Bourbaki

Exercise 2, § 6

Lie Groups and Lie Algebras, V

Bourbaki marks this one as harder than the rest.

Assume that WW is irreducible. Let CC be a chamber relative to WW, (H1,…,Hl)(H_1, \ldots, H_l) its walls, and eie_i a non-zero vector orthogonal to HiH_i. Assume that e1,…,ere_1, \ldots, e_r (resp. er+1,…,ele_{r+1}, \ldots, e_l) are pairwise orthogonal (cf. no. 2). For all u∈Zu \in \mathbf{Z}, define HuH_u and sus_u by Hu=HkH_u = H_k if u≡k(modl)u \equiv k \pmod{l} and su=sHus_u = s_{H_u}.

a) Show that the elements of H\mathfrak{H} are the s1s2…su−1Hus_1 s_2 \ldots s_{u-1} H_u for u=1,2,…,lh/2u = 1, 2, \ldots, lh/2.

b) Let s′=s1…srs' = s_1 \ldots s_r and s′′=sr+1…sls'' = s_{r+1} \ldots s_l, so that c=s′s′′c = s's'' is the Coxeter transformation associated to the ordered chamber C. Let w0w_0 be the element of W that transforms C to −C-C (cf. § 4, Exerc. 2). Show that, if hh is odd,

w0=s′s′′s′…s′′s′⏟h terms=s′′s′s′′…s′s′′⏟h terms=s′′c(h−1)/2=c(h−1)/2s′. w_0 = \underbrace{s' s'' s' \ldots s'' s'}_{h \text{ terms}} = \underbrace{s'' s' s'' \ldots s' s''}_{h \text{ terms}} = s'' c^{(h-1)/2} = c^{(h-1)/2} s'.

c) Put S={s1,…,sl}S = \{ s_1, \ldots, s_l \}. The pair (W,S)(W, S) is a Coxeter system. If w∈Ww \in W, denote by lS(w)l_S(w) the length of ww with respect to S (Chap. IV, § 1, no. 1). Show that

lS(s′)=r,lS(s′′)=l−r,lS(c)=l. l_S(s') = r, \quad l_S(s'') = l - r, \quad l_S(c) = l.

Deduce that, with the assumptions in b),

lS(w0)≤r+h−12landlS(w0)≤(l−r)+h−12l. l_S(w_0) \leq r + \frac{h-1}{2} l \quad \text{and} \quad l_S(w_0) \leq (l-r) + \frac{h-1}{2} l.

Show on the other hand that lS(w0)=Card⁡(H)=hl/2l_S(w_0) = \operatorname{Card}(\mathfrak{H}) = hl/2 (use Exerc. 22 of Chap. IV, § 1). Deduce that r=l/2r = l/2 and that (s1,s2,…,slh/2)(s_1, s_2, \ldots, s_{lh/2}) is a reduced decomposition of w0w_0.

d) Show that, if hh is even, (s1,…,slh/2)(s_1, \ldots, s_{lh/2}) is a reduced decomposition of w0=ch/2w_0 = c^{h/2}. (Same method.)

Solution

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Transcribed from the printed volume. Not translated.