MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm looking for references for the following fact, to pass on to a grad student. I think I can prove it, but it is a bit sloppy and I'd rather have something which is already written up.

Let $X$ be a complex manifold. Let $E$ be a smooth, complex vector bundle. Let $C^{\infty}(E)$ be the sheaf of smooth sections of $E$. A dee-bar connection is a $\mathbb{C}$-linear map $\nabla : C^{\infty}(E) \to C^{\infty}(E) \otimes \Omega^{0,1}$, obeying the Liebnitz rule $$\nabla(f \sigma) = f \nabla(\sigma) + \bar{\partial}f \otimes \sigma$$ Just as with connection in the $C^{\infty}$ setting, we can define the curvature; it is a section of $\Omega^{0,2} \otimes \mathrm{End}(E)$. Suppose that the curvature is zero.

Let $\mathrm{Ker}(\nabla)$ be the kernel of $\nabla : C^{\infty}(E) \to C^{\infty}(E) \otimes \Omega^{0,1}$. Let $\mathcal{O}$ be the sheaf of homolomorphic functions. It is easy to see that $\mathrm{Ker}(\nabla)$ is a $\mathcal{O}$-module. I want a reference for the fact that:

$\mathrm{Ker}(\nabla)$ is a locally free $\mathcal{O}$-module of rank $\dim E$.

The analogous fact for smooth connections is (one direction of) the Riemann-Hilbert correspondence; see, for example, Theorem 2.6 here.

share|cite|improve this question
up vote 3 down vote accepted

This is Theorem 2.1.53 in Donaldson--Kronheimer's Geometry of 4-manifolds. They prove it from scratch via the contraction mapping principle but there are arguments from the Newlander--Nirenberg theorem also.

share|cite|improve this answer
Thanks! This isn't really a proof from scratch; large parts of it are copying the proof of Dolbeault's lemma, so I can just think "plug in the proof of Dolbeault here". – David Speyer Jul 17 '13 at 19:55

Kobayashi, "Differential geometry of complex vector bundles", Proposition 3.7. He does it using the Frobenius theorem.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.