Bourbaki

Exercise 24, § 1

Lie Groups and Lie Algebras, VI

We retain the notation and assumptions of the preceding exercise.

a) Let p∈P(R)p \in P(R), and let W.pW.p be the orbit of pp under WW. Show that the following conditions are equivalent:

(i) S(p)=W.pS(p) = W.p.

(ii) ⟨p,α∨⟩=0,1\langle p, \alpha^\vee \rangle = 0, 1 or −1-1 for all α∈R\alpha \in R.

(If (ii) is not satisfied, construct an element q∈S(p)q \in S(p) such that (q∣q)<(p∣p)(q|q) < (p|p), and hence q∉W.pq \notin W.p.)

b) Let XX be a non-empty saturated subset of P(R)P(R). Show that XX contains an element pp satisfying conditions (i) and (ii) above. (Take pp to be an element of XX of minimal length.)

c) Assume that RR is irreducible. Let CC be a chamber of RR, let B={αi}i∈IB = \{\alpha_i\}_{i \in I} be the corresponding basis, and let

γ∗=∑iniαi \gamma^* = \sum_i n_i \alpha_i

be the highest root of RR. Let JJ be the set of i∈Ii \in I such that ni=1n_i = 1. Let p≠0p \neq 0 be a dominant weight. Show that conditions (i) and (ii) of a) are equivalent to each of the following conditions:

² This exercise, hitherto unpublished, was communicated to us by R. Steinberg.

(iii) ⟨p,γ∗⟩=1\langle p, \gamma^* \rangle = 1.

(iv) There exists i∈Ji \in J such that pp is equal to the corresponding fundamental weight.

A weight pp satisfying these conditions is called minuscule. Show that every non-empty saturated subset of P(R)P(R) contains 0 or a minuscule weight.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.