Bourbaki

Lemma 3

Tag 0A2D, permanent. Cite it as [Bourbaki, Tag 0A2D].

Lie Groups and Lie Algebras, chapter II, § 1. Enveloping bigebra of a Lie algebra

If a family of scalars (λα,β)(\lambda_{\alpha, \beta}) of finite support (with α,β\alpha, \beta in N(Ω)−{0}\mathbf{N}^{(\Omega)} - \{0\}) satisfies relations (20) and (21), there exists a family (μα)∣α∣≥2(\mu_\alpha)_{|\alpha| \geq 2} of finite support such that μα+β=λα,β\mu_{\alpha + \beta} = \lambda_{\alpha, \beta} for α,β\alpha, \beta non-zero.

It suffices to prove that

(23)

α+β=γ+δ \alpha + \beta = \gamma + \delta

implies λα,β=λγ,δ\lambda_{\alpha, \beta} = \lambda_{\gamma, \delta} for α,β,γ,δ\alpha, \beta, \gamma, \delta non-zero. By Riesz's Decomposition Lemma (Algebra, Chapter VI, § 1, no. 10, Theorem 1) there exist π,ρ,σ,τ\pi, \rho, \sigma, \tau in N(Ω)\mathbf{N}^{(\Omega)} such that

α=π+σ,β=ρ+τ,γ=π+ρ,δ=σ+τ. \alpha = \pi + \sigma, \qquad \beta = \rho + \tau, \qquad \gamma = \pi + \rho, \qquad \delta = \sigma + \tau.

Suppose π≠0\pi \neq 0; as σ+β=ρ+δ\sigma + \beta = \rho + \delta, relation (20) implies

λα,β=λπ+σ,β=λπ,σ+β=λπ,ρ+δ=λπ+ρ,δ=λγ,δ. \lambda_{\alpha, \beta} = \lambda_{\pi + \sigma, \beta} = \lambda_{\pi, \sigma + \beta} = \lambda_{\pi, \rho + \delta} = \lambda_{\pi + \rho, \delta} = \lambda_{\gamma, \delta}.
If on the other hand π=0\pi = 0, then β=γ+τ\beta = \gamma + \tau and δ=α+τ\delta = \alpha + \tau, whence
λα,β=λα,γ+τ=λα+τ,γ=λδ,γ \lambda_{\alpha, \beta} = \lambda_{\alpha, \gamma + \tau} = \lambda_{\alpha + \tau, \gamma} = \lambda_{\delta, \gamma}
by (20), but also λδ,γ=λγ,δ\lambda_{\delta, \gamma} = \lambda_{\gamma, \delta} by (21), whence λα,β=λγ,δ\lambda_{\alpha, \beta} = \lambda_{\gamma, \delta}.

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

Cited by

Nothing.

Cites

Transcribed from the printed volume. Not translated.