Bourbaki

Example

Tag 0A7M, permanent. Cite it as [Bourbaki, Tag 0A7M].

Lie Groups and Lie Algebras, chapter III, § 1. Lie groups

Let EE be a normable space, G=GL(E)G = \mathbf{GL}(E), TT the translation group of EE and AA the permutation group of EE generated by GG and TT. The group AA is algebraically the semi-direct product of GG by TT. (If EE is finite-dimensional, AA is the affine group of EE, cf. Algebra, Chapter II, § 9, no. 4). Let σ\sigma be the identity linear representation of GG on EE and SS the external semi-direct product of GG by EE relative to σ\sigma. For all x∈Ex \in E, let txt_x be the translation of EE defined by xx. The mapping (x,u)↦tx∘u(x, u) \mapsto t_x \circ u is isomorphism Φ\Phi of the group SS onto the group AA. The mapping (x,u)↦σ(u)x=u(x)(x, u) \mapsto \sigma(u)x = u(x) of E×L(E)E \times \mathscr{L}(E) into EE is continuous and bilinear and hence analytic; its restriction to E×GE \times G is therefore analytic. Thus the group SS, with the product manifold structure of EE and GG, is a Lie group. We transport this structure to AA by means of Φ\Phi. Then AA becomes a Lie group, the internal semi-direct product of GG by TT as a Lie group.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

Transcribed from the printed volume. Not translated.