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Coupled Penrose vanishing on the round four-sphere

Statement

Assume full AC. Let S4 have the unit round metric and the positive stereographic orientation for z1=x0+ix1, z2=x2+ix3. Let E→S4 be a smooth finite-rank complex vector bundle with a positive Hermitian metric h and a smooth connection ∇A preserving h. Its curvature is FA=∇A2∈Ω2(S4,End⁡CE). For real two-forms η,ζ, let ∗g be characterized by η∧∗gζ=⟨η,ζ⟩g dvolg, and extend it complex-linearly on the form factor of FA. Set FA+=(FA+∗gFA)/2 and assume FA+=0, the anti-self-duality condition in this orientation. On Z=CP3 use homogeneous coordinates [λ1:λ2:μ1:μ2] and the incidence equations μ1=z1λ1−zˉ2λ2, μ2=z2λ1+zˉ1λ2, with infinity fibre λ1=λ2=0; these fibres define the incidence projection π:Z→S4. On λ1≠0, put t=λ2/λ1, H1=∂zˉ1−t∂z2, and H2=∂zˉ2+t∂z1. The projected complex alpha-plane at [λ]∈π−1(x) is span⁡C{dπ(H1),dπ(H2)}; the second projective chart covers t=∞. Let OZ(−1) be the tautological holomorphic line bundle and OZ(−2)=OZ(−1)⊗2. Suppose F is a finite-rank holomorphic bundle (locally trivial with holomorphic transition maps), smoothly identified with π∗E, and its Dolbeault operator is the coefficientwise ordinary vertical operator and the connection ∇A on the projected complex alpha-planes. For a finite-rank holomorphic bundle G→Z, write H1(Z,G) for analytic sheaf cohomology of its holomorphic-section sheaf; under the analytic Dolbeault comparison it is canonically ker⁡(∂ˉG:Ω0,1(Z,G)→Ω0,2(Z,G))/im⁡(∂ˉG:Ω0,0(Z,G)→Ω0,1(Z,G)), where Ω0,q(Z,G) denotes the global smooth G-valued (0,q)-forms. Write F∨ for the fibrewise complex dual. Then

H1(Z,F⊗OZ(−2))=H1(Z,F∨⊗OZ(−2))=0.

This applies to arbitrary reducible and flat summands. It assumes no algebraic-to-analytic GAGA, monad, or inverse ADHM normal form as an independent premise.

Facts & Assumptions

Given: Full AC; the unit round sphere with the specified positive orientation and incidence; a smooth finite-rank Hermitian bundle with unitary ASD connection; and the holomorphic pullback bundle with the stated vertical and horizontal Dolbeault operator.

[A1]

The actual batch-67 carrier states that under full AC, for each r∈{0,1,2,3} and every finite-rank holomorphic vector bundle G on Pr(C), the canonical comparison Hq(X,O(G))≅H∂ˉq(X,G) holds for every q≥0; for the trivial line it also states H0(X,O)=C and Hq(X,O)=0 for q>0. This matches the comparison used here. This is the local analytic Dolbeault-resolution comparison, whose hypotheses match the present holomorphic-bundle use; it asserts no algebraic-to-analytic GAGA. Projective Dolbeault comparison and scalar structure-sheaf cohomology

[F1]

The incidence fixes the positive orientation, the unit round metric and the alpha-plane distribution; its horizontal curvature equation detects exactly FA+=0. Positive incidence geometry for the coupled Penrose transform, Incidence alpha planes detect positive-orientation anti-self-duality

[F2]

Fibrewise harmonic projection defines a well-defined injective map from the analytic cohomology group to Γ(S4,E), conditional on the comparison in [A1]. Injective fibre-harmonic edge for coupled Penrose cohomology

[F3]

The image of that map solves (∇A∗∇A+2)s=0, and the same local calculation applies to the dual unitary ASD connection. Round coupled Penrose correction gives the R-over-six operator

[F4]

The scalar curvature in the fixed unit-radius convention is R=12, so 2=R/6. Scalar curvature of the unit round four-sphere

[F5]

Full AC is inherited through the analytic and harmonic comparison inputs; no stronger choice principle is used in the integration-by-parts argument. The Axiom of Choice

Proof

Given: The theorem hypotheses and the conditional analytic comparison [A1].

1.1givenA1F2F3F5algebra

Let ξ∈H1(Z,F⊗OZ(−2)) and set s=PA(ξ). By [F2] the map is injective, and [F3] gives (∇A∗∇A+2)s=0. Since S4 is compact without boundary and ∇A∗ is the formal adjoint, integration gives 0=∫S4⟨(∇A∗∇A+2)s,s⟩ dvolg=∫S4(∣∇As∣2+2∣s∣2) dvolg. Both integrands are nonnegative, so ∫S4∣s∣2=0.

2.1F2step 1.1algebra

The smooth function ∣s∣2 is nonnegative. If it were positive at a point, continuity would make it bounded below by a positive number on a nonempty open set, which has positive Riemannian volume and contradicts the zero integral in step 1.1. Thus s=0 everywhere, and injectivity in [F2] gives ξ=0. Hence H1(Z,F⊗OZ(−2))=0.

3.1F1F2F3F4F5step 1.1step 2.1algebra

The dual connection A∨ is unitary and has curvature FA∨=−FAT; transposition acts only on bundle coefficients and commutes with the Hodge star, so A∨ is ASD in the same orientation. The holomorphic dual F∨ has the induced vertical and horizontal Dolbeault operator. Apply [F2] and [F3] to this dual data and repeat the integration in step 1.1: every term is nonnegative and the scalar term is again 2∣s∨∣2. The continuity argument in step 2.1 gives s∨=0, and injectivity gives H1(Z,F∨⊗OZ(−2))=0. This uses no irreducibility or nonflatness condition.

4.1A1F1F2F3step 1.1step 2.1step 3.1∎

The analytic comparison in [A1] is supplied by its local Dolbeault-resolution theorem for the holomorphic bundles on CP3 used above. The fibre groups, global edge and operator coefficient have been obtained without any algebraic GAGA, monad or inverse-normal-form premise. All conclusions therefore retain the stated arbitrary finite-rank unitary ASD scope, with the analytic inputs and full AC hypothesis explicit.

Remarks

Rawnsley’s theorem on printed p. 49, proved on p. 50, identifies the coupled cohomology with the kernel of D∗D+R/6; the detailed construction occupies §§2–3, pp. 46–56. His notation calls the sphere bundle F and its twistor pullback E, and uses self-duality for his standard orientation. Here the incidence and curvature suppliers specify the orientation in which the connection is anti-self-dual. Rawnsley’s introduction describes positive-charge instanton bundles; the dual, reducible and flat-summand scope asserted here is supplied by the local edge and operator lemmas and the energy argument above. The full cited article is a scan: its article pages have no searchable text layer, so text extraction alone does not display these passages.

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