Bourbaki

Corollary

Tag 0132, permanent. Cite it as [Bourbaki, Tag 0132].

Lie Groups and Lie Algebras, chapter VIII, § 6. Modules over a split semi-simple Lie algebra

Let ⟨⋅,⋅⟩\langle \cdot ,\cdot \rangle be a non-degenerate invariant symmetric bilinear form on g\mathfrak{g}, C the Casimir element associated to ⟨⋅,⋅⟩\langle \cdot ,\cdot \rangle. Denote also by ⟨⋅,⋅⟩\langle \cdot ,\cdot \rangle the inverse form on h∗\mathfrak{h}^* of the restriction of ⟨⋅,⋅⟩\langle \cdot ,\cdot \rangle to h(\mathfrak{h}(§2, no. 3, Prop. 5). Then χ(C)=⟨λ,λ+2ρ⟩\chi (C) =\langle \lambda , \lambda + 2\rho \rangle, where ρ=12∑α∈R+α\rho =\frac{1}{2}\sum_{\alpha\in R_+}\alpha.

We recall the notations of §2, no. 3, Prop. 6. We have

1 1

C = ∑XαX−α+∑X−αXα\sum X_{\alpha}X_{-\alpha}+\sum X_{-\alpha}X_{\alpha}

α∈R−⟨Xα,X−α⟩‾αR+∈⟨Xα,X−α⟩‾ _{\alpha\in R_-}\overline{\langle X_{\alpha}, X_{-\alpha}\rangle}\alpha^{\in}_{R_+}\overline{\langle X_{\alpha}, X_{-\alpha}\rangle}

1 ′'

+∑[Xα,X−α]+∑HiH +\sum[X_{\alpha}, X_{-\alpha}] +\sum H_iH
α∈R+⟨Xα,X−α⟩‾i∈Ii _{\alpha\in R_+}\overline{\langle X_{\alpha}, X_{-\alpha}\rangle}i_{\in I}^i

so

φ(C)=∑α∈R+1⟨Xα,X−α⟩[Xα,X−α]+∑i∈IHiHi′ \varphi (C) =\sum_{\alpha\in R_+}\frac{1}{\langle X_{\alpha}, X_{-\alpha}\rangle}[X_{\alpha}, X_{-\alpha}] +\sum_{i\in I}H_iH'_i

By Prop. 7,

1 ′'

χ(C)=∑λ([Xα,X−α])+∑λ(Hi)λ(H) \chi (C) =\sum\lambda ([X_{\alpha}, X_{-\alpha}]) +\sum\lambda (H_i)\lambda (H)
α∈R+⟨Xα,X−α⟩‾i∈Ii _{\alpha\in R_+}\overline{\langle X_{\alpha}, X_{-\alpha}\rangle}i_{\in I}i

Let hλh_{\lambda} be the element of h\mathfrak{h} such that ⟨hλ,h⟩=λ(h)\langle h_{\lambda}, h\rangle =\lambda (h) for all h∈hh\in \mathfrak{h}. By §2, no. 2, Prop. 1,

(( 1 )⟨)\langle 1 ⟩\rangle

λ[Xα,X−α]\lambda [X_{\alpha}, X_{-\alpha}] = hλ,[Xα,X−α]=α(hλ)=⟨λ,α⟩h_{\lambda},[X_{\alpha}, X_{-\alpha}]=\alpha (h_{\lambda}) =\langle \lambda , \alpha \rangle.

⟨Xα,X−α⟩‾⟨Xα,X−α⟩‾ \overline{\langle X_{\alpha}, X_{-\alpha}\rangle}\overline{\langle X_{\alpha}, X_{-\alpha}\rangle}

Hence

χ(C)=((∑α∈R+⟨λ,α⟩)+⟨λ,λ⟩=⟨λ,λ+2ρ⟩ \chi (C) =((\sum_{\alpha\in R_+}\langle \lambda , \alpha \rangle )+\langle \lambda , \lambda \rangle =\langle \lambda , \lambda + 2\rho \rangle

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Cites

Transcribed from the printed volume. Not translated.