Bourbaki

Exercise 19, § 4

Tag 01DR, permanent. Cite it as [Bourbaki, Tag 01DR].

Lie Groups and Lie Algebras, IX, 404

We retain the notations of Exerc. 16 to 18; assume that the subgroup H is principal. Give X(T)⊗RX(T)\otimes \mathbf{R} (resp. X(S)⊗RX(S)\otimes \mathbf{R}) a scalar product invariant under WG(T)W_G(T) (resp. under WH(S)W_H(S)). Let λ,μ\lambda , \mu be two roots in BHB_H.

a) Show that, if λ\lambda and μ\mu are orthogonal, the sets R(λ)R(\lambda ) and R(μ)R(\mu) are orthogonal. Deduce that if G is almost simple then so is H.

b) Assume from now on that n(λ,μ)=−1n(\lambda , \mu) =-1. Show that there exists a surjective map u:R(λ)→R(μ)u: R(\lambda )\rightarrow R(\mu) such that, for α∈R(λ),u(α)\alpha \in R(\lambda ),u(\alpha ) is the unique root of R(μ)R(\mu) linked (that is, not orthogonal) to α\alpha; we have Kμ=∑β∈R(μ)KβK_\mu=\sum_{\beta\in R(\mu)}K_{\beta}, and the roots

in R(μ)R(\mu) are pairwise orthogonal (write KμK_\mu and KλK_{\lambda} as linear combinations of the KαK_{\alpha}, then identify the coefficients).

c) If n(μ,λ)=−1n(\mu, \lambda ) =-1, the map uu is bijective, and the only pairs of linked roots in R(λ)∪R(μ)R(\lambda )\cup R(\mu) are the pairs (α,u(α))(\alpha , u(\alpha )) for α∈R(λ)\alpha \in R(\lambda ).

d) Assume that n(μ,λ)=−2n(\mu, \lambda ) =-2; let β∈R(μ)\beta \in R(\mu). Show that u−1(β)u^{-1}(\beta ) contains either one or two elements; if u−1(β)={α1,α2}u^{-1}(\beta ) =\{\alpha_1, \alpha_2\}, the root αi(i=1,2)\alpha_i(i= 1,2) is orthogonal to the other roots in R(λ)R(\lambda ), and has the same length as β\beta; if u−1(β)={α}u^{-1}(\beta ) =\{\alpha \} and ∥α∥=∥β∥,α\|\alpha \|=\|\beta \|,\alpha is linked to a unique root α′∈R(λ)\alpha '\in R(\lambda ), of the same length as α\alpha, and {α,α′}\{\alpha , \alpha '\} is orthogonal to the remainder of R(λ)R(\lambda ); if u−1(β)={α}u^{-1}(\beta ) =\{\alpha \} and ∥α∥≠∥β∥,α\|\alpha \| \not=\|\beta \|,\alpha is orthogonal to the other roots in R(λ)R(\lambda ).

e) Study similarly the case n(μ,λ)=−3n(\mu, \lambda ) =-3.

Solution

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