Bourbaki

Exercise 17, § 1

Lie Groups and Lie Algebras, VI

Let RR be a reduced root system and BB a basis of RR. For any subset JJ of BB, let WJW_J be the subgroup of W(R)W(R) generated by the reflections sαs_\alpha for α∈J\alpha \in J, and let wJw_J be the longest element of WJW_J. Let Φ\Phi be the set of wJw_J.

a) Show that an element w∈W(R)w \in W(R) belongs to Φ\Phi if and only if

W(B)⊆R+(B)∪(−B). W(B) \subseteq R_+(B) \cup (-B).
(To show that the condition is sufficient, denote by J1J_1 the set of α∈B\alpha \in B such that w(α)∈−Bw(\alpha) \in -B and put J=−w(J1)J = -w(J_1). If α∈J1\alpha \in J_1, w(α)∈−Jw(\alpha) \in -J and wJw(α)∈Bw_J w(\alpha) \in B;

if α∈B−J1\alpha \in B - J_1, wJw(α)∈R+w_J w(\alpha) \in R_+: otherwise w(α)w(\alpha) would be positive and wJw(α)w_J w(\alpha) negative, which would imply that w(α)w(\alpha) belongs to the subsystem generated by the β∈J\beta \in J and that α\alpha belongs to the subsystem generated by the β∈J1\beta \in J_1, which is absurd. Deduce that wJw=1.w_J w = 1.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.