# $E$ is a holomorphic vector bundle if and only if there is a Dolbeault operator $\bar{\partial}_E$

I am looking for a reference which shows that the following statements are equivalent for a complex vector bundle $E$:

• $E$ is a holomorphic vector bundle.
• There is a Dolbeault operator $\bar{\partial}_E$, i.e. a $\mathbb{C}$ linear operator $\bar{\partial}_E : \Omega^{0,0}(E) \to \Omega^{0,1}(E)$ which satisfies the Leibniz rule and $\bar{\partial}_E^2 = 0$.

This is stated without proof in Huybrechts' Complex Geometry: An Introduction.

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Over Riemann surfaces there is a proof of this fact which does not use Newlander-Nirenberg: Atiyah, Bott: The Yang-Mills equations over Riemann surfaces. –  Sebastian Jul 22 '13 at 7:36
–  David Speyer Jul 22 '13 at 15:38

A. Moroianu gives a detailed proof on pp. 72-74 of his Lectures on Kähler geometry (Theorem 9.2), available on the internet. (The preprint has it as Theorem 3.2.)

He attributes that proof to S. Kobayashi, Differential geometry of complex vector bundles. I guess he means Proposition I.3.7 there, which Kobayashi couches in a different language involving connections.

P.S. Note that Moroianu gives himself a whole complex of operators $\bar\partial_E:\Omega^{p,q}(E)\to\Omega^{p,q+1}(E)$, and Donaldson-Kronheimer at least the $(0,q)$ ones, plus Leibniz. I'm not 100% clear that just the $(0,0)$ one plus Leibniz suffice to determine everything, as the question seems to imply.

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There is a proof in section 2.2 of The geometry of four-manifolds, by Donaldson and Kronheimer. They point out the similarities between the proofs of the integrability theorems for Dolbeault operators and for flat connections.

(While it's possible to prove this integrability theorem by quoting Newlander-Nirenberg, the PDE problem that underlies it is considerably easier to solve than the N-N problem, because the bundle is decoupled from the coordinates on the base.)

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Dear Tim, I think it is the same complexity to show that flat connections are locally trivial and that flat Riemannian manifolds are locally euclidean: Take a ON parallel frame of the Riemannian manifold, because the connection is torsion free these vectorfields are commuting, so you can integrate them up to obtain a local isometry to euclidean space. –  Sebastian Jul 22 '13 at 7:24
Sebastian, good point. –  Tim Perutz Jul 22 '13 at 13:11

Maybe R.O. Wells' book ?