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For a complex manifold $M$, the complexified tangent space $\Omega^1(M)$ splits into a direct sum $\Omega^1(M) = \Omega^{(1,0)}(M) \oplus \Omega^{(0,1)}(M)$. As is well-known $\Omega^{(1,0)}(M)$ is canonically a holomorphic vector bundle over $M$.

Question: When can $\Omega^{(0,1)}(M)$ be given a holomorphic vector bundle structure? When will it be unique?

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  • $\begingroup$ You might want to have a look at David Speyer's answer to MO-question mathoverflow.net/questions/8484 $\endgroup$
    – Matthias Wendt
    Commented May 5, 2015 at 15:36
  • $\begingroup$ a slight variation: if $\Omega^{(0,1)}(M)$ happens to be trivial as a complex vector bundle, then obviously it is possible to put the trivial holomorphic structure on it. This happens e.g. for an elliptic curve. Maybe I got something wrong, but the "can never be given a holomorphic structure" seems to weak to be true. $\endgroup$
    – Matthias Wendt
    Commented May 5, 2015 at 15:38
  • $\begingroup$ So what about: "When can $\Omega^{(0,1)}$ be given a holomorphic structure"? $\endgroup$
    – Falertu Jopanos
    Commented May 5, 2015 at 15:41
  • $\begingroup$ Well, David Speyer's answer linked above seems to claim that $\Omega^{(0,1)}$ can always be given a holomorphic structure. This should not be unique; when $M$ is an elliptic curve, $\Omega^{(0,1)}$ is trivial, but there are many non-trivial holomorphic line bundles whose underlying complex line bundle is trivial. $\endgroup$
    – Matthias Wendt
    Commented May 5, 2015 at 16:02
  • $\begingroup$ So David Speyer's answer is exactly what I'm looking for. How does one go about accepting an answer from another question? Sorry, I'm new here. $\endgroup$
    – Falertu Jopanos
    Commented May 6, 2015 at 13:47

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