Bourbaki marks this one as harder than the rest.
We 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 from g into U(g), and by S(g) the symmetric algebra of the K-module g. There exists a linear mapping
η(g):S(g)→U(g) and one such mapping only, such that
η(g)(xn)=σ(x)n for all
x∈g and all
n∈N. We have:
η(g)(x1…xn)=n!1s∈Sn∑σ(xs(1))…σ(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. We denote by
sh (resp.
uh) the ideal of
S(g) (resp. the two-sided ideal of
U(g)) generated by
h (resp. by
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 lines are exact. Deduce that the image
u~h of
sh by
η(g) is contained in
uh and that one has
u~h=uh when
η(g) and
η(g/h) are injective (in particular when the modules
g and
g/h are free). (c) Take for
g a free Lie algebra, with basic family
(X1,…,Xn,H), with
n≥0, and take for
h the ideal of
g generated by
H. Show that
g and
g/h are free modules. Deduce that, in this case, one has
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 (notice 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 from a free Lie algebra into
g). Deduce that
u~h=uh. (e) Show that, if
η(g) is bijective, the same is true of
η(g/h). Deduce finally that, for every Lie algebra
g,
η(g) is bijective (write
g as a quotient of a free Lie algebra), and that the canonical homomorphism
ω:S(g)→gr U(g)(cf. chap. I, § 2, n° 6) is an isomorphism (« Poincaré–Birkhoff–Witt theorem » for Lie algebras over
Q-algebras).