Dolbeault-Poincaré lemma

On a sufficiently small polydisc in a complex manifold, every $\bar{\partial }$-closed $(p,q)$-form with $q>0$ is locally $\bar{\partial }$-exact. A coefficientwise one-variable integral operator and induction over the coordinates prove the result.

For a $\bar{\partial }$-closed positive-antiholomorphic-degree differential form of type (p, q) on a polydisc, write $\phi =d{\bar{z}}_j\wedge \alpha +\beta$ with no $d{\bar{z}}_j$ in $\alpha ,\beta$. Apply the Cauchy-Green operator in coordinate $j$ to the coefficients of $\alpha$. Subtracting its Dolbeault operator derivative removes every $d{\bar{z}}_j$ term. Closedness then makes the remaining coefficients holomorphic in $z_j$. Eliminating successive coordinates preserves holomorphicity in earlier coordinates and produces a local primitive after finitely many steps.

On a polydisc, a $\partial$-closed differential form of type (p, q) with $p>0$ is $\partial$-exact. Apply the Dolbeault-Poincaré lemma to its complex conjugate, whose antiholomorphic degree is $p$.

For a smooth function $g$ compactly supported in a disc $D\subset \mathbb{C}$, the Cauchy-Green operator isHere $w=x+iy$ and $dw\wedge d\bar{w}=-2idx\wedge dy$, fixing the sign between the two forms of the integral. The kernel is locally integrable, andIndeed, the distribution $1/(\pi z)$ is a fundamental solution of ${\partial }_{\bar{z}}$, so differentiating its convolution with $g$ gives ${\delta }_0\ast g=g$. Smooth compactly supported input makes the convolution smooth and the identity classical.^[ref]

For local data without compact support, multiply by a smooth cutoff function equal to one on a smaller disc. The resulting operator solves the original equation there. Applying this construction to coefficients in one coordinate at a time gives the local primitives in the Dolbeault-Poincaré lemma. The solution need not itself have compact support.