Exercise 9, § 4
Lie Groups and Lie Algebras, III
Bourbaki marks this one as harder than the rest.
Let be a compact Lie group over , of dimension . We suppose that the Lie algebra is simple. For every , every integer , and every neighbourhood of in , we denote by the set of elements of of the form , for in .
a) Prove that if is an element of whose centralizer is not open, if is an integer , and if is a neighbourhood of , then is a neighbourhood of . (There exists 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 (where ). We have . There exist in V such that { (Ad )(b), ..., (Ad )(b) } contains a basis of L(G).
We consider the mapping from into G. Prove that the tangent mapping at is surjective.)
b) Let U be a neighbourhood of e in G. Let m be an integer . 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 be a bijective homomorphism of abstract groups. Then is continuous. (Apply b) to G’ and a) to G.)
Solution
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