Flat summands in coupled Penrose vanishing

The trivial connection $d$ preserves the standard Hermitian metric and has zero curvature. Therefore $A=A_0\oplus d$ preserves the direct-sum metric and has curvature $F_A=F_{A_0}\oplus 0$, so $F_A^{+}=0$. The smooth isometric identifications give $F=F_0\oplus {\mathcal{O}}_Z^{\oplus m}\cong {\pi }^{\ast }E$; its vertical and horizontal Dolbeault components are respectively the coefficientwise vertical operator and ${\mathrm{\nabla }}_{A_0}\oplus d={\mathrm{\nabla }}_A$ on the projected alpha-planes. Thus the entire direct-sum data meet [F1]'s hypotheses, including full AC. Applying [F1] to this $F$ gives both asserted vanishings for $F$ and $F^{\vee }=F_0^{\vee }\oplus {\mathcal{O}}_Z^{\oplus m}$. When $m=0$ this is precisely the theorem for $(E_0,A_0,F_0)$, with no assumption that $E_0$ has rank zero.

The global Dolbeault differential acts componentwise on a finite direct sum, so its closed forms and exact forms split, giving the corresponding direct-sum decomposition of the cohomology quotient in [A1]. On each fibre the Dolbeault operator and its adjoint preserve the orthogonal summands, hence their Laplacian, its harmonic kernel and its orthogonal projection split. The normalized scalar harmonic generator in [F2] is the same for every bundle component; therefore the coefficient extraction gives ${\mathcal{P}}_A={\mathcal{P}}_{A_0}\oplus {\mathcal{P}}_d$. The unique vertical primitive and mixed correction in [F2] also split because their equations split and each has a unique solution. Applying the same argument to $F^{\vee }=F_0^{\vee }\oplus {\mathcal{O}}_Z^{\oplus m}$ gives ${\mathcal{P}}_{A^{\vee }}={\mathcal{P}}_{A_0^{\vee }}\oplus {\mathcal{P}}_{d^{\vee }}$. In a standard constant frame of the trivial summand each operation acts on its $m$ scalar components separately, including the empty summand when $m=0$. Together with step 1.1 this verifies all the claims.