Bourbaki

Exercise 9, § 4

Lie Groups and Lie Algebras, III

Bourbaki marks this one as harder than the rest.

Let GG be a compact Lie group over KK, of dimension nn. We suppose that the Lie algebra L(G)L(G) is simple. For every g∈Gg \in G, every integer m≥1m \geq 1, and every neighbourhood VV of ee in GG, we denote by M(g,m,V)M(g, m, V) the set of elements of GG of the form ∏i=1mxi(yi,g)xi−1\prod_{i=1}^{m} x_i(y_i, g)x_i^{-1}, for x1,…,xm,y1,…,ymx_1, \ldots, x_m, y_1, \ldots, y_m in VV.

a) Prove that if gg is an element of GG whose centralizer is not open, if mm is an integer ≥n\geq n, and if VV is a neighbourhood of ee, then M(g,m,V)M(g, m, V) is a neighbourhood of ee. (There exists a∈L(G)a \in L(G) such that (Ad g)(a) ≠ a. Let φ be an exponential mapping of G. Let y be the image of 1 by the tangent mapping at 0 to λ↦(φ(λa),g)\lambda \mapsto (\varphi(\lambda a), g) (where λ∈K\lambda \in K). We have 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 x1x_1)(b), ..., (Ad xmx_m)(b) } contains a basis of L(G).

We consider the mapping (x1′,…,xm′,y1,…,ym)↦∏i=1mxi′(yi,g)xi′−1(x'_1, \ldots, x'_m, y_1, \ldots, y_m) \mapsto \prod_{i=1}^{m} x'_i(y_i, g){x'_i}^{-1} from G2mG^{2m} into G. Prove 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\geq 1. There exists an element g of G whose centralizer is not open, such that M(g, m, G) ⊂ U. (Reason by contradiction using the compactness of G.)

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

Solution

No solution has been written yet.

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.