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A point-supported SL2 module gives the shifted Verma module

Example

Use G=SL2, X=P1, and κ=cρ with c∈Z. Let x0 be the Borel-fixed point z=0 in the U0 chart of The SL2 flag charts realize the stated rho shift, and let δx0 be the left Dcρ-module supported at x0. Then Γ(X,δx0)≅M((c−1)ρ)=U(sl2)⊗U(b)C(c−1)ρ. For the Casimir C=ef+fe+12h2, both this Verma module and the central character χcρ have scalar 12(c2−1).

Facts & Assumptions

Given: The SL2 basis and chart formulas, integral c, the delta module at z=0, and the Harish–Chandra shift of Flag-variety twisted differential operators and their modules.

[F1]

On U0, the infinitesimal generators are e=z2∂z+(c+1)z, f=−∂z, h=2z∂z+(c+1), and Dcρ is Diff⁡(O(−c−1)) (The SL2 flag charts realize the stated rho shift).

[F2]

The BGG Verma module of highest weight λ is M(λ)=U(g)⊗U(b)Cλ (Verma modules).

[F3]

For a cyclic highest-weight vector v of weight λ, a central element acts by pr⁡(z)(λ) (The Harish-Chandra projection computes the highest-weight scalar).

[F4]

The shifted central character is χκ(z)=pr⁡(z)(κ−ρ) (Flag-variety twisted differential operators and their modules, The Harish-Chandra projection).

Verification

1.1F1givenalgebra

On the affine chart U0, the module supported at z=0 is the Weyl algebra quotient C[z]⟨∂z⟩/C[z]⟨∂z⟩z. Ordered monomials ∂znzm show that its global sections have basis δ,∂zδ,∂z2δ,…; sections vanish off the support point. The relation zδ=0 gives z∂zδ=−δ. Therefore [F1] gives eδ=0,hδ=(c+1−2)δ=(c−1)δ,fδ=−∂zδ.

2.1F2step 1.1

The universal property in [F2] gives a surjection from the Verma module of highest weight (c−1)ρ to Γ(X,δx0), sending its highest vector to δ. Its PBW basis is fnv for n≥0, and fnδ=(−1)n∂znδ is the basis in step 1.1. The map is therefore an isomorphism.

3.1F3F4step 1.1algebra∎

Since eδ=0 and hδ=(c−1)δ, the Casimir acts by efδ+feδ+12h2δ=((c−1)+12(c−1)2)δ=12(c2−1)δ. In the chosen PBW order, ef=fe+h and the Harish–Chandra projection of C is h+12h2. Evaluating it at κ−ρ=(c−1)ρ gives the same scalar by [F4]. Thus the module has the exact central character stated.

Depends on

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