Bourbaki

Corollary 1

Tag 0AB5, permanent. Cite it as [Bourbaki, Tag 0AB5].

Lie Groups and Lie Algebras, chapter III, § 3. Passage from a Lie group to its Lie algebra

Let GG be a Lie group, E1,…,En+1E_1, \ldots, E_{n+1} complete normable spaces and π1,…,πn+1\pi_1, \ldots, \pi_{n+1} analytic linear representations of GG on E1,…,En+1E_1, \ldots, E_{n+1}. Let

E=L(E1,…,En;En+1) E = \mathscr{L}(E_1, \ldots, E_n; E_{n+1})
the complete normable space of continuous multilinear mappings of E1×⋯×EnE_1 \times \cdots \times E_n into En+1E_{n+1} (General Topology, Chapter X, § 3, no. 2). For all g∈Gg \in G, let π(g)\pi(g) be the automorphism of EE defined by
(π(g)u)(x1,…,xn)=πn+1(g)(u(π1(g)−1x1,…,πn(g)−1xn)). (\pi(g)u)(x_1, \ldots, x_n) = \pi_{n+1}(g)(u(\pi_1(g)^{-1}x_1, \ldots, \pi_n(g)^{-1}x_n)).

Then π\pi is an analytic linear representation of GG on EE and

((L(π)a)u)(x1,…,xn)=−∑i=1nu(x1,…,xi−1,(L(πi)a)xi,xi+1,…,xn) ((L(\pi)a)u)(x_1, \ldots, x_n) = -\sum_{i=1}^n u(x_1, \ldots, x_{i-1}, (L(\pi_i)a)x_i, x_{i+1}, \ldots, x_n)
+(L(πn+1)a)(u(x1,…,xn)) \quad + (L(\pi_{n+1})a)(u(x_1, \ldots, x_n))
for all a∈L(G)a \in L(G), u∈Eu \in E, x1∈E1,…,xn∈Enx_1 \in E_1, \ldots, x_n \in E_n.

Every element (A1,…,An+1)(A_1, \ldots, A_{n+1}) of L(E1)×⋯×L(En+1)\mathscr{L}(E_1) \times \cdots \times \mathscr{L}(E_{n+1}) defines a continuous endomorphism θ(A1,…,An+1)\theta(A_1, \ldots, A_{n+1}) of EE by the formula

(θ(A1,…,An+1)u)(x1,…,xn)=An+1(u(A1x1,…,Anxn)). (\theta(A_1, \ldots, A_{n+1})u)(x_1, \ldots, x_n) = A_{n+1}(u(A_1x_1, \ldots, A_nx_n)).
The mapping θ\theta of L(E1)×⋯×L(En+1)\mathscr{L}(E_1) \times \cdots \times \mathscr{L}(E_{n+1}) into L(E)\mathscr{L}(E) is continuous and multilinear. Then, for all g∈Gg \in G,
π(g)=θ(π1(g−1),…,πn(g−1),πn+1(g)) \pi(g) = \theta(\pi_1(g^{-1}), \ldots, \pi_n(g^{-1}), \pi_{n+1}(g))
and hence π\pi is analytic. We apply Proposition 41 to the mapping
(x1,…,xn,u)↦u(x1,…,xn) (x_1, \ldots, x_n, u) \mapsto u(x_1, \ldots, x_n)
of E1×⋯×En×EE_1 \times \cdots \times E_n \times E into En+1E_{n+1}. Then
πn+1(g)(u(x1,…,xn))=(π(g)u)(π1(g)x1,…,πn(g)xn) \pi_{n+1}(g)(u(x_1, \ldots, x_n)) = (\pi(g)u)(\pi_1(g)x_1, \ldots, \pi_n(g)x_n)
and hence
(L(πn+1)a)(u(x1,…,xn))=∑i=1nu(x1,…,(L(πi)a)xi,…,xn) (L(\pi_{n+1})a)(u(x_1, \ldots, x_n)) = \sum_{i=1}^n u(x_1, \ldots, (L(\pi_i)a)x_i, \ldots, x_n)
+((L(π)a)u)(x1,…,xn). \quad + ((L(\pi)a)u)(x_1, \ldots, x_n).

When the EiE_i are finite-dimensional, the representation L(π)L(\pi) of L(G)L(G) is derived from the representations L(π1),…,L(πn+1)L(\pi_1), \ldots, L(\pi_{n+1}) under the procedure of Chapter I, § 3, Proposition 3.

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.