Let E be a finite dimensional simple g-module, ω its highest weight, and B the vector space of g-invariant bilinear forms on E. Let m be the integer ∑α∈R+ω(Hα), so that m/2 is the sum of the coordinates of ω with respect to B (Chap. VI, §1, no. 10, Cor. of Prop. 29). Let w0 be the element of W such that w0(B)=−B.
(i) If w0(ω)=−ω, then B=0.
(ii) Assume that w0(ω)=−ω. Then B is of dimension 1, and every non-zero element of B is non-degenerate. If m is even (resp. odd ), every element of B is symmetric (resp. alternating).
a) Let Φ∈B. The map φ from E to E∗ defined, for x,y∈E, by φ(x)(y)=Φ(x,y) is a homomorphism of g-modules. If Φ=0, then φ=0, so φ is an isomorphism by Schur’s lemma, and hence Φ is non-degenerate. Consequently, the g-module E is isomorphic to the g-module E∗, so that w0(ω)=−ω. We have thus proved (i).
b) Assume from now on that w0(ω)=−ω. Then E is isomorphic to E∗. The vector space B is isomorphic to Homg(E,E∗), and hence to Homg(E,E) which is of dimension 1 (§6, no. 1, Prop. 1 (iii)). Hence dimB=1. Every non-zero element Φ of B is non-degenerate by a). Put Φ1(x,y)=Φ(y,x) for x,y∈E. By the preceding, there exists λ∈k such that Φ1(x,y)=λΦ(x,y) for all x,y∈E. Then Φ(y,x)=λΦ(x,y)=λ2Φ(y,x), so λ2=1 and λ=±1. Thus, Φ is either symmetric or alternating.
c) By Chap. VII, §1, no. 3, Prop. 9 (v), Eλ and Eμ are orthogonal with respect to Φ if λ+μ=0. Since Φ is non-degenerate, it follows that Eω,E−ω are not orthogonal with respect to Φ.
d) There exists a homomorphism φ from sl(2,k) onto a subalgebra of g that takes H to
α∈R+∑Hα
(Lemma 2). Consider E as an sl(2,k)-module via this homomorphism. Then the elements of Eλ are of weight λ(∑α∈R+Hα). If λ∈P is such that Eλ=0 and λ=ω,λ=−ω, then −ω<λ<ω, so
−m=−ωα∈R+∑Hα<λα∈R+∑Hα<ωα∈R+∑Hα=m.
Let G be the isotypical component of type V(m) of the sl(2,k)-module E. By the preceding, G is of length 1 and contains Eω,E−ω. By c), the restriction of Φ to G is non-zero. By §1, no. 3, Remark 3, π is even or odd according as this restriction is symmetric or alternating. In view of b), this completes the proof.
Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.