Bourbaki

Exercise 4, § 4

Lie Groups and Lie Algebras, VI

Bourbaki marks this one as harder than the rest.

Let RR be an irreducible reduced root system in VV. (α1,…,αl)(\alpha_1, \ldots, \alpha_l) a basis of RR, α~\tilde{\alpha} the highest root. Put α~=n1α1+⋯+nlαl\tilde{\alpha} = n_1 \alpha_1 + \cdots + n_l \alpha_l. We are going to determine the maximal closed, symmetric subsets of RR, distinct from RR.

a) Let i∈{1,2,…,l}i \in \{1, 2, \ldots, l\}. Let RiR_i be the set of α∈R\alpha \in R which are linear combinations of the αj\alpha_j for j≠ij \neq i. Show that RiR_i is maximal if and only if ni=1n_i = 1.

(1) M. Kneser has shown that all the "exceptional isomorphisms" between the finite classical groups can be obtained by using an analogous method. Cf. Über die Ausnahme-Isomorphismen zwischen klassischen Gruppen, Hamburger Abh., Vol. XXXI (1967), p. 136-140.

b) Let i∈{1,2,…,l}i \in \{1, 2, \ldots, l\} and assume that ni>1n_i > 1. Let SiS_i be the set of roots ∑j=1lmjαj\sum_{j=1}^l m_j \alpha_j with mi≡0(modni)m_i \equiv 0 \pmod{n_i}. Show that SiS_i is maximal if and only if nin_i is prime. (If ni=abn_i = ab with a>1,b>1a > 1, b > 1, consider the subset S′S' of RR consisting of the roots ∑jmjαj\sum_j m_j \alpha_j with mi≡0(moda)m_i \equiv 0 \pmod{a}, and show that S′S' strictly contains SiS_i.) Show that the roots −α~,αj (j≠i)-\tilde{\alpha}, \alpha_j \ (j \neq i) form a basis of SiS_i (which is of rank ll). Deduce the Dynkin graph of SiS_i.

c) Every maximal closed, symmetric subset of RR is transformed by an element of W(R)W(R) into one of the subsets described in aa or bb. (Let Σ\Sigma be such a subset. Then Σ=R∩H\Sigma = R \cap H, where HH is a subgroup of Q(R)Q(R) of finite index (§1§ 1, Exerc. 6 b)); we can assume that Q(R)/HQ(R)/H is cyclic. Then there exists u∗∈V∗u^* \in V^* such that Σ\Sigma is the set of α∈R\alpha \in R such that ⟨u∗,α⟩∈Z\langle u^*, \alpha \rangle \in \mathbf{Z}.

d) List the maximal, closed, symmetric subsets of RR for the different types of irreducible reduced root systems.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.