(a) Existence of F: let h be the homomorphism of Lib(X) into $g$ extending f (no. 1). For all $a$ in Lib(X), $h(Q(a)) = h(a.a) = [h(a), h(a)] = 0$; similarly, the Jacobi identity satisfied by $g$ implies that $h(J(a, b, c)) = 0$ for $a, b, c$ in Lib(X). It follows that $h(a) = 0$, whence there is a homomorphism F of $L(X)$ into $g$ such that $h = F \circ \psi$. By restricting to X, we obtain $f = F \circ \phi$.

(b) Uniqueness of F: let $F': L(X) \to g$ be a homomorphism such that $f = F' \circ \phi$. The homomorphisms $F \circ \psi$ and $F' \circ \psi$ of Lib(X) into $g$ coincide on X and hence are equal; as $\psi$ is surjective, $F = F'$.

Let $x_1, x_2, \ldots, x_n$ be distinct elements of X and $\lambda_1, \ldots, \lambda_n$ be elements of K such that

Let $g$ be the commutative Lie algebra with K as underlying module. For $i = 1, 2, \ldots, n$, there exists a homomorphism $F_i$ of $L(X)$ into $g$ such that $F_i(\phi(x_i)) = 1$ and $F_i(\phi(x)) = 0$ for $x \neq x_i$ (Proposition 1); applying $F_i$ to relation (3), we obtain $\lambda_i = 0$.

Let $a \xrightarrow{\lambda} g \xrightarrow{\mu} L(X)$ be such an extension (Chapter I, § 1, no. 7). As $\mu$ is surjective, there exists a mapping f of X into $g$ such that $\phi = \mu \circ f$. Let F be the homomorphism of $L(X)$ into $g$ such that $f = F \circ \phi$ (Proposition 1). Then $(\mu \circ F) \circ \phi = \mu \circ f = \phi$ and Proposition 1 shows that $\mu \circ F$ is the identity automorphism of $L(X)$. The given extension is therefore inessential (Chapter I, § 1, no. 7, Proposition 6 and Definition 6).

As the ring K is non-zero, Corollary 1 to Proposition 1 shows that $\phi$ is injective. Hence the set X can be identified by means of $\phi$ with its image in $L(X)$; with this convention, X generates $L(X)$ and every mapping of X into a Lie algebra $g$ can be extended to a Lie algebra homomorphism of $L(X)$ into $g$.

The existence and uniqueness of $L(u)$ follow from Proposition 1 of no. 2. The homomorphisms $L(v \circ u)$ and $L(v) \circ L(u)$ have the same restriction to X and hence are equal (Proposition 1).

Since the assertion is trivial for $X = \varnothing$, we assume $X \neq \varnothing$. If u is injective there exists a mapping v of Y into X such that $v \circ u$ is the identity mapping of X; by Proposition 2, $L(v) \circ L(u)$ is the identity of automorphism of $L(X)$ and hence $L(u)$ is injective. When u is surjective, there exists a mapping w of Y into X such that $u \circ w$ is the identity mapping of Y; then $L(u) \circ L(w)$ is the identity mapping of $L(Y)$, which proves that $L(u)$ is surjective.

† Recall (Algebra, Chapter IV, § 5, no. 10) the definition of polynomial mappings of a free module M into a module N: if q is an integer $\geq 0$, a mapping $f : M \to N$ is called homogeneous polynomial of degree q if there exists a multilinear mapping u of $M^q$ into N such that

Let X be a set and S a subset of X. The above corollary shows that the canonical injection of S into X can be extended to an isomorphism $\alpha$ of $L(S)$ onto the Lie subalgebra $L'(S)$ of $L(X)$ generated by S; we shall identify $L(S)$ and $L'(S)$ by means of $\alpha$.

Let $(S_\alpha)_{\alpha \in I}$ be a right directed family of subsets of X with union S. The relation $S_\alpha \subset S_\beta$ implies $L(S_\alpha) \subset L(S_\beta)$ and hence the family of Lie subalgebras $L(S_\alpha)$ of $L(X)$ is right directed. Therefore $g = \bigcup_{\alpha \in I} L(S_\alpha)$ is a Lie subalgebra of $L(X)$; then $S \subset g$, whence $L(S) \subset g$, and, as $L(S_\alpha) \subset L(S)$ for all $\alpha \in I$, $g \subset L(S)$. Hence

for every right directed family $(S_\alpha)_{\alpha \in I}$ of subsets of X.

Applying the above to the family of finite subsets of X, we see that every element of $L(X)$ is of the form $P(x_1, \ldots, x_n)$ where P is a Lie polynomial in n indeterminates and $x_1, \ldots, x_n$ are elements of X.

such that $v(x \otimes 1) = x$ for $x \in X$. Further, v is an isomorphism.

Applying Proposition 1 to $g = L_{K'}(X)$ considered as a Lie K-algebra and the mapping $x \mapsto x$ of X into g, we obtain a K-homomorphism $L_K(X) \to L_{K'}(X)$, whence there is a $K'$-homomorphism $v : L_K(X) \otimes K' \to L_{K'}(X)$. The fact that v is unique and is an isomorphism follows from the fact that the ordered pair $(L_K(X) \otimes K', x \mapsto x \otimes 1)$ is a solution of the same universal mapping problem as the ordered pair $(L_{K'}(X), x \mapsto x)$.

Let the Lie algebra $L(X) = \mathrm{Lib}(X)/a$ be given the quotient graduation. The homogeneous component of $L(X)$ of degree $\delta$ is denoted by $L^\delta(X)$; it is the submodule of $L(X)$ generated by the images of the elements $w \in M(X)$ such that $\phi(w) = \delta$.

We shall make special use of the following two cases: (a) Total graduation: we take $\Delta = \mathbf{N}$ and $\phi_0(x) = 1$ for all $x \in X$, whence $\phi(w) = l(w)$ for $w$ in $M(X)$. The K-module $L^n(X)$ is generated by the images of the elements of length $n$ in $M(X)$, which we shall call alternants of degree n. We shall see later that the module $L^n(X)$ is free and admits a basis consisting of alternants of degree $n$ (no. 11, Theorem 1). Then $L(X) = \bigoplus_{n \geq 1} L^n(X)$ and $L^1(X)$ admits $X$ as basis (no. 2, Corollary 1 to Proposition 1). By the construction of $M(X)$,

(b) Multigraduation: we take $\Delta$ to be the free commutative monoid $\mathbf{N}^{(X)}$ constructed on $X$. The mapping $\phi_0$ of $X$ into $\Delta$ is defined by $(\phi_0(x))(x') = \delta_{xx'}$, where $\delta_{xx'}$ is the Kronecker symbol. For $w \in M(X)$ and $x \in X$, the integer $(\phi(w))(x)$ is "the number of occurrences of the letter $x$ in $w$". For $\alpha$ in $\mathbf{N}^{(X)}$, we write $|\alpha| = \sum_{x \in X} \alpha(x)$, whence $|\phi(w)| = l(w)$ for all $w$ in $M(X)$. It follows that

Let $\alpha \in \mathbf{N}^{(S)}$. The module $L^\alpha(S)$ is generated by the images in $L(X)$ of the elements $w$ in $M(S)$ such that $\phi(w) = \alpha$, that is (Algebra, § 7, no. 9, formulae (23) and (24)) the set of $w$ in $M(X)$ such that $\phi(w) = \alpha$. Hence $L^\alpha(S) = L^\alpha(X)$.

The proposition follows from this and the relation $L(S) = \sum_{\alpha \in \mathbf{N}^{(S)}} L^\alpha(S)$.

When we take $P = g$, the sequence $(P_n)$ is the lower central series ($\mathscr{C}^n g$) of $g$ (Chapter I, § 1, no. 5). Hence:

Generalizing Definition 1 of Chapter I, § 4, no. 1, we shall say that a Lie algebra $g$ is nilpotent if $\mathscr{C}^n g = \{0\}$ for $n$ sufficiently large. The nilpotency class of a nilpotent Lie algebra $g$ is the smallest integer $n$ such that $\mathscr{C}^{n+1} g = \{0\}$.

(a) We apply Proposition 5 with $g = L(X)$ and $P = L^1(X)$. By induction on $n$, we deduce from (12) (no. 6) and (19) the equality $P_n = L^n(X)$. The desired relation is then equivalent to the definition $[P, P_n] = P_{n+1}$. (b) This follows from (a) by induction on $n$. (c) Let $g = L(X)$ and $g_n = \sum_{p \geq n} L_p(X)$. Then $g = g_1$ and formula (13) of no. 6 implies $[g_n, g_m] \subset g_{n+m}$ and in particular $[g, g_n] \subset g_{n+1}$. By induction on $n$, $\mathscr{C}^n g \subset g_n$. On the other hand, from (a) we deduce that $L^n(X) \subset \mathscr{C}^n g$ by induction on $n$. As $\mathscr{C}^n g$ is an ideal of $g$, the relation $L^p(X) \subset \mathscr{C}^n g$ implies that

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 $D' = D$.

When $M$ is equal to $L(X)$ with the adjoint representation, relation (20) means that $D$ is a derivation.

(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