Corollary 1

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

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

Let $w_1, \ldots, w_q \in W$ and let $w = w_1 \ldots w_q$ . If

l_S(w) = l_S(w_1) + \cdots + l_S(w_q),

then

C(w) = C(w_1) \ldots C(w_q).

On taking reduced decompositions of the $w_i$ , one is reduced to the case of a reduced decomposition

w = s_1 \ldots s_q, \quad \text{with } s_i \in S.

If $u = s_2 \ldots s_q$ , then $w = s_1 u$ and $l_S(s_1 u) > l_S(u)$ , so $C(w) = C(s_1).C(u)$ by the theorem. The required formula follows from this by induction on $q$ .

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

Cited by

Exercise 10, § 2
Lemma 2, § 2
Corollary 2, § 2
Corollary 3, § 2
Theorem 3, § 2

Cites

Nothing.