Bourbaki

Lemma 4

Tag 0A0L, permanent. Cite it as [Bourbaki, Tag 0A0L].

Lie Groups and Lie Algebras, chapter I, § 6. Semi-simple Lie algebras

Let ρ,σ,τ\rho, \sigma, \tau be representations of gg on vector spaces M,N,PM, N, P. Suppose that we are given a KK-bilinear mapping (m,n)↦m.n(m, n) \mapsto m.n of M×NM \times N into PP such that

(ρ(x)m).n+m.(σ(x)n)=τ(x)(m.n) (\rho(x)m).n + m.(\sigma(x)n) = \tau(x)(m.n)

for all m∈M,n∈N,x∈gm \in M, n \in N, x \in g.

(a) If m0∈M0m_0 \in M_0, the mapping n↦m0.nn \mapsto m_0.n is a gg-module homomorphism.

(b) If n∈Nδn \in N_\delta, then m0.n∈Pδm_0.n \in P_\delta.

(c) If MM is a (not necessarily associative) algebra and the ρ(x)\rho(x) are derivations of MM, M0M_0 is a subalgebra of MM and each MδM_\delta is a right and left M0M_0-module.

For m0∈M0,n∈Nm_0 \in M_0, n \in N and x∈gx \in g,

τ(x)(m0.n)=m0.(σ(x)n), \tau(x)(m_0.n) = m_0.(\sigma(x)n),

whence (a). Assertion (b) follows from (a) (Algebra, Chapter VIII, § 3, no. 4, Proposition 10). If N=P=MN = P = M and σ=τ=ρ\sigma = \tau = \rho, assertion (b) gives assertion (c) as a special case.

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

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Cites

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Transcribed from the printed volume. Not translated.