Bourbaki

Exercise 25, § 7

Tag 0151, permanent. Cite it as [Bourbaki, Tag 0151].

Lie Groups and Lie Algebras, VIII, 247

Bourbaki marks this one as harder than the rest.

Let a\mathfrak{a} be a Lie algebra and U its enveloping algebra. Apply the definitions and results of Exerc. 24 to the algebra U. In particular, U′⊂U∗U'\subset U^* and the dual of U′U' can be identified with the algebra U^=\widehat{U} = lim←−U/m_{\leftarrow -}U/\mathfrak{m}; the canonical map U→U^U\rightarrow \widehat{U} is injective (Chap. I, §7, Exerc. 3).

a) The coalgebra structure of U (Chap. II, §1, no. 4) defines an algebra structure on the dual U∗U^* (cf. Chap. II, §1, no. 5, Prop. 10, and Algebra, Chap. III, §11, no. 2). If E,FE,F are finite dimensional a\mathfrak{a}-modules,

θa,b∘θc,d=θa⊗c,b⊗d\theta_{a,b}\circ \theta_{c,d}=\theta_{a\otimes c,b\otimes d} for a∈E,b∈E∗,c∈F,d∈F∗a\in E, b\in E^*, c\in F, d\in F^*.

Deduce that U′U' is a subalgebra of U∗U^*. The coalgebra and algebra structures of U′U' make it a commutative bigebra (Algebra, Chap. III, §11, no. 4).

b) Let xx be an element of U^\widehat{U}; identify xx with a linear form U′→kU'\rightarrow k. Prove the equivalence of the following properties:

(i) xx is a homomorphism of algebras from U′U' to kk.

(ii) xE⊗F=xE⊗xFx_{E\otimes F}=x_E\otimes x_F for all finite dimensional a\mathfrak{a}-modules E and F.

(Prove first that (ii) is equivalent to

(ii′')x(θa⊗c,b⊗d)=x(θa,b)x(θc,d)x(\theta_{a\otimes c,b\otimes d}) =x(\theta_{a,b})x(\theta_{c,d}) if a∈E,b∈E∗,c∈F,d∈F∗a\in E, b\in E^*, c\in F, d\in F^*,

and use the fact that the θa,b\theta_{a,b} generate U′.)U'.)

c) Let xx be an element of U^\widehat{U} satisfying conditions (i) and (ii) of b). Show the equivalence of the following conditions:

(iii) xx takes the unit element of U′U' to the unit element of kk.

(iv) x≠0x\not= 0.

(v) If kk is given the trivial a\mathfrak{a}-module structure, we have xk=x_k= Id.

(vi) xEx_E is invertible for all E.

(The equivalence (iii) ⇐⇒\Leftarrow \Rightarrow (iv) follows from the fact that xx is a homomorphism of algebras. On the other hand, xk=λx_k=\lambda Id, with λ∈k\lambda \in k. Using the a\mathfrak{a}-isomorphism k⊗E→Ek\otimes E\rightarrow E, deduce that λxE=xE\lambda x_E=x_E for all E, and, in particular, that λ2=λ\lambda^2=\lambda by taking E=kE =k. The case λ=1\lambda = 1 corresponds to x≠0x\not= 0, hence (iv) ⇐⇒\Leftarrow \Rightarrow (v), and (vi) =⇒=\Rightarrow (v). To prove that (v) =⇒=\Rightarrow (vi), show that, if F is the dual of E, then txF∘xE=λ^tx_F\circ x_E=\lambda IdE._E.)

d) Let G be the set of elements of U^\widehat{U} satisfying conditions (i) to (vi) above. Show that G is a subgroup of the group of invertible elements of U^\widehat{U}.

Let x∈Gx\in G. If E is a finite dimensional a\mathfrak{a}-module, then xE∈GL(E)x_E\in \mathbf{G}\mathbf{L}(E). This applies in particular to E=aE =\mathfrak{a}, equipped with the adjoint representation; this gives an element xa∈GL(a)x_{\mathfrak{a}}\in \mathbf{G}\mathbf{L}(\mathfrak{a}). Show that xax_{\mathfrak{a}} is an automorphism of a\mathfrak{a} (use the a\mathfrak{a}-homomorphism a⊗a→a\mathfrak{a}\otimes \mathfrak{a}\rightarrow \mathfrak{a} given by the bracket). This gives a homomorphism v:G→v: G\rightarrow Aut(a\mathfrak{a}). If E is a finite dimensional a\mathfrak{a}-module,

xE(y.e)=v(x)(y).xE(e)x_E(y.e) =v(x)(y).x_E(e) if y∈a,e∈Ey\in \mathfrak{a}, e\in E

(use the a\mathfrak{a}-homomorphism a⊗E→E\mathfrak{a}\otimes E\rightarrow E given by the operation of a\mathfrak{a} on E).

e) The principal anti-automorphism σ\sigma of U extends by continuity to U^\widehat{U}. Its transpose leaves U′U' stable and induces an inversion on U′(U'(Algebra, Chap. III, §11, Exerc. 4). If x∈Gx\in G, then σ(x)=x−1\sigma (x) =x^{-1}.

f) Assume that a\mathfrak{a} is semi-simple16^{16}. Let nn be a nilpotent element of a\mathfrak{a}. Then there exists a unique element ene^n of G such that (en)E=(e^n)_E= exp(nEn_E) for every finite dimensional a\mathfrak{a}-module E. We have v(en)=v(e^n) = exp(ad nn)∈\in Aut(a\mathfrak{a}), so Aute(a)⊂v(G)_e(\mathfrak{a})\subset v(G).

Show that, if b\mathfrak{b} is a subalgebra of a\mathfrak{a} consisting of nilpotent elements, then

en.em=eH(n,m)e^n.e^m=e^{H(n,m)} for n,m∈bn, m\in \mathfrak{b},

where H denotes the Hausdorff series (Chap. II, §6).

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