Bourbaki

Corollary 3

Tag 0AB7, permanent. Cite it as [Bourbaki, Tag 0AB7].

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

Let GG be a Lie group and π1,…,πn\pi_1, \ldots, \pi_n analytic linear representations of GG on finite-dimensional vector spaces E1,…,EnE_1, \ldots, E_n. Then the representation π1⊗⋯⊗πn\pi_1 \otimes \cdots \otimes \pi_n of GG (Appendix) is analytic and L(π1⊗⋯⊗πn)L(\pi_1 \otimes \cdots \otimes \pi_n) is the tensor product of L(π1),…,L(πn)L(\pi_1), \ldots, L(\pi_n).

The mapping (A1,…,An)↦A1⊗⋯⊗An(A_1, \ldots, A_n) \mapsto A_1 \otimes \cdots \otimes A_n of L(E1)×⋯×L(En)\mathscr{L}(E_1) \times \cdots \times \mathscr{L}(E_n) into L(E1⊗⋯⊗En)\mathscr{L}(E_1 \otimes \cdots \otimes E_n) is multilinear, whence the fact that π\pi is analytic. Consider the mapping (x1,…,xn)↦x1⊗⋯⊗xn(x_1, \ldots, x_n) \mapsto x_1 \otimes \cdots \otimes x_n of E1×⋯×EnE_1 \times \cdots \times E_n into

E1⊗⋯⊗En. E_1 \otimes \cdots \otimes E_n.
By Proposition 41, we see that
(L(π)a)(x1⊗⋯⊗xn)=∑i=1nx1⊗⋯⊗(L(πi)a)xi⊗⋯⊗xn (L(\pi)a)(x_1 \otimes \cdots \otimes x_n) = \sum_{i=1}^n x_1 \otimes \cdots \otimes (L(\pi_i)a)x_i \otimes \cdots \otimes x_n
for all a∈L(G)a \in L(G), xi∈Eix_i \in E_i for 1≤i≤n1 \leq i \leq n. Hence L(π)L(\pi) is the tensor product of the L(πi)L(\pi_i).

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

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Transcribed from the printed volume. Not translated.