Bourbaki

Exercise 4, § 2

Tag 00ZF, permanent. Cite it as [Bourbaki, Tag 00ZF].

Lie Groups and Lie Algebras, VIII, 227

Let (g,h)(\mathfrak{g},\mathfrak{h}) be a split semi-simple Lie algebra, and (Xα)α∈R(X_{\alpha})_{\alpha\in R} a family of elements satisfying the conditions of Lemma 2. If α,β∈R\alpha , \beta \in R and α+β∈R\alpha +\beta \in R, define NαβN_{\alpha \beta} by the formula [Xα,Xβ]=NαβXα+β[X_{\alpha}, X_{\beta}] = N_{\alpha \beta}X_{\alpha+\beta}; if α+β∉R\alpha +\beta \notin R, put Nαβ=0N_{\alpha \beta}= 0. Prove the following formulas:

a)Nαβ=−Nβα a) N_{\alpha \beta}=-N_{\beta \alpha}

b) If α,β,γ∈R\alpha , \beta , \gamma \in R are such that α+β+γ=0\alpha +\beta +\gamma = 0, then

NαβNβγNγα N_{\alpha \beta}N_{\beta \gamma}N_{\gamma \alpha}
⟨γ,γ⟩‾=⟨α,α⟩‾=⟨β,β⟩‾ \overline{\langle\gamma , \gamma\rangle}=\overline{\langle\alpha , \alpha\rangle}=\overline{\langle\beta , \beta\rangle}

c) If α,β,γ,δ∈R\alpha , \beta , \gamma , \delta \in R are such that α+β+γ+δ=0\alpha +\beta +\gamma +\delta = 0, and if no pair has sum zero, we have:

NαβNγδNβγNαδNγαNβδ N_{\alpha \beta}N_{\gamma \delta}N_{\beta \gamma}N_{\alpha \delta}N_{\gamma \alpha}N_{\beta \delta}

+ + =0= 0.

⟨γ+δ,γ+δ⟩‾⟨α+δ,α+δ⟩‾⟨β+δ,β+δ⟩‾ \overline{\langle\gamma + \delta , \gamma + \delta\rangle}\overline{\langle\alpha + \delta , \alpha +\delta\rangle}\overline{\langle\beta + \delta , \beta + \delta\rangle}
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Cites

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