ in $X - S$, we see that $\alpha \circ \beta$ coincides with the identity on X, whence $\alpha \circ \beta = \mathrm{Id}_{L(X)}$. On the other hand, $[s, t] = d(s, t)$ in $g$ for $s \in S, t \in T$ by construction; it follows that $t = (s_1, \ldots, s_n, x)$ is equal in $g$ to $(\operatorname{ad} s_1 \circ \cdots \circ \operatorname{ad} s_2)(x)$ whence $t = \beta(\alpha(t))$. As $\beta(\alpha(s)) = s$ for $s \in S$ and $S \cup T$ generates $g$, $\beta \circ \alpha = \mathrm{Id}_g$.

As $\alpha$ is an isomorphism of $g$ onto $L(X)$, Proposition 9 shows that the restric-

† For $n = 0$, we obtain the elements of $X - S$, whence $X - S \subset T$.

tion of $\alpha$ to $L(T)$ is an isomorphism $\phi$ of $L(T)$ onto an ideal $b$ of $L(X)$ such that the module $L(X)$ is the direct sum of $L(S)$ and $b$. Obviously

for $(s_1, \ldots, s_n, x)$ in $T$.

Hence $\phi(T) \subset a$, whereas $b \subset a$ since $\phi(T)$ generates the subalgebra $b$ of $L(X)$. But $b$ is an ideal and $X - S \subset \phi(T) \subset b$, whence $a \subset b$.