Conditional Penrose vanishing from positive scalar curvature

Let $\xi \in H^1(Z,\mathcal{E}(-2))$. By the assumed injective map ${\mathcal{P}}_{\mathcal{E}}$, its image is a smooth section solving the coupled equation in the Statement. Since $R\mathrm{/}6>0$, [F2] makes that image zero, and injectivity gives $\xi =0$. Thus $H^1(Z,\mathcal{E}(-2))=0$.

For the additional application assume full AC, and construct the forward holomorphic bundle directly, without invoking [F1]'s full correspondence. On the smooth pullback ${\pi }^{\ast }E$, take the ordinary vertical Dolbeault derivative and the connection derivative along the projected complex alpha-planes in [A1]. In the incidence chart $t={\lambda }_2\mathrm{/}{\lambda }_1$, these directions are ${\mathrm{\partial }}_{\bar{t}}$, $H_1={\mathrm{\partial }}_{{\bar{z}}_1}-t{\mathrm{\partial }}_{z_2}$ and $H_2={\mathrm{\partial }}_{{\bar{z}}_2}+t{\mathrm{\partial }}_{z_1}$. They annihilate the holomorphic coordinates $t$, ${\mu }_1=z_1-{\bar{z}}_{2}t$, ${\mu }_2=z_2+{\bar{z}}_{1}t$, and commute. The mixed vertical/horizontal squares vanish because the horizontal coefficients are holomorphic in $t$ and $A$ is pulled back from the base. The horizontal square is $F_A(H_1,H_2)=F_{\bar{1}\bar{2}}-t(F_{1\bar{1}}+F_{2\bar{2}})+t^2{F}_{12}=0$: in the specified positive orientation the three coefficients are the self-dual curvature components, so ASD makes them zero. The same calculation in the second projective chart covers $t=\mathrm{\infty }$; using the other stereographic base chart covers the infinity fibre. The vertical/alpha-plane prescription is intrinsic to the stated incidence and connection, so its chart expressions agree. Thus it is a global smooth square-zero Dolbeault operator. By [F3] it defines a holomorphic bundle $F$ on $Z$, smoothly equal to ${\pi }^{\ast }E$, with exactly this vertical/horizontal operator. This is the forward Ward bundle by its defining operator, without any inverse Ward or algebraization claim.

Apply the same argument to ${\xi }^{\vee }\in H^1(Z,{\mathcal{E}}^{\vee }(-2))$ and the assumed injective map ${\mathcal{P}}_{{\mathcal{E}}^{\vee }}$; [F2] gives a zero image and injectivity gives ${\xi }^{\vee }=0$. Hence $H^1(Z,{\mathcal{E}}^{\vee }(-2))=0$, proving both asserted vanishings conditional on the specified transforms.

The sphere, orientation, smooth pullback and holomorphic operator of step 1.2 are precisely those required by [A1]. That analytic coupled-vanishing theorem therefore gives both $H^1(Z,F(-2))=0$ and $H^1(Z,F^{\vee }(-2))=0$, without an extra injective-map premise or irreducibility assumption. The dual connection is unitary and ASD since its curvature is $-F_A^t$ and Hodge star commutes with transpose; its induced holomorphic operator is the dual vertical/horizontal operator. Hence the dual clause matches as well. This additional conclusion assumes full AC but no projective GAGA. Steps 1.1–2.1 prove the original conditional Statement independently of this construction and without full AC.