Round coupled Penrose correction gives the R-over-six operator

For the global metric splitting, the mixed component of $\bar{\mathrm{\partial }}(s\sigma )$ is a vertical degree-one form with coefficient $\mathcal{O}(-1{)}^{\oplus 2}\otimes E_x$. On a fibre it has a unique vertical primitive by [F4]. The fixed-fibre $B={\bar{\mathrm{\partial }}}^{\ast }G$ construction and smooth-parameter difference-quotient argument in [F2], with the estimates [F5], make that primitive smooth over the base. The same fibrewise uniqueness makes it agree across frame and chart changes. Denote the horizontal correction by $\varphi (s)$; then $\psi (s)=s\sigma +\varphi (s)$ has zero mixed differential. If $s$ is the image of a closed representative $\alpha$, [F2]'s decomposition and closedness say that its horizontal remainder solves this same mixed equation. The difference is fibrewise holomorphic in $\mathcal{O}(-1{)}^{\oplus 2}\otimes E_x$, hence zero by [F4]. Thus $\psi (s)=\alpha -{\bar{\mathrm{\partial }}}_F\gamma$ is closed for a class image. The differential can still have a horizontal degree-two component for arbitrary $s$; the curvature component of the metric splitting is retained.

Compute that residual in the simpler constant-$t$ local splitting, without transplanting metric-frame coefficient formulas. Let $D_j=D_{z_j}$, $D_{\bar{j}}=D_{{\bar{z}}_j}$ be the covariant derivatives in a local unitary frame of $E$ and set $u=s\mathrm{/}a$. The holomorphic tautological frame $e$ is killed by the scalar $\bar{\mathrm{\partial }}$. Let ${\theta }^1,{\theta }^2$ be the $(0,1)$ covectors dual to $H_1,H_2$ and vanishing on ${\mathrm{\partial }}_{\bar{t}}$. These three vector fields commute, so scalar Dolbeault differentiation is evaluated by them without bracket terms. The form with the prescribed vertical part and zero mixed differential is ${\psi }_{\mathrm{l}\mathrm{o}\mathrm{c}}(s)=e(\frac{cu}{q^2}d\bar{t}+h_1{\theta }^1+h_2{\theta }^2),h_1=\frac{c}{q}(\bar{t}{D}_{\bar{1}}u+D_{2}u),h_2=\frac{c}{q}(\bar{t}{D}_{\bar{2}}u-D_{1}u)\mathrm{.}$ Indeed ${\mathrm{\partial }}_{\bar{t}}(\bar{t}\mathrm{/}q)=q^{-2}$ and ${\mathrm{\partial }}_{\bar{t}}(1\mathrm{/}q)=-t{q}^{-2}$, so ${\mathrm{\partial }}_{\bar{t}}{h}_i$ equals the corresponding $D_{H_i}(cu\mathrm{/}{q}^2)$ exactly. The factor $u=s\mathrm{/}(1+r^2)$ is essential: it is the integral-normalized harmonic coefficient from [F2], not an adjustable conformal ansatz.

These local forms are regular on the complete projective fibre. For $w=1\mathrm{/}t$, the horizontal vectors are ${\tilde{H}}_i=w{H}_i$, ${\tilde{\theta }}^i=w^{-1}{\theta }^i$, and $\tilde{e}=w^{2}e$. Thus the new horizontal coefficients are ${\tilde{h}}_i=t{h}_i$. They become $c(D_{\bar{1}}u+\bar{w}{D}_{2}u)\mathrm{/}(1+\mathrm{\mid }w{\mathrm{\mid }}^2)$ and $c(D_{\bar{2}}u-\bar{w}{D}_{1}u)\mathrm{/}(1+\mathrm{\mid }w{\mathrm{\mid }}^2)$, both smooth. The vertical coefficient changes to $-cud\bar{w}\mathrm{/}(1+\mathrm{\mid }w{\mathrm{\mid }}^2{)}^2$ in frame $\tilde{e}$, also smooth. The sign is the canonical-coordinate transition, not a change of harmonic normalization. Therefore ${\psi }_{\mathrm{l}\mathrm{o}\mathrm{c}}$ exists over every entire fibre above this base chart. Its difference from the intrinsic form of step 1.1 has zero vertical part and zero mixed differential. By [F4] that difference is zero. This proves that the coordinate calculation computes the actual global correction, despite using a different horizontal splitting for its displayed coefficients.

Put $S=D_{\bar{1}}{D}_1+D_{\bar{2}}{D}_2$ and $T=D_1{D}_{\bar{1}}+D_2{D}_{\bar{2}}$. Directly evaluating the remaining differential on $H_1,H_2$ gives $\frac{q}{c}{\bar{\mathrm{\partial }}}_F{\psi }_{\mathrm{l}\mathrm{o}\mathrm{c}}(H_1,H_2)\mathrm{/}e=\bar{t}{F}_{\bar{1}\bar{2}}u-t{F}_{12}u-Su-\mathrm{\mid }t{\mathrm{\mid }}^{2}Tu\mathrm{.}$ Here, for example, $D_{H_1}{h}_2=c{q}^{-1}(\bar{t}{D}_{\bar{1}}{D}_{\bar{2}}-D_{\bar{1}}{D}_1-t\bar{t}{D}_2{D}_{\bar{2}}+t{D}_2{D}_1)u$, and subtracting $D_{H_2}{h}_1$ gives precisely the displayed expression. By [F3], $T-S=F_{1\bar{1}}+F_{2\bar{2}}=0$ and the first two terms vanish. Since $D_j=(D_{2j-2}-i{D}_{2j-1})\mathrm{/}2$ and $D_{\bar{j}}=(D_{2j-2}+i{D}_{2j-1})\mathrm{/}2$, $S+T=\frac{1}{2}{\sum }_{\mu =0}^3{D}_{\mu }^2$ and hence $S=T=\frac{1}{4}{\sum }_{\mu }{D}_{\mu }^2$. The residual is therefore exactly $-c{\sum }_{\mu }{D}_{\mu }^2(s\mathrm{/}a)\mathrm{/}4$ in the nonvanishing local coefficient frame $e{\theta }^1\wedge {\theta }^2$. No uncomputed curvature cancellation or generic Penrose-operator theorem is used.

For $g={\mathrm{\Omega }}^2\delta$ in four dimensions, integration against its volume ${\mathrm{\Omega }}^{4}dx$ and the unitary connection gives ${\mathrm{\nabla }}_A^{\ast }{\mathrm{\nabla }}_{A}s=-{\mathrm{\Omega }}^{-4}{\sum }_{\mu }{D}_{\mu }({\mathrm{\Omega }}^2{D}_{\mu }s)=-{\mathrm{\Omega }}^{-2}{\sum }_{\mu }{D}_{\mu }^{2}s-2{\mathrm{\Omega }}^{-3}{\sum }_{\mu }({\mathrm{\partial }}_{\mu }\mathrm{\Omega })D_{\mu }s\mathrm{.}$ With $\mathrm{\Omega }=2\mathrm{/}a$, differentiation gives ${\mathrm{\partial }}_{\mu }\mathrm{\Omega }=-4x_{\mu }\mathrm{/}{a}^2$ and ${\sum }_{\mu }{\mathrm{\partial }}_{\mu }^2\mathrm{\Omega }=-16\mathrm{/}{a}^3=-2{\mathrm{\Omega }}^3$. The product rule now proves, for this very connection, ${\mathrm{\Omega }}^{-3}[-{\sum }_{\mu }{D}_{\mu }^2(\mathrm{\Omega }s)]=({\mathrm{\nabla }}_A^{\ast }{\mathrm{\nabla }}_A+2)s\mathrm{.}$ Since $s\mathrm{/}a=\mathrm{\Omega }s\mathrm{/}2$, step 4.1 is ${\bar{\mathrm{\partial }}}_F\psi (s)=\tau ({\mathrm{\nabla }}_A^{\ast }{\mathrm{\nabla }}_A+2)s,\tau =\frac{c{\mathrm{\Omega }}^3}{8}e{\theta }^1\wedge {\theta }^2\mathrm{.}$ The factor $\tau$ is nonzero on the local chart. The zeroth-order coefficient is derived by $\mathrm{\Delta }\mathrm{\Omega }=-2{\mathrm{\Omega }}^3$; it is not inferred from uniqueness. In particular a constant scalar with flat connection gives residual $2c\mathrm{/}{a}^3$, agreeing with the displayed $+2$. By [F6] this is $R\mathrm{/}6$.

The local expression equals the intrinsically constructed differential by step 3.1. The same construction and calculation apply in any stereographic chart with the unit round metric. Thus it patches smoothly across the base overlap and infinity; a chartwise solution satisfies the invariant operator equation globally. For the class image $s$, the residual vanishes by step 1.1, and the nonzero frame in step 5.1 gives $({\mathrm{\nabla }}_A^{\ast }{\mathrm{\nabla }}_A+2)s=0$. Gauge changes preserve the fibre normalization and transport the unique mixed correction, while $D_{\mu }$ and the final operator conjugate by the same unitary frame change. On the dual, $F_{A^{\vee }}=-F_A^{\mathsf{T}}$ remains ASD; the identical construction applies. More generally a connection-preserving smooth bundle map commutes with the componentwise vertical projection and inverse, and transports the unique mixed correction; thus the edge and residual construction are natural for such maps, including direct sums. Nothing used irreducibility, absence of parallel sections or nonzero curvature.