Exercise 5, § 2
Lie Groups and Lie Algebras, VI
Bourbaki marks this one as harder than the rest.
Assume that is irreducible, and choose a chamber of in ; denote by the corresponding basis of .
a) Show that the minuscule weights (§ 1, Exerc. 24) of form a system of representatives in of the non-zero elements of . (Apply to the corollary of Prop. 6.)
b) Let be a saturated subset of (§ 1, Exerc. 23). Assume that is non-empty and not reduced to . Let be a non-zero element of of minimal length. Show that . (In view of a), does not satisfy condition (ii) of Exerc. 24 of § 1; hence, there exists such that , so p - α ∈ X. Since the length of p - α is strictly less than that of p, p - α = 0 so p ∈ R.)
c) Let p be a dominant weight not belonging to Q(R). Show that the saturated subset S(p) generated by p (cf. § 1, Exerc. 23) contains a unique minuscule weight. (Remark that S(p) is contained in a non-trivial class mod. Q(R); conclude by applying a) and Exerc. 24 c) of § 1.)
d) Let p be a dominant weight not belonging to Q(R). Prove that the following two properties are equivalent:
(i) p is minuscule.
(v) There is no dominant weight q ≠ p such that p - q is a linear combination of elements of B with non-negative integer coefficients.
(If q satisfies the conditions of (v), let p₁ be a minuscule weight belonging to S(q). Then, p₁ ≡ p mod. Q(R), p₁ < p, and a) shows that p is not minuscule. Hence, (i) ⇒ (v). Conversely, if p is not minuscule, let q ∈ S(p) - W.p; transforming q by an element of W if necessary, we can assume that q ∈ C̄; by Exerc. 23 of § 1, q < p, q ≠ p, and q ≡ p mod. Q(R). Hence, (v) ⇒ (i).)
Solution
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