Bourbaki

Exercise 16, § 2

Lie Groups and Lie Algebras, II

Bourbaki marks this one as harder than the rest.

We 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 from gg into U(g)U(g), and by S(g)S(g) the symmetric algebra of the KK-module gg. There exists a linear mapping

η(g):S(g)→U(g) \eta(g): S(g) \to U(g)
and one such mapping only, 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}. We have:
η(g)(x1…xn)=1n!∑s∈Snσ(xs(1))…σ(xs(n)),xi∈g. \eta(g)(x_1 \ldots x_n) = \frac{1}{n!} \sum_{s \in S_n} \sigma(x_{s(1)}) \ldots \sigma(x_{s(n)}), \quad 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. We denote by shs_{\mathfrak{h}} (resp. uhu_{\mathfrak{h}}) the ideal of S(g)S(g) (resp. the two-sided ideal of U(g)U(g)) generated by h\mathfrak{h} (resp. by 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) \quad \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 lines are exact. Deduce that the image u~h\tilde{u}_{\mathfrak{h}} of shs_{\mathfrak{h}} by η(g)\eta(g) is contained in uhu_{\mathfrak{h}} and that one has 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 modules gg and g/hg/\mathfrak{h} are free). (c) Take for gg a free Lie algebra, with basic family (X1,…,Xn,H)(X_1, \ldots, X_n, H), with n≥0n \geq 0, and take for h\mathfrak{h} the ideal of gg generated by HH. Show that gg and g/hg/\mathfrak{h} are free modules. Deduce that, in this case, one has u~h=uh\tilde{u}_{\mathfrak{h}} = u_{\mathfrak{h}}. In particular, there exists x∈shx \in s_{\mathfrak{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_{\mathfrak{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 \mathfrak{h} (notice that uhu_{\mathfrak{h}} coincides with the left ideal generated by h\mathfrak{h}). Show, using (c), that such an element belongs to u~h\tilde{u}_{\mathfrak{h}} (use a suitable homomorphism from a free Lie algebra into gg). Deduce that u~h=uh\tilde{u}_{\mathfrak{h}} = u_{\mathfrak{h}}. (e) Show that, if η(g)\eta(g) is bijective, the same is true of η(g/h)\eta(g/\mathfrak{h}). Deduce finally that, for every Lie algebra gg, η(g)\eta(g) is bijective (write gg 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) \omega: S(g) \to \mathrm{gr}\ U(g) \tag{cf. chap. I, § 2, n° 6}
is an isomorphism (« Poincaré–Birkhoff–Witt theorem » for Lie algebras over Q\mathbf{Q}-algebras).

Solution

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Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.