Proposition 3

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

Lie Groups and Lie Algebras, chapter VI, § 1. Root systems

For $x, y \in V$ , put

(x|y) = \sum_{\alpha \in R} \langle \alpha \tilde{}, x \rangle \langle \alpha \tilde{}, y \rangle.

Then

(x|y)

is a non-degenerate symmetric bilinear form on

V

, invariant under

A(R)

. For

x, y \in V_{\mathbf{Q}}

we have

(x|y) \in \mathbf{Q}

. The canonical extension of

(x|y)

V_{\mathbf{R}} = V_{\mathbf{Q}} \otimes_{\mathbf{Q}} \mathbf{R}

is non-degenerate and positive.

It is clear that $(x|y)$ is a symmetric bilinear form on $V$ . If $g \in A(R)$ ,

(g(x)|g(y)) = \sum_{\alpha \in R} \langle {}^t g(\alpha \tilde{}), x \rangle \langle {}^t g(\alpha \tilde{}), y \rangle = (x|y)

since

{}^t g)(R \tilde{}) = R \tilde{}

. If

x, y \in V_{\mathbf{Q}}

, then

(x|y) \in \mathbf{Q}

by (RS_{III}). If

z \in V_{\mathbf{R}}

, then

(z|z) = \sum_{\alpha \in R} \langle \alpha \tilde{}, z \rangle^2 \geq 0

, and

(z|z) > 0

z \neq 0

by Prop. 1, so the canonical extension of

(x|y)

V_{\mathbf{R}}

is positive and non-degenerate. The restriction of

(x|y)

V_{\mathbf{Q}}

is thus non-degenerate, and hence the form

(x|y)

V

is non-degenerate.

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

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