Bourbaki

Proposition 7

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

Lie Groups and Lie Algebras, chapter VI, § 2. Affine Weyl group

Assume that RR is irreducible. Let B={α1,…,αl}B = \{ \alpha_1, \ldots, \alpha_l \} be a basis of RR, ff the connection index of RR (§ 1, no. 9) and α~=n1α1+⋯+nlαl\tilde{\alpha} = n_1 \alpha_1 + \cdots + n_l \alpha_l the highest root of RR (for the order defined by BB). Then the order of WW is equal to

(l!)n1n2…nlf. (l!) n_1 n_2 \ldots n_l f.

Let (ω‾1,…,ω‾l)(\overline{\omega}_1, \ldots, \overline{\omega}_l) be the basis of P(R−)P(R^-) dual to BB. By the Cor. of Prop. 5, the open simplex CC with vertices 0,n1−1ω‾1,…,nl−1ω‾l0, n_1^{-1} \overline{\omega}_1, \ldots, n_l^{-1} \overline{\omega}_l is an alcove of EE. Choose a Haar measure μ\mu on the additive group V∗V^*. Let AA be the set of elements of V∗V^* of the form ξ1ω‾1+⋯+ξlω‾l\xi_1 \overline{\omega}_1 + \cdots + \xi_l \overline{\omega}_l, with 0<ξi<10 < \xi_i < 1 for i=1,…,li = 1, \ldots, l. By Cor. 2 of Prop. 15 of Integration, Chap. VII, § 1, no. 10,

μ(A)/μ(C)=(l!)n1n2…nl.(5) \mu(A)/\mu(C) = (l!) n_1 n_2 \ldots n_l. \tag{5}
On the other hand, let A′A' be the set of elements of V∗V^* of the form
ξ1α1~+⋯+ξlαl~, \xi_1 \alpha_{\tilde{1}} + \cdots + \xi_l \alpha_{\tilde{l}},
with 0<ξi<10 < \xi_i < 1 for i=1,…,li = 1, \ldots, l. Since (α1~,…,αl~)(\alpha_{\tilde{1}}, \ldots, \alpha_{\tilde{l}}) is a basis of the Z\mathbf{Z}-module Q(R−)Q(R^-), we can apply Lemma 1 with X=V∗X = V^*, G=WaG = W_a, G′=QG' = Q, U=CU = C and U′=A′U' = A'. This gives
μ(A′)/μ(C)=(Wa:Q)=Card W.(6) \mu(A')/\mu(C) = (W_a : Q) = \text{Card } W. \tag{6}

Finally, Lemma 1 can be applied again, taking X = V*, G = P, G' = Q, U = A and U' = A'. This gives

μ(A′)/μ(A)=(P:Q)=(P(R−):Q(R−))=f. \mu(A')/\mu(A) = (P : Q) = (P(R^-) : Q(R^-)) = f.

The proposition now follows by comparing formulas (5), (6) and (7).

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.