Local holomorphic frames for an integrable smooth Dolbeault operator

First solve one equation with all other coordinates regarded as parameters $p$. Write $z_j=\rho w$ on a disk slightly larger than $\mathrm{\mid }w\mathrm{\mid }\le 2$, choose a fixed smooth cutoff $\chi$ supported in $\mathrm{\mid }w\mathrm{\mid }<2$ and equal to one for $\mathrm{\mid }w\mathrm{\mid }\le 1$, and put $a_{\rho }(w,p)=\rho A_j(\rho w,p)$. For fixed $0<\alpha <1$, its $C_w^{\alpha }$ norm is bounded uniformly on a compact parameter polydisc by $C\rho (M_0+\rho M_1)$, where $M_0,M_1$ bound $A_j$ and its first derivatives. Multiplication is bounded in the matrix Hölder norm by the product rule for increments, and [F1] bounds $K(p)u=T(\chi a_{\rho }u)$ on $C^{\alpha }({\bar{D}}_2)$. Choose one positive $\rho$ making $\mathrm{\parallel }K(p)\mathrm{\parallel }<1\mathrm{/}4$ for every parameter. By [F3], $g=1-K(p)g$ has a unique fixed point; $\mathrm{\parallel }g-1\mathrm{\parallel }\le \mathrm{\parallel }K\mathrm{\parallel }\mathrm{/}(1-\mathrm{\parallel }K\mathrm{\parallel })<1\mathrm{/}3$, so $g$ is invertible everywhere. On $\mathrm{\mid }w\mathrm{\mid }<1$, [F1] gives ${\mathrm{\partial }}_{\bar{w}}g=-a_{\rho }g$.

The coefficients depend smoothly on parameters as maps into $C_w^{\alpha }$, since all their parameter derivatives have uniform bounds on compact subsets. The inverse $(1+K(p){)}^{-1}$ is the uniformly convergent Neumann series; the identity $L(p+h)-L(p)=-L(p+h)(K(p+h)-K(p))L(p)$ for $L=(1+K{)}^{-1}$ gives its first derivative in operator norm, and induction gives derivatives of every order. Hence $g=L(p)1$ is smoothly parameter-dependent in $C_w^{\alpha }$. For each such derivative, the fixed-point equation and [F1] gain one $w$ derivative; induction gives all mixed $C_w^{m,\alpha }$ derivatives on the same radius, because only the inverse in the initial $C^{\alpha }$ space needs smallness. Thus $g$ is jointly smooth. If $A_j$ is holomorphic in an untouched coordinate, $K$ has zero antiholomorphic parameter derivative by [F2], and differentiating the inverse identity gives the same for $g$.

Gauge by this $g$ on the smaller disk. The transformed coefficient is $A_i^{\mathrm{\prime }}=g^{-1}({\mathrm{\partial }}_{{\bar{z}}_i}g+A_{i}g)$, and $A_j^{\mathrm{\prime }}=0$. The square-zero condition in a smooth frame is ${\mathrm{\partial }}_{{\bar{z}}_i}{A}_j-{\mathrm{\partial }}_{{\bar{z}}_j}{A}_i+[A_i,A_j]=0$, and it is preserved by the gauge transformation by direct conjugation of $(D^{\mathrm{\prime }\mathrm{\prime }}{)}^2$. Therefore every remaining $A_i^{\mathrm{\prime }}$ is holomorphic in $z_j$. Kill the next coordinate by steps 1.1–2.1 with a cutoff only in that coordinate. Their holomorphic-parameter assertion preserves $A_j^{\mathrm{\prime }}=0$. Continue through the finitely many coordinates, shrinking the respective coordinate disks as needed. The product of the resulting smooth invertible gauges has all transformed coefficients zero, proving the required local frame.

If $G$ and $H$ are two such frames, $D^{\mathrm{\prime }\mathrm{\prime }}G=D^{\mathrm{\prime }\mathrm{\prime }}H=0$ gives $\bar{\mathrm{\partial }}(G^{-1}H)=0$. For smooth scalar functions this implies holomorphy: the one-variable Cauchy formula applies with the other variables fixed, and its iterated integral on a product of small circles gives a locally convergent joint power series. Thus the transition matrices are holomorphic, giving [F4]'s bundle structure; any other structure with the same $D^{\mathrm{\prime }\mathrm{\prime }}$ has the same class of holomorphic frames, hence agrees. In these frames an intertwining smooth bundle map has matrix $M$ with $\bar{\mathrm{\partial }}M=0$, so is holomorphic by the same calculation, and the converse follows by the Leibniz rule. Rank zero is immediate and the finite gauge induction covers every positive finite rank.