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The sl2 singular weight has no cohomology

Example

Assume the Axiom of Choice (The Axiom of Choice). For G=SL2(C) the weight λ=−ρ=−ω1 has λ+ρ=0 on the Weyl wall, and L−ω1≅O(−1) on X≅P1. All cohomology of L−ω1 vanishes: H0(X,L−ω1)=H1(X,L−ω1)=0. More generally the wall-crossing isomorphism Hi(X,L−ω1)≅Hi+1(X,L−ω1) of Rank-one cohomology shifts across a simple wall together with the vanishing above degree 1 forces every cohomology group to vanish.

Facts & Assumptions

Given: The Axiom of Choice, G=SL2(C), its upper triangular Borel, the flag variety X=G/B≅P1, the fundamental weight ω1 and Weyl vector ρ=ω1, the simple root α with reflection s, and the bundle L−ω1.

[F1]

In rank one α=2ω1 and ρ=α/2=ω1. Thus for λ=−ω1 one has λ+ρ=0, s⋅λ=s(0)−ρ=−ω1, and ⟨λ,α∨⟩=−1. The shifted weight is therefore not regular (Fundamental weights for a chosen simple root system, The Weyl vector, Rank-one cohomology shifts across a simple wall).

[F2]

For G=SL2 the unique simple root α has Pα=G, so the fibre F=Pα/B is all of X=G/B, and under the fixed identification of the fibre with the two-affine projective line P1 the restriction Lλ∣F is isomorphic to OP1(⟨λ,α∨⟩); in particular L−ω1≅O(−1) (A minimal-parabolic flag projection is a projective-line bundle, Flag line-bundle degree on a minimal-parabolic fiber, Two-affine projective line and its twists).

[F3]

On P1 over C one has Hq(O(d))=0 unless q=0 or q=1; H0(O(d))=0 for d<0, and H1(O(d)) is nonzero exactly for d≤−2. In particular at d=−1 both H0 and H1 vanish (Cohomology of O(d) on projective space).

[F4]

The wall-crossing isomorphism: since the pairing is −1, Hi(X,L−ω1)≅Hi+1(X,Ls⋅(−ω1))=Hi+1(X,L−ω1) for all i≥0; and all cohomology vanishes by the singular-weight lemma (Rank-one cohomology shifts across a simple wall, Singular dot weights have zero line-bundle cohomology, The Borel-Weil-Bott theorem).

[F5]

X is smooth projective of dimension 1, so Hq(X,L−ω1)=0 for q>1 (Serre duality for locally free sheaves on a smooth projective variety).

Verification

technique · direct
1.1F2F3F5

By [F2] the bundle is O(−1), so [F3] gives H0(X,L−ω1)=H1(X,L−ω1)=0 and Hq=0 for q>1.

2.1F4F5step 1.1

The consistency argument is independent of the table: [F4] gives Hi≅Hi+1 for all i≥0, while Hq=0 for q>1 by [F5]; starting from H2=0 gives H1≅H2=0 and then H0≅H1=0.

3.1F1step 1.1step 2.1∎

Both computations agree: for λ=−ω1, the shifted weight λ+ρ=0 lies on the Weyl wall (the boundary value n=−1 of the rank-one shift), and all cohomology of Lλ vanishes, as asserted.

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