Bourbaki

Proposition 12

Tag 0144, permanent. Cite it as [Bourbaki, Tag 0144].

Lie Groups and Lie Algebras, chapter VIII, § 7. Finite dimensional modules over a split semi-simple Lie algebra

Let E be a finite dimensional simple g\mathfrak{g}-module, ω\omega its highest weight, and B\mathscr{B} the vector space of g\mathfrak{g}-invariant bilinear forms on E. Let mm be the integer ∑α∈R+ω(Hα)\sum_{\alpha\in R_+}\omega (H_{\alpha}), so that m/2m/2 is the sum of the coordinates of ω\omega with respect to B (Chap. VI, §1, no. 10, Cor. of Prop. 29). Let w0w_0 be the element of W such that w0(B)=−Bw_0(B) =-B.

(i) If w0(ω)≠−ωw_0(\omega )\not=-\omega, then B=0\mathscr{B}= 0.

(ii) Assume that w0(ω)=−ωw_0(\omega ) =-\omega. Then B\mathscr{B} is of dimension 1, and every non-zero element of B\mathscr{B} is non-degenerate. If mm is even (resp. odd ), every element of B\mathscr{B} is symmetric (resp. alternating).

a) Let Φ∈B\Phi \in \mathscr{B}. The map φ\varphi from E to E∗E^* defined, for x,y∈Ex, y\in E, by φ(x)(y)=\varphi (x)(y) = Φ(x,y)\Phi (x, y) is a homomorphism of g\mathfrak{g}-modules. If Φ≠0\Phi \not= 0, then φ≠0\varphi \not= 0, so φ\varphi is an isomorphism by Schur’s lemma, and hence Φ\Phi is non-degenerate. Consequently, the g\mathfrak{g}-module E is isomorphic to the g\mathfrak{g}-module E∗E^*, so that w0(ω)=−ωw_0(\omega ) =-\omega. We have thus proved (i).

b) Assume from now on that w0(ω)=−ωw_0(\omega)=-\omega. Then E is isomorphic to E∗E^*. The vector space B\mathscr B is isomorphic to Hom⁡g(E,E∗)\operatorname{Hom}_{\mathfrak g}(E,E^*), and hence to Hom⁡g(E,E)\operatorname{Hom}_{\mathfrak g}(E,E) which is of dimension 1 (§6, no. 1, Prop. 1 (iii)). Hence dim⁡B=1\dim \mathscr B=1. Every non-zero element Φ\Phi of B\mathscr B is non-degenerate by a). Put Φ1(x,y)=Φ(y,x)\Phi_1(x,y)=\Phi(y,x) for x,y∈Ex,y\in E. By the preceding, there exists λ∈k\lambda\in k such that Φ1(x,y)=λΦ(x,y)\Phi_1(x,y)=\lambda\Phi(x,y) for all x,y∈Ex,y\in E. Then Φ(y,x)=λΦ(x,y)=λ2Φ(y,x)\Phi(y,x)=\lambda\Phi(x,y)=\lambda^2\Phi(y,x), so λ2=1\lambda^2=1 and λ=±1\lambda=\pm1. Thus, Φ\Phi is either symmetric or alternating.

c) By Chap. VII, §1, no. 3, Prop. 9 (v), EλE^\lambda and EμE^\mu are orthogonal with respect to Φ\Phi if λ+μ≠0\lambda+\mu\ne0. Since Φ\Phi is non-degenerate, it follows that Eω,E−ωE^\omega,E^{-\omega} are not orthogonal with respect to Φ\Phi.

d) There exists a homomorphism φ\varphi from sl(2,k)\mathfrak{sl}(2,k) onto a subalgebra of g\mathfrak g that takes HH to

∑α∈R+Hα \sum_{\alpha\in R_+}H_\alpha
(Lemma 2). Consider E as an sl(2,k)\mathfrak{sl}(2,k)-module via this homomorphism. Then the elements of EλE^\lambda are of weight λ(∑α∈R+Hα)\lambda\left(\sum_{\alpha\in R_+}H_\alpha\right). If λ∈P\lambda\in P is such that Eλ≠0E^\lambda\ne0 and λ≠ω,λ≠−ω\lambda\ne\omega,\lambda\ne-\omega, then −ω<λ<ω-\omega<\lambda<\omega, so
−m=−ω(∑α∈R+Hα)<λ(∑α∈R+Hα)<ω(∑α∈R+Hα)=m. -m=-\omega\left(\sum_{\alpha\in R_+}H_\alpha\right)<\lambda\left(\sum_{\alpha\in R_+}H_\alpha\right)<\omega\left(\sum_{\alpha\in R_+}H_\alpha\right)=m.
Let G be the isotypical component of type V(m)(m) of the sl(2,k)\mathfrak{sl}(2,k)-module E. By the preceding, G is of length 1 and contains Eω,E−ωE^\omega,E^{-\omega}. By c), the restriction of Φ\Phi to G is non-zero. By §1, no. 3, Remark 3, π\pi 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.

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