Bourbaki

Theorem 2

Tag 015Y, permanent. Cite it as [Bourbaki, Tag 015Y].

Lie Groups and Lie Algebras, chapter VIII, § 9. The formula of Hermann Weyl

Let E be a finite dimensional simple g\mathfrak{g}-module, λ\lambda its highest weight and (⋅∣⋅)(\cdot | \cdot ) a W-invariant non-degenerate positive symmetric bilinear form on h∗\mathfrak{h}^*. Then:

∏⟨λ+ρ,Hα⟩∏((λ∣α)) \prod\langle \lambda +\rho , H_{\alpha}\rangle \prod((\lambda |\alpha ))

dim E = = 1 +.

α∈R+⟨ρ,Hα⟩α∈R+(ρ∣α)‾ _{\alpha\in R_+}\langle \rho , H_{\alpha}\rangle_{\alpha\in R_+}\overline{(\rho|\alpha)}

Let T be an indeterminate. For all ν∈P\nu \in P, denote by fνf_{\nu} the homomorphism from Z[P]\mathbf{Z}[P] to R[[T]]\mathbf{R}[[T]] that takes eμe^\mu to e(ν∣μ)Te^{(\nu|\mu)T} for all μ∈P\mu\in P. Then dim E is the constant term of the series fνf_{\nu}(ch E).

For all μ,ν∈P\mu, \nu \in P, we have

fν(J(eμ))=∑w∈Wε(w)e(ν∣wμ)T f_{\nu}(J(e^\mu)) =\sum_{w\in W}\varepsilon (w)e^{(\nu|w\mu)T}
=∑w∈Wε(w)e(w−1ν∣μ)T=fμ(J(eν)) =\sum_{w\in W}\varepsilon (w)e^{(w^{-1}\nu|\mu)T}=f_\mu(J(e^{\nu}))

In particular, in view of Chap. VI, §3, no. 3, formula (3),

fρ(J(eμ))=fμ(J(eρ))=e(μ∣ρ)T∏α∈R+(1−e−(μ∣α)T) f_{\rho}(J(e^\mu)) =f_\mu(J(e^{\rho})) =e^{(\mu|\rho)T}\prod_{\alpha\in R_+}(1-e^{-(\mu|\alpha)T})

Hence, setting Card(R+_+) =N= N,

fρ(J(eμ))≡TN∏α∈R(μ∣α)f_{\rho}(J(e^\mu))\equiv T^N\prod_{\alpha\in R}(\mu|\alpha ) (mod TN+1R[[T]]T^{N+1}\mathbf{R}[[T]]).

The equality J(eλ+ρ)=J(e^{\lambda+\rho}) = ch(E).J(eρ).J(e^{\rho}) (Th. 1) thus implies that

TN∏α∈R+(λ+ρ∣α)≡fρT^N\prod_{\alpha\in R_+}(\lambda +\rho |\alpha )\equiv f_{\rho}(ch EE).TN∏α∈R+(ρ∣α).T^N\prod_{\alpha\in R_+}(\rho |\alpha ) (mod TN+1R[[T]]T^{N+1}\mathbf{R}[[T]])

so

dim E=((∏α∈R+(λ+ρ∣α))/((∏α∈R+(ρ∣α))=∏α∈R+(1+(λ∣α)(ρ∣α))E =((\prod_{\alpha\in R_+}(\lambda +\rho |\alpha ))/((\prod_{\alpha\in R_+}(\rho |\alpha ))=\prod_{\alpha\in R_+}(1 +\frac{(\lambda|\alpha)}{(\rho|\alpha)}).

Now, if α∈R+,α\alpha \in R_+,\alpha can be identified with an element of hR\mathfrak{h}_{\mathbf{R}} proportional to HαH_{\alpha}, so

(λ+ρ∣α)/(ρ∣α)=⟨λ+ρ,Hα⟩/⟨ρ,Hα⟩ (\lambda +\rho |\alpha )/(\rho |\alpha ) =\langle \lambda +\rho , H_{\alpha}\rangle /\langle \rho , H_{\alpha}\rangle

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.

Cited by

Cites

Transcribed from the printed volume. Not translated.