Bourbaki

Exercise 6, § 4

Lie Groups and Lie Algebras, VI

Bourbaki marks this one as harder than the rest.

Let RR be an irreducible reduced root system, (α1,…,αl)(\alpha_1, \ldots, \alpha_l) a basis of RR, p1α1+⋯+plαlp_1 \alpha_1 + \cdots + p_l \alpha_l the highest root, a1α1+⋯+alαla_1 \alpha_1 + \cdots + a_l \alpha_l the sum of the positive roots, m1,…,mlm_1, \ldots, m_l the exponents of W(R)W(R), gg its order, ff the connection index.

a) Verify that, in each case,

l!p1p2⋯plm1m2⋯ml=a1a2⋯al. l! p_1 p_2 \cdots p_l m_1 m_2 \cdots m_l = a_1 a_2 \cdots a_l.

b) Show that

gm1m2⋯ml=fa1a2⋯al. g m_1 m_2 \cdots m_l = f a_1 a_2 \cdots a_l.

(Use aa and Prop. 7 of §2§ 2, no. 4.)

c) For any positive root α=∑iciφi\alpha = \sum_i c_i \varphi_i, let e(α)=∑icie(\alpha) = \sum_i c_i. Calculate in each case the polynomial

P(t)=∑α>0te(α). P(t) = \sum_{\alpha > 0} t^{e(\alpha)}.

Verify5^5 that P(t)=t∑i=1l(1+t+⋯+tmi−1)P(t) = t \sum_{i=1}^l (1 + t + \cdots + t^{m_i-1}).

5^5 For a proof that does not use the classification, see: B. KOSTANT, The principal three-dimensional subgroup and the Betti numbers of a complex simple Lie group, Amer. J. of Math.s., Vol. LXXXI (1959), p. 973-1032.

Solution

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Transcribed from the printed volume. Not translated.