Bourbaki marks this one as harder than the rest.
Suppose that K is a Q-algebra. If g is a Lie algebra, we denote by U(g) its enveloping algebra, by σ the canonical mapping of g into U(g) and by S(g) the symmetric algebra of the K-module g. There exists one and only one linear mapping
η(g):S(g)→U(g) such that
η(g)(xn)=σ(x)n for all
x∈g and all
n∈N. Then
η(g)(x1…xn)=n!1s∈Sn∑σ(xs(1)…xs(n)),xi∈g. We propose to show that
η(g) is bijective.
(a) Prove that η(g) is surjective and that it is bijective when the K-module g is free (use the Poincaré–Birkhoff–Witt Theorem).
(b) Let h be an ideal of g. Let sh (resp. uh) denote the ideal of S(g) (resp. the two-sided ideal of U(g)) generated by h (resp. σ(h)). The diagram
00→→sh→S(g)→S(g/h)→0↓η(g)↓η(g/h)uh→U(g)→U(g/h)→0 is commutative and its rows are exact. Deduce that the image
u~h of
sh under
η(g) is contained in
uh and that
u~h=uh when
η(g) and
η(g/h) are injective (in particular, when the module
g and
g/h are free).
(c) Take g to be a free Lie algebra with basic family (X1,…,Xn,H), where n≥0, and take h to be the ideal of g generated by H. Show that g and g/h are free modules. Deduce that, in that case, u~h=uh. In particular, there exists x∈sh such that (η(g))(x)=σ(X1)…σ(Xn)σ(H).
(d) We return to the general case. Show that uh is generated over K by the elements of the form σ(x1)…σ(xn)σ(h), with n≥0, xi∈g and h∈h (note that uh coincides with the left ideal generated by h). Show, using (c), that such an element belongs to u~h (use a suitable homomorphism of a free Lie algebra into g). Deduce that u~h=uh.
(e) Show that, if η(g) is bijective, so is η(g/h). Deduce finally that, for every Lie algebra g, η(g) is bijective (write g as the quotient of a free Lie algebra) and that the canonical homomorphism
ω:S(g)→grU(g)(cf. Chapter I, § 2, no. 6)
is an isomorphism ("Poincaré–Birkhoff–Witt Theorem" for Lie algebras over Q-algebras).