Injective fibre-harmonic edge for coupled Penrose cohomology

Write $V^{0,1}$ for the vertical tangent and use [F1]'s global orthogonal complement. Filter the smooth Dolbeault complex of $F(-2)$ by the number of horizontal covectors. Vertical restriction is the ordinary fibre Dolbeault operator. The differential has components of horizontal/vertical bidegrees $(0,1)$, $(1,0)$ and potentially $(2,-1)$; the last is the curvature of the splitting, so a bicomplex with only two differentials is not presumed. On a form annihilating vertical tangent, the mixed component is the intrinsic vertical Bott/Dolbeault operator on the horizontal coefficient bundle: it is obtained by evaluating $\bar{\mathrm{\partial }}$ on one vertical and one horizontal vector. This description is independent of the lift of the horizontal vector. In the constant-$t$ local frame the vectors ${\mathrm{\partial }}_{\bar{t}},H_1,H_2$ commute; hence that coefficient operator is ordinary ${\mathrm{\partial }}_{\bar{t}}$.

Let $Q_x\subset {\mathbb{C}}^4$ be the quaternionic two-plane over $x$. Its Hermitian quaternionic structure gives a canonical complex alternating volume form ${\delta }_x$: every frame $(v,jv)$ of unit norm has determinant one for this form, and the changes are in $SU(2)$. On $L_x=\mathbb{P}(Q_x)$, the identity $K_{L_x}=\det (Q_x{)}^{\ast }\otimes \mathcal{O}(-2){\mathrm{\mid }}_{L_x}$ therefore gives an intrinsic isomorphism ${\iota }_x:\mathcal{O}(-2){\mathrm{\mid }}_{L_x}\to K_{L_x}$. In the finite incidence chart put $a=1+r^2$, $q=1+\mathrm{\mid }t{\mathrm{\mid }}^2$, $v_1=(1,0,z_1,z_2)$, $v_2=(0,1,-{\bar{z}}_2,{\bar{z}}_1)$ and $e=(v_1+t{v}_2{)}^{\otimes 2}$. The two $v_i$ are orthogonal with squared norm $a$, so ${\delta }_x(v_1,v_2)=a$ and ${\iota }_x(e)=adt$. The tautological metric is $\mathrm{\parallel }e{\mathrm{\parallel }}^2=a^2{q}^2$. The harmonic vertical form is ${\sigma }_x=\frac{i}{2\pi }\frac{ed\bar{t}}{a{q}^2}\mathrm{.}$ Indeed its weighted coefficient $\mathrm{\parallel }e{\mathrm{\parallel }}^2\mathrm{/}(a{q}^2)=a$ is holomorphic in $t$, so the local formal-adjoint equation is zero; [F2]–[F3] make the harmonic line one-dimensional. Moreover ${\int }_{L_x}{\iota }_x({\sigma }_x)=\frac{i}{2\pi }{\int }_{\mathbb{C}}dt\wedge d\bar{t}\mathrm{/}{q}^2=1$. This fixes both scale and phase. The intrinsic metric/volume construction proves agreement on projective and base-chart overlaps, including infinity, and smooth dependence on $x$. Extend this vertical form by zero on the global orthogonal horizontal complement to obtain $\sigma$. This extension need not be Dolbeault closed.

Smoothness of the needed primitive does not require a theorem about varying complex structures. Choose smooth unit quaternionic frames of $Q_x$ on a base neighborhood and a smooth unitary frame of $E$. These identify the vertical problem exactly with the fixed $SU(2)$-invariant Hermitian $\mathcal{O}(-2)$ problem on $\mathbb{C}{P}^1$, componentwise in $E_x$. Set $B={\bar{\mathrm{\partial }}}^{\ast }G$ on vertical degree one. By [F3], $\bar{\mathrm{\partial }}B=I-\mathrm{\Pi }$, and by [F2] its degree-zero kernel is zero. The estimates [F4] and the first-order nature of ${\bar{\mathrm{\partial }}}^{\ast }$ give $B:H^m\to H^{m+1}$ bounded for each $m$. For jointly smooth $f(x,t)$, base difference quotients converge in every fibre $H^m$ uniformly on compact base sets, by the Taylor formula and compactness of the fibre. Applying the fixed bounded $B$ shows ${\mathrm{\partial }}_x^{\beta }(Bf)=B({\mathrm{\partial }}_x^{\beta }f)$ at every order in every $H^{m+1}$. Sobolev embedding on the fixed two-dimensional compact fibre, with arbitrarily large $m$, gives joint smoothness in $(x,t)$. For the horizontal coefficient problem, choose also a smooth unit quaternionic frame of $Q_x^{\perp }$. Together the two quaternionic frames give a smooth $Sp(2)$ transformation from a fixed pair of planes to $(Q_x,Q_x^{\perp })$. Its complex-linear action on $\mathbb{C}{P}^3$ is holomorphic and preserves the Fubini–Study and tautological metrics. It therefore transports the horizontal quotient, its vertical Bott operator and its induced metric to the fixed $\mathcal{O}(-1{)}^{\oplus 2}$ problem identified below. That inverse is likewise one fixed bounded operator componentwise, so the same derivative argument applies. Thus no moving-cokernel inverse or unproved parametric Green assertion is used.

By [A1], a sheaf class has a global smooth closed representative $\alpha$. Its vertical restriction has harmonic part $s(x){\sigma }_x$, where $s(x)={\int }_{L_x}{\iota }_x({\alpha }_V)$ componentwise in $E_x$. This is smooth by integration over compact fibres. Solve ${\gamma }_x=B({\alpha }_V-s{\sigma }_x)$; then ${\bar{\mathrm{\partial }}}_V{\gamma }_x={\alpha }_V-s{\sigma }_x$. Step 2.1 makes $\gamma$ smooth. A change of unit quaternionic frame acts by an $SU(2)$ fibre isometry and a change of unitary $E$ frame acts by a matrix constant on each fibre; both preserve the equation. Uniqueness from $H^{0,0}(\mathcal{O}(-2))=0$ makes these local solutions agree. Hence $\gamma$ is global and ${\alpha }_H:=\alpha -s\sigma -{\bar{\mathrm{\partial }}}_F\gamma$ annihilates vertical tangent, proving the asserted decomposition. The harmonic/exact orthogonality and the same zero kernel prove its uniqueness.

Replacing $\alpha$ by $\alpha +{\bar{\mathrm{\partial }}}_F\beta$ changes its vertical restriction by ${\bar{\mathrm{\partial }}}_V\beta$; its harmonic projection and integral against the fibre trace are zero. Thus $s$ depends only on the cohomology class, and the construction is linear. It is a section of $E$, not of an unspecified conformal-weight line, because the metric quaternionic volume in step 1.2 supplies the precise canonical identification of the harmonic line with $\mathbb{C}$.

In the second projective chart $u=1\mathrm{/}t$, the constant-coordinate horizontal vectors satisfy ${\tilde{H}}_i=u{H}_i$. The intrinsic horizontal cotangent quotient therefore restricts to $\mathcal{O}(1{)}^{\oplus 2}$; after the $\mathcal{O}(-2)$ twist its coefficient bundle is $\mathcal{O}(-1{)}^{\oplus 2}\otimes E_x$. This transition determines the vertical Bott holomorphic structure from step 1.1, rather than assuming a normal-bundle or generic Penrose theorem. If $s=0$, closedness gives ${\bar{\mathrm{\partial }}}_F{\alpha }_H=0$, since ${\bar{\mathrm{\partial }}}_F^2=0$. Its mixed component says that ${\alpha }_H{\mathrm{\mid }}_{L_x}$ is holomorphic for precisely this coefficient bundle. By [F2] it is zero on each fibre. Thus $\alpha ={\bar{\mathrm{\partial }}}_F\gamma$ globally, proving injectivity, including zero rank.