Bourbaki

Exercise 26, § 6

Lie Groups and Lie Algebras, III, 390

Let G be an n-dimensional connected compact complex Lie group, X a finite-dimensional connected complex analytic manifold and (g,x)↦gx(g, x) \mapsto gx a law of analytic left operation of G on X. For all g∈Gg \in G, let φ(g)\varphi(g) be the mapping x↦gxx \mapsto gx of X into X. Suppose, that, for all g∈Gg \in G distinct from e, φ(g)≠IdX\varphi(g) \neq \mathrm{Id}_X. Then every orbit of G in X is an n-dimensional closed submanifold of X. (Let x∈Xx \in X, H be the stabilizer of x and H' the identity component of H. For all g∈H′g \in H', let u(g)u(g) be the tangent mapping at x to φ(g)\varphi(g). Arguing as in Proposition 6 of no. 3, show that u(g)u(g) is the identity for all g∈H′g \in H'. Using § 1, Exercise 4 and the connectedness of X, deduce that H′={e}H' = \{e\}.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.