Bourbaki

Exercise 4, § 9

Tag 0168, permanent. Cite it as [Bourbaki, Tag 0168].

Lie Groups and Lie Algebras, VIII, 254

Bourbaki marks this one as harder than the rest.

Assume that g\mathfrak{g} is simple, and denote by γ\gamma the element of R+R_+ such that HγH_{\gamma} is the highest root of R∨R^{\vee}; write Hγ=∑α∈BnαHαH_{\gamma}=\sum_{\alpha\in B}n_{\alpha}H_{\alpha}. We have ⟨ρ,Hγ⟩=\langle \rho , H_{\gamma}\rangle = ∑nα=h−1\sum n_{\alpha}=h-1, where hh is the Coxeter number of R (Chap. VI, §1, no. 11, Prop. 31).

a) Let α∈B\alpha \in B. Show that, for all β∈R+\beta \in R_+,

⟨ϖα+ρ,Hβ⟩≤h+nα−1 \langle \varpi_{\alpha}+\rho , H_{\beta}\rangle \leq h+n_{\alpha}-1

and that equality holds if β=γ\beta =\gamma. Deduce that every prime factor of dim E(ϖα)E(\varpi_{\alpha}) is ≤h+nα−1\leq h+n_{\alpha}-1.

b) Assume that ϖα\varpi_{\alpha} is not minuscule, i.e. nα≥2n_{\alpha}\geq 2. Let m∈2,nαm\in 2, n_{\alpha} and p=h+m−1p=h+m-1. Verify (cf. Chap. VI, Plates) that there exists β∈R+\beta \in R_+ such that ⟨ϖα,Hβ⟩=nα\langle \varpi_{\alpha}, H_{\beta}\rangle =n_{\alpha} and ⟨ρ,Hβ⟩=h−1−(nα−m)\langle \rho , H_{\beta}\rangle =h-1-(n_{\alpha}-m), hence ⟨ϖα+ρ,Hβ⟩=p\langle \varpi_{\alpha}+\rho , H_{\beta}\rangle =p. Deduce that, if pp is prime, pp divides dim E(ϖα)E(\varpi_{\alpha}). (Remark that pp does not divide any of the ⟨ρ,Hβ⟩\langle \rho , H_{\beta}\rangle for β∈R+\beta \in R_+, cf. Exerc. 3.)

c) When g\mathfrak{g} is of type G2G_2 (resp. F4,E8F_4,E_8), we have h=6h= 6 (resp. 12,3012,30), and dim E(ϖα)E(\varpi_{\alpha}) is divisible by 7 (resp. 13,3113,31), When g\mathfrak{g} is of type E6E_6 (resp. E7E_7), and ϖα\varpi_{\alpha} is not minuscule, dim E(ϖα)E(\varpi_{\alpha}) is divisible by 13 (resp. 19).

Solution

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Cited by

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Cites

  • Chap. VI, §1, no. 11, Prop. 31 not in this corpus
  • Exerc. 3
Transcribed from the printed volume. Not translated.