Lemma 3

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

Lie Groups and Lie Algebras, chapter VI, § 3. Exponential invariants

The $S(e^p)$ for $p \in P \cap \overline{C}$ form a basis of the $A$ -module $A[P]^W$ .

e_{i_0} = x_{i_0} - \sum_{j < i_0} a_{i_0 j} e_j,

from which (b) follows. Hence

J' \in \mathfrak{S}

, a contradiction. Thus,

J = I

and the lemma is proved.

We now prove Prop. 3. We apply Lemma 4 with $I = P \cap \overline{C}$ . Let $q \in I$ , and let $I_q$ be the set of $p \in I$ such that $p \leq q$ . If $p \in I_q$ , the relations

q - p \geq 0,\quad p \in \overline{C},\quad q \in \overline{C}

imply that

(q - p|p) \geq 0 \quad \text{and} \quad (q - p|q) \geq 0,

and hence that

(p|p) \leq (p|q) \leq (q|q).

The set

I_q

is thus bounded. Since

I

is discrete, it follows that

I_q

is finite, and it is clear that

I

satisfies the condition (MIN). On the other hand, for all

p \in I

x_p = e^p + \sum_{q < p} c_{pq} e^q

so by (6).

x_p = S(e^p) + \sum_{q < p, q \in I} c_{pq} S(e^q).

The proposition now follows from Lemmas 3 and 4.

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

Cited by

Corollary 2, § 8

Cites

Nothing.