Bourbaki

Lemma 3

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

Lie Groups and Lie Algebras, chapter V, § 3. Groups of displacements generated by reflections

Let q be a positive quadratic form on a real vector space V and let B be the associated symmetric bilinear form. Let a1,…,ana_1, \ldots, a_n be elements of V such that B(ai,aj)≤0B(a_i, a_j) \leq 0 for i≠ji \neq j.

(i) If c1,…,cnc_1, \ldots, c_n are real numbers such that q(∑iciai)=0q \left( \sum_i c_i a_i \right) = 0, then

q(∑i∣ci∣ai)=0. q \left( \sum_i |c_i| a_i \right) = 0.

(ii) If q is non-degenerate and if there exists a linear form f on V such that f(ai)>0f(a_i) > 0 for all i, the vectors a1,…,ana_1, \ldots, a_n are linearly independent.

The relation B(ai,aj)≤0B(a_i, a_j) \leq 0 for i≠ji \neq j immediately implies that

q(∑i∣ci∣ai)≤q(∑iciai), q \left( \sum_i |c_i| a_i \right) \leq q \left( \sum_i c_i a_i \right),
hence (i). If q is non-degenerate, the relation ∑iciai=0\sum_i c_i a_i = 0 thus implies that
∑i∣ci∣ai=0; \sum_i |c_i| a_i = 0;
it follows that, for any linear form f on V, we have ∑i∣ci∣f(ai)=0\sum_i |c_i| f(a_i) = 0, and hence ci=0c_i = 0 for all i if we also have f(ai)>0f(a_i) > 0 for all i. This proves (ii).

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

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Cites

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Transcribed from the printed volume. Not translated.