Dolbeault–Grothendieck lemma

Remarks

The key case is $n = 1$, $q = 1$: for a smooth function $g$ with compact support in $\C$, the function $$u(z) = \frac{1}{2\pi i} \int_\C \frac{g(w)}{w - z}\, dw \wedge d\bar w$$ is smooth and satisfies $\partial u / \partial \bar z = g$. In general one first proves the result on a slightly smaller polydisc, by induction on the largest index $k$ such that some $d\bar z_k$ occurs in $\alpha$. The coefficients involving $d\bar z_k$ are holomorphic in $z_{k+1}, \dots, z_n$, and solving the one-variable equation in $z_k$ with a cutoff removes $d\bar z_k$ without destroying this holomorphy. An exhaustion of $\Delta$ by smaller polydiscs, together with Runge approximation to correct the successive solutions, gives the result on $\Delta$ itself. Example: on $\C$, the $(0,1)$-form $\bar z \, d\bar z$ is $\bar\partial$ of $\bar z^2/2$. The condition $q \ge 1$ is necessary to make the statement meaningful; for $q = 0$ a $\bar\partial$-closed $(p,0)$-form is a holomorphic $p$-form and need not vanish. The conclusion fails on general open sets: on $\C^2 \setminus \set{0}$ the Dolbeault group $H^{0,1}$ is not zero.

Going deeper

For a student

The lemma is the local input to the Dolbeault isomorphism. It says that the sequence of sheaves $0 \to \Omega^p \to \mathcal{A}^{p,0} \to \mathcal{A}^{p,1} \to \cdots$ is exact, where $\Omega^p$ is the sheaf of holomorphic $p$-forms and $\mathcal{A}^{p,q}$ is the sheaf of smooth $(p,q)$-forms. So every Dolbeault cohomology class is locally trivial, and the global groups measure only the topology and the complex structure of the manifold as a whole.