Bourbaki

Theorem 2

Tag 0ANE, permanent. Cite it as [Bourbaki, Tag 0ANE].

Lie Groups and Lie Algebras, chapter IV, § 2. Tits Systems

The pair (W,S)(W, S) is a Coxeter system. Moreover, for s∈Ss \in S and w∈Ww \in W, the relations C(sw)=C(s).C(w)C(sw) = C(s).C(w) and lS(sw)>lS(w)l_S(sw) > l_S(w) are equivalent.

For any s∈Ss \in S, let PsP_s be the set of elements w∈Ww \in W such that

C(s).C(w)=C(sw). C(s).C(w) = C(sw).
We are going to verify that the PsP_s satisfy conditions (A'), (B') and (C) of § 1, no. 7; the two assertions of the theorem will then follow from Prop. 6 of § 1, no. 7.

Condition (A') is clear.

We verify (B'). If PsP_s and sPssP_s had an element ww in common, we would have w∈Psw \in P_s and sw∈Pssw \in P_s, and hence

C(s).C(w)=C(sw),C(s).C(sw)=C(w). C(s).C(w) = C(sw), \qquad C(s).C(sw) = C(w).
It would follow that C(s).C(s).C(w)=C(w)C(s).C(s).C(w) = C(w) and, by formula (5), this would imply that C(w)=C(sw)C(w) = C(sw), which would contradict Th. 1.

We verify (C). Let s,s′∈Ss, s' \in S and w,w′∈Ww, w' \in W with w′=ws′w' = ws'. The assumption that w∈Psw \in P_s and w′∉Psw' \notin P_s implies that

C(sw)=C(s).C(w)(9) C(sw) = C(s).C(w) \tag{9}
C(w′)⊂C(s).C(w′)(10) C(w') \subset C(s).C(w') \tag{10}
by (3).

From (9) and the relation w=w′s′w = w's', it follows that

C(s)w′s′B=C(sw).(11) C(s)w's'B = C(sw). \tag{11}
By formula (2'), C(w′).C(s′)⊂C(w′)∪C(w′s′)C(w').C(s') \subset C(w') \cup C(w's'), which immediately implies that
C(w′)s′B⊂C(ws′)∪C(w).(12) C(w')s'B \subset C(ws') \cup C(w). \tag{12}
Since C(w′)C(w') is a union of left cosets gBgB and since
C(s).C(w′)=C(s)w′B, C(s).C(w') = C(s)w'B,
formula (10) shows that C(s)w′C(s)w' meets C(w′)C(w') and a fortiori that C(s)w′s′BC(s)w's'B meets C(w′)s′BC(w')s'B. It follows from formulas (11) and (12) that the double coset C(sw)C(sw) is equal to one of the double cosets C(ws′)C(ws') and C(w)C(w); since sw≠wsw \neq w, Th. 1 allows us to conclude that sw=ws′sw = ws'.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Transcribed from the printed volume. Not translated.