Bourbaki

Exercise 17, § 2

Lie Groups and Lie Algebras, IV

We retain the hypotheses and notation of Exerc. 15 and 16. Choose an element φ∈J\varphi \in \mathscr{J} taking the canonical chamber CC of A0A_0 (§1§ 1, Exerc. 16) to the canonical chamber CeC_e of II and, as in Exerc. 15 dd), denote by NN the stabilizer of the apartment φ(A0)∈U\varphi(A_0) \in \mathscr{U} in the group GG. For any subset DD of A0A_0, denote by BDB_D the subgroup of GG leaving fixed all the points of φ(D)\varphi(D): we have B=BCB = B_C and B∩N=BA0B \cap N = B_{A_0}.

a) Let α\alpha be a half of A0A_0 containing CC. Show that BαB_\alpha acts transitively on the semi-apartments of II distinct from φ(α)\varphi(\alpha) whose wall is the wall LL of φ(α)\varphi(\alpha) (let XX be such a semi-apartment, FF a panel contained in LL and C′C' a chamber of φ(α)\varphi(\alpha) containing FF; show, by using Exerc. 16 cc) and Exerc. 13 aa), that there exists g∈Gg \in G such that g(X∪φ(α))=φ(A0)g(X \cup \varphi(\alpha)) = \varphi(A_0) and g(C′)=C′g(C') = C'; show that g(F)=Fg(F) = F and deduce from Exerc. 22 cc of §1§ 1 that g∈Bαg \in B_\alpha). Deduce that BLB_L acts doubly transitively on the semi-apartments with wall LL and that (BL,Bα,BL∩N)(B_L, B_\alpha, B_L \cap N) is a Tits system whose Weyl group is of order 2 (cf. Exerc. 7).

b) Let D1D_1 and D2D_2 be two convex subsets of A0A_0 such that C⊂D1⊂D2C \subset D_1 \subset D_2 and that there exists a unique half α\alpha of A0A_0 with D1⊂αD_1 \subset \alpha and D2⊄αD_2 \not\subset \alpha: then D1=D2∩αD_1 = D_2 \cap \alpha (§1§ 1, Exerc. 21 cc). Show that BD1=BαBD2B_{D_1} = B_\alpha B_{D_2} and that Bα∩BD2=B∩NB_\alpha \cap B_{D_2} = B \cap N (by considering a gallery of smallest possible length having one extremity equal to CC and the other containing a point a∈D2−D1a \in D_2 - D_1); show that there exist two neighbouring chambers C1′C'_1 and C2′C'_2 of A0A_0 such that C1′⊂D1,C2′⊂D2,C2′⊄D1C'_1 \subset D_1, C'_2 \subset D_2, C'_2 \not\subset D_1 and C1′∩C2′C'_1 \cap C'_2 contained in the wall L′L' of α\alpha. Put Ci=φ(Ci′),F=φ(C1′∩C2′)C_i = \varphi(C'_i), F = \varphi(C'_1 \cap C'_2) and L=φ(L′)L = \varphi(L'). Show that the convex hull of D1∪C2′D_1 \cup C'_2 is equal to D2D_2. Let b∈BD1b \in B_{D_1}: then b(C1)=C1b(C_1) = C_1, so b(F)=Fb(F) = F and b(C2)≠C1b(C_2) \neq C_1. Deduce that the convex hull of b(C2)∪Lb(C_2) \cup L is a semi-apartment XX distinct from φ(α)\varphi(\alpha) and with wall LL, and that there exists b′∈Bαb' \in B_\alpha such that

b′(φ(A0))=X∪φ(α). b'(\varphi(A_0)) = X \cup \varphi(\alpha).

Show that b′b(a)=ab'b(a) = a for all a∈φ(D1∪C2′)a \in \varphi(D_1 \cup C'_2) and that b′b∈BD2b'b \in B_{D_2}.

c) Let (αi)1≤i≤q(\alpha_i)_{1 \leq i \leq q} be the halves of A0A_0 containing CC, numbered so that the map

j↦⋂1≤i≤jαi j \mapsto \bigcap_{1 \leq i \leq j} \alpha_i

is strictly decreasing (cf. §1§ 1, Exerc. 17). Show that B=Bα1…BαqB = B_{\alpha_1} \ldots B_{\alpha_q} and that if bi,bi′∈Bαib_i, b'_i \in B_{\alpha_i} with b1…bq=b1′…bq′b_1 \ldots b_q = b'_1 \ldots b'_q, then bi′∈bi(B∩N)b'_i \in b_i(B \cap N) for 1≤i≤q1 \leq i \leq q.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.