Bourbaki

Exercise 16, § 2

Lie Groups and Lie Algebras, II, 185

Bourbaki marks this one as harder than the rest.

Suppose that KK is a Q\mathbf{Q}-algebra. If gg is a Lie algebra, we denote by U(g)U(g) its enveloping algebra, by σ\sigma the canonical mapping of gg into U(g)U(g) and by S(g)S(g) the symmetric algebra of the KK-module gg. There exists one and only one linear mapping

η(g):S(g)→U(g) \eta(g): S(g) \to U(g)
such that η(g)(xn)=σ(x)n\eta(g)(x^n) = \sigma(x)^n for all x∈gx \in g and all n∈Nn \in \mathbf{N}. Then
η(g)(x1…xn)=1n!∑s∈Snσ(xs(1)…xs(n)),xi∈g. \eta(g)(x_1 \ldots x_n) = \frac{1}{n!} \sum_{s \in \mathfrak{S}_n} \sigma(x_{s(1)} \ldots x_{s(n)}), \qquad x_i \in g.
We propose to show that η(g)\eta(g) is bijective.

(a) Prove that η(g)\eta(g) is surjective and that it is bijective when the KK-module gg is free (use the Poincaré–Birkhoff–Witt Theorem).

(b) Let h\mathfrak{h} be an ideal of gg. Let shs_\mathfrak{h} (resp. uhu_\mathfrak{h}) denote the ideal of S(g)S(g) (resp. the two-sided ideal of U(g)U(g)) generated by h\mathfrak{h} (resp. σ(h)\sigma(\mathfrak{h})). The diagram

0→sh→S(g)→S(g/h)→0↓η(g)↓η(g/h)0→uh→U(g)→U(g/h)→0 \begin{array}{ccc} 0 & \to & s_\mathfrak{h} \to S(g) \to S(g/\mathfrak{h}) \to 0 \\ & & \downarrow \eta(g) \qquad \downarrow \eta(g/\mathfrak{h}) \\ 0 & \to & u_\mathfrak{h} \to U(g) \to U(g/\mathfrak{h}) \to 0 \end{array}
is commutative and its rows are exact. Deduce that the image u~h\tilde{u}_\mathfrak{h} of shs_\mathfrak{h} under η(g)\eta(g) is contained in uhu_\mathfrak{h} and that u~h=uh\tilde{u}_\mathfrak{h} = u_\mathfrak{h} when η(g)\eta(g) and η(g/h)\eta(g/\mathfrak{h}) are injective (in particular, when the module gg and g/hg/\mathfrak{h} are free).

(c) Take gg to be a free Lie algebra with basic family (X1,…,Xn,H)(X_1, \ldots, X_n, H), where n≥0n \geq 0, and take hh to be the ideal of gg generated by HH. Show that gg and g/hg/h are free modules. Deduce that, in that case, u~h=uh\tilde{u}_h = u_h. In particular, there exists x∈shx \in s_h such that (η(g))(x)=σ(X1)…σ(Xn)σ(H)(\eta(g))(x) = \sigma(X_1) \ldots \sigma(X_n) \sigma(H).

(d) We return to the general case. Show that uhu_h is generated over KK by the elements of the form σ(x1)…σ(xn)σ(h)\sigma(x_1) \ldots \sigma(x_n) \sigma(h), with n≥0n \geq 0, xi∈gx_i \in g and h∈hh \in h (note that uhu_h coincides with the left ideal generated by hh). Show, using (c), that such an element belongs to u~h\tilde{u}_h (use a suitable homomorphism of a free Lie algebra into gg). Deduce that u~h=uh\tilde{u}_h = u_h.

(e) Show that, if η(g)\eta(g) is bijective, so is η(g/h)\eta(g/h). Deduce finally that, for every Lie algebra gg, η(g)\eta(g) is bijective (write gg as the quotient of a free Lie algebra) and that the canonical homomorphism

ω:S(g)→gr⁡U(g)(cf. Chapter I, § 2, no. 6) \omega : S(g) \to \operatorname{gr} U(g) \quad \text{(cf. Chapter I, § 2, no. 6)}

is an isomorphism ("Poincaré–Birkhoff–Witt Theorem" for Lie algebras over Q\mathbf{Q}-algebras).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.