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Hello!

We know that we have an alternative way to define a complex structure on manifold, by means of an integrable almost complex structure. The two points of view are equivalent, this is the content of the Newlander-Nirenberg theorem (which is very difficult to prove).

I think that there is the same kind of notion for vector bundles (just replace the bundle map between the tangent bundle and itself by a bundle map from your vector bundle and itself). My question is:

Do we have also an equivalence theorem in the spirit of Newlander-Nirenberg for general vector bundles?

If somebody could give me some reference towards an article, it would be great.

Thank you!

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3 Answers 3

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I think I do not exactly understand the question. A complex vector bundle on a manifold is the same as a real vector bundle, together with an endomorphism $J$ with $J^2 =-1$. This is elementary linear algebra and does not deserve to be called an analog of Nirenberg-Newlander theorem.

A different question is whether one can characterize holomorphic bundles on complex manifolds in a similar fashion. If $V$ is a holomorphic vector bundle, then there is the Cauchy-Riemann (Dolbeault or Wirtinger) operator $\overline{D}:\Gamma (M;V) \to \mathcal{A}^{0,1}(M,V)$ and it satisfies $\overline{D} (fs) = (\overline{\partial}f)s +f \overline{D}s$ for each function $f$ and each section $s$ (I wrote $D$ to distinguish it from the operator on functions). Clearly, $\overline{D}$ encodes the holomorphic structure (holomorphic sections are solutions of $Ds=0$). It is a reasonable question whether for any such operator $D$ on $V$, there exists a holomorphic structure whose Cauchy-Riemann operator is $D$. This amounts, for each $x \in M$, to the existence of local solutions that form a basis at $x$. At least in complex dimension $1$, the answer is ''yes''; see Atiyah, Bott: ''The Yang-Mills equation on Riemann surfaces'', p.555. They give a reference to a different proof, which might well generalize to higher dimensions.

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  • $\begingroup$ @Johannes Ebert: Thank you very much for your answer, this is exactly what I needed, and I am sorry for the confusion. Thank you again! $\endgroup$
    – Benjamin
    Commented May 13, 2011 at 11:19
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Let me supplement Johannes' answer a bit. In higher dimensions a holomorphic vector bundle $V$ is equivalent to a $C^\infty$ bundle equipped with a Cauchy-Riemann operator $\bar D$ (as explained in his answer) which is integrable in the sense that $(\bar D)^2=0$. In one direction, this is easy. For the converse see theorem 2.1.53 in the book by Donaldson and Kronheimer.

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  • $\begingroup$ And to further supplement Johannes' answer: In dimension 1 we automatically have $\overline{D}^2 = 0$ because $\mathcal{A}^{0,2} = 0$. $\endgroup$
    – Kevin H. Lin
    Commented May 14, 2011 at 0:39
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Another great reference is S.Kobayashi, "Differential Geometry of Complex Vector Bundles", Chapter I, Proposition 3.7.

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