Bourbaki

Exercise 7, § 1

Lie Groups and Lie Algebras, III, 371

Bourbaki marks this one as harder than the rest.

Let E be a complete normable space over R\mathbf{R} and F a closed vector subspace of E such that there exists no bicontinuous linear bijection of E onto F×(E/F)F \times (E/F).‡ Let X be the analytic manifold R×E×F\mathbf{R} \times E \times F. Let G be the

† For more details, see R. S. Palais, On the existence of slices for actions of non-compact Lie groups, Ann. of Math., 72 (1961), pp. 295–323. ‡ For an example of such an ordered pair (E, F), cf. for example A. Douady, Le problème des modules pour les sous-espaces analytiques compacts d’un espace analytique donné, Ann. Inst. Fourier, 16 (1966), p. 16.

additive group of F, which operates on X by the mapping

(g,(λ,e,f))↦(λ,e+g,f+λg) (g, (\lambda, e, f)) \mapsto (\lambda, e + g, f + \lambda g)
of G×XG \times X into X. For all x∈Xx \in X, let RxR_x be the G-orbit of x, which is a quasi-submanifold of X. Then condition (a) of Proposition 10 is satisfied, but there exists no ordered pair (Y,π)(Y, \pi) with the following properties: Y is an analytic manifold, π\pi is a morphism of X into Y and, for all x∈Xx \in X,
Tx(π):Tx(X)→Tπ(x)(Y) T_x(\pi): T_x(X) \to T_{\pi(x)}(Y)
is surjective with kernel Tx(Rx)T_x(R_x).

(Suppose that (Y,π)(Y, \pi) exists. Let H be the tangent space to Y at π(0,0,0)\pi(0, 0, 0). Then H is isomorphic to R×F×(E/F)\mathbf{R} \times F \times (E/F). On the other hand H is isomorphic to Tπ(x)YT_{\pi(x)}Y for x sufficiently close to (0,0,0)(0, 0, 0) and is hence isomorphic to R×E\mathbf{R} \times E.)

Solution

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