Newlander-Nirenberg theorem for general vector bundles - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T04:57:06Z http://mathoverflow.net/feeds/question/64873 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/64873/newlander-nirenberg-theorem-for-general-vector-bundles Newlander-Nirenberg theorem for general vector bundles Benjamin 2011-05-13T05:57:43Z 2011-05-13T14:37:18Z <p>Hello!</p> <p>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).</p> <p>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:</p> <p>Do we have also an equivalence theorem in the spirit of Newlander-Nirenberg for general vector bundles?</p> <p>If somebody could give me some reference towards an article, it would be great.</p> <p>Thank you!</p> http://mathoverflow.net/questions/64873/newlander-nirenberg-theorem-for-general-vector-bundles/64882#64882 Answer by Johannes Ebert for Newlander-Nirenberg theorem for general vector bundles Johannes Ebert 2011-05-13T09:10:24Z 2011-05-13T09:10:24Z <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/64873/newlander-nirenberg-theorem-for-general-vector-bundles/64894#64894 Answer by Donu Arapura for Newlander-Nirenberg theorem for general vector bundles Donu Arapura 2011-05-13T12:55:09Z 2011-05-13T12:55:09Z <p>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.</p> http://mathoverflow.net/questions/64873/newlander-nirenberg-theorem-for-general-vector-bundles/64899#64899 Answer by Peter Dalakov for Newlander-Nirenberg theorem for general vector bundles Peter Dalakov 2011-05-13T14:37:18Z 2011-05-13T14:37:18Z <p>Another great reference is S.Kobayashi, "Differential Geometry of Complex Vector Bundles", Chapter I, Proposition 3.7.</p>