Bourbaki

Exercise 14, § 9

Tag 016I, permanent. Cite it as [Bourbaki, Tag 016I].

Lie Groups and Lie Algebras, VIII, 258

Bourbaki marks this one as harder than the rest.

a) Let μ,ν∈P++\mu, \nu \in P_{++} and let γ\gamma be a weight of E(μ)E(\mu). Let ρμ\rho_\mu be the representation of g\mathfrak{g} on E(μ)E(\mu). Let Xα∈gαX_{\alpha}\in \mathfrak{g}^{\alpha} -- {0},Yα∈g−α\{0\}, Y_{\alpha}\in \mathfrak{g}^{-\alpha} -- {0}\{0\}. If α∈B\alpha \in B, put να=ν(Hα)\nu_{\alpha}=\nu (H_{\alpha}), and

E+(μ,γ,ν)=E(μ)γ∩⋂α∈BE^+(\mu, \gamma , \nu ) = E(\mu)^{\gamma}\cap \bigcap_{\alpha\in B} Ker ρμ(Xα)να+1\rho_\mu(X_{\alpha})^{\nu_{\alpha}+1},

E−(μ,γ,ν)=E(μ)γ∩⋂α∈BE^-(\mu, \gamma , \nu ) = E(\mu)^{\gamma}\cap \bigcap_{\alpha\in B} Ker ρμ(Yα)να+1\rho_\mu(Y_{\alpha})^{\nu_{\alpha}+1},

d+(μ,γ,ν)=d^+(\mu, \gamma , \nu ) = dim E+(μ,γ,ν),d−(μ,γ,ν)=E^+(\mu, \gamma , \nu ),d^-(\mu, \gamma , \nu ) = dim E−(μ,γ,ν)E^-(\mu, \gamma , \nu ).

For all λ∈h∗\lambda \in \mathfrak{h}^*, put λ∗=−w0λ\lambda^*=-w_0\lambda, where w0w_0 is the element of W that takes B to −B-B. Show that

d+(μ,γ,ν)=d−(μ,−γ∗,ν∗) d^+(\mu, \gamma , \nu ) =d^-(\mu,-\gamma^*, \nu^*)

b) Let λ1,λ2∈P++,V\lambda_1, \lambda_2\in P_{++}, V the g\mathfrak{g}-module Homk(E(λ1∗),E(λ2)),U_k(E(\lambda^*_1),E(\lambda_2)), U the set of φ∈V\varphi \in V such that Yα.φ=0Y_{\alpha}.\varphi = 0 for all α∈B\alpha \in B, and ww a primitive vector in E(λ1∗)E(\lambda^*_1). Show that φ→φ(w)\varphi \rightarrow \varphi (w) is an isomorphism from U to the set of v∈E(λ2)v\in E(\lambda_2) such that Yαλ1∗(Hα)+1.v=0Y_{\alpha}^{\lambda^*_1(H_{\alpha})+1}.v= 0 for all α∈B\alpha \in B. (Use Exerc. 15 of §7 to prove surjectivity.) c) With the notations of Prop. 2, prove that, for λ1,λ2,λ∈P++\lambda_1, \lambda_2, \lambda \in P_{++},

m(λ1,λ2,λ)=d+(λ,λ2−λ1∗,λ1∗)=d−(λ,λ1−λ2∗,λ1) m(\lambda_1, \lambda_2, \lambda ) =d^+(\lambda , \lambda_2-\lambda^*_1, \lambda^*_1) =d^-(\lambda , \lambda_1-\lambda^*_2, \lambda_1)
=d+(λ1,λ−λ2,λ2)=d−(λ1,λ2∗−λ∗,λ2∗) =d^+(\lambda_1, \lambda -\lambda_2, \lambda_2) =d^-(\lambda_1, \lambda^*_2-\lambda^*, \lambda^*_2)

(Observe that m(λ1,λ2,λ)m(\lambda_1, \lambda_2, \lambda ) is the dimension of the space of g\mathfrak{g}-invariant elements of

E(λ)∗⊗E(λ1)⊗E(λ2) E(\lambda )^*\otimes E(\lambda_1)\otimes E(\lambda_2)

and hence is equal to m(λ1∗,λ,λ2).)m(\lambda^*_1, \lambda , \lambda_2).)

d) Let λ1,λ2∈P++\lambda_1, \lambda_2\in P_{++}, and let λ\lambda be the unique element of P++∩W.(λ1−λ2∗)P_{++}\cap W.(\lambda_1-\lambda^*_2). We have

m(λ1,λ2,λ)=1 m(\lambda_1, \lambda_2, \lambda ) = 1

(Use cc).).) Deduce that E(λ1)⊗E(λ2)E(\lambda_1)\otimes E(\lambda_2) contains a unique submodule isomorphic to E(λ)E(\lambda ).

e) We retain the notations of d). Show the equivalence of the following conditions:

(i) λ=λ1+λ2\lambda =\lambda_1+\lambda_2

(ii) ∥λ∥=∥λ1+λ2∥\|\lambda \|=\|\lambda_1+\lambda_2\|

(iii) λ1\lambda_1 and λ2∗\lambda^*_2 are orthogonal

(iv) λ1\lambda_1 and λ2∗\lambda^*_2 are disjoint (Exerc. 13)

(v) λ1\lambda_1 and λ2\lambda_2 are disjoint.

Conclude that E(λ1)⊗E(λ2)E(\lambda_1)\otimes E(\lambda_2) is not a simple module unless λ1\lambda_1 and λ2\lambda_2 are disjoint (and hence another proof of Exerc. 18f18f) of §7).

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