7
$\begingroup$

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.

Improve this question
$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Sebastian
    Commented Jul 22, 2013 at 7:36
  • $\begingroup$ See mathoverflow.net/questions/136999 $\endgroup$
    – David E Speyer
    Commented Jul 22, 2013 at 15:38

4 Answers 4

Reset to default
6
$\begingroup$

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.

Improve this answer
$\endgroup$
8
$\begingroup$

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.)

Improve this answer
$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Sebastian
    Commented Jul 22, 2013 at 7:24
  • $\begingroup$ Sebastian, good point. $\endgroup$
    – Tim Perutz
    Commented Jul 22, 2013 at 13:11
4
$\begingroup$

Maybe R.O. Wells' book ?

http://www.amazon.com/Differential-Analysis-Manifolds-Graduate-Mathematics/dp/0387738916

Sorry I'm not in the office so I can't check.

I don't think it's particularly difficult. Certainly if the bundle is holomorphic you can construct the Dolbeault operator by just acting on the components of a section written as a linear combination of holomorphic sections. This patches to give a global Dolbeault operator because the clutching functions are holomorphic. To go the other way you need an integrability theorem such as Newlander-Nirenberg to show there are enough sections in the kernel of the Dolbeault operator to span the fibre of the bundle and trivialise it locally.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ I don't think that Wells proves the equivalence, but Chapter III, Section 2 does go into a lot of detail on such connections. $\endgroup$
    – MTS
    Commented Jul 20, 2013 at 19:45
0
$\begingroup$

I believe that this is the content of the paper "Flat partial connections and holomorphic structures in $C^\infty$ vector bundles" by J.Rawnsley.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .