Bourbaki

Exercise 9, § 4

Lie Groups and Lie Algebras, III, 380

Bourbaki marks this one as harder than the rest.

Let G be a compact Lie group over K of dimension n. Suppose that the Lie algebra L(G) is simple. For all g ∈ G, every integer m ≥ 1 and every neighbourhood V of e in G, let M(g, m, V) denote the set of elements of G of the form ∏i=1mxi(yi,g)xi−1\prod_{i=1}^m x_i(y_i, g)x_i^{-1} with x1,…,xm,y1,…,ymx_1, \ldots, x_m, y_1, \ldots, y_m in V.

(a) Show that if g is an element of G whose centralizer is not open, if m is an integer ≥ n and if V is a neighbourhood of e, then M(g, m, V) is a neighbourhood of e. (There exists a∈L(G)a \in L(G) such that (Ad g)(a)≠a(\mathrm{Ad}\,g)(a) \neq a. Let φ be an exponential mapping of G. Let y be the image of 1 under the tangent mapping at 0 to λ↦(ϕ(λa),g)\lambda \mapsto (\phi(\lambda a), g) (where λ∈K\lambda \in K). Then b=(Ad g−1)(a)≠0b = (\mathrm{Ad}\,g - 1)(a) \neq 0. There exist x1,…,xmx_1, \ldots, x_m in V such that {(Ad x1)(b),…,(Ad xm)(b)}\{(\mathrm{Ad}\,x_1)(b), \ldots, (\mathrm{Ad}\,x_m)(b)\} contains a basis of L(G). Consider the mapping

(x1′,…,xm′,y1,…,ym)↦∏i=1nxi′(yi,g)xi′−1 (x'_1, \ldots, x'_m, y_1, \ldots, y_m) \mapsto \prod_{i=1}^n x'_i(y_i, g){x'_i}^{-1}

of G2mG^{2m} into G. Show that the tangent mapping at (x1,…,xm,e,…,e)(x_1, \ldots, x_m, e, \ldots, e) is surjective.)

(b) Let U be a neighbourhood of e in G. Let m be an integer ≥ 1. There exists an element g of G whose centralizer is not open, such that M(g,m,G)⊂UM(g, m, G) \subset U. (Argue by reductio ad absurdum using the compactness of G.)

(c) Let G' be a compact Lie group over K whose Lie algebra is simple. Let ϕ:G→G′\phi : G \to G' be a bijective homomorphism of abstract groups. Then ϕ\phi is continuous. (Apply (b) to G' and (a) to G.)

Solution

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Transcribed from the printed volume. Not translated.