A vector bundle over a complex manifold is said to be holomorphic if its trivialization maps are biholomorphic maps. What is a "natural" example example of a vector bundle over compact complex manifold which is not holomorphic? I guess by "natural" I mean that one would be interested in such examples for reasons besides them being a counter example.

TO CLARIFY: I am interested in



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    $\begingroup$ Related: mathoverflow.net/questions/7304/… $\endgroup$ – user347489 Apr 4 '20 at 21:11
  • $\begingroup$ @Thomas: So what are interesting examples of bundles that do not admit almost complex structures? $\endgroup$ – Nadia SUSY Apr 5 '20 at 12:44
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    $\begingroup$ The introduction to the paper paper arxiv.org/pdf/1506.08111.pdf contains a good survey of the problem. $\endgroup$ – Angelo Apr 6 '20 at 12:34
  • $\begingroup$ Nadia, if you are interested in real vector bundles that don't admit the structure of a complex vector bundle, then you should say so. Because, while not trivial, this can be investigated by methods of algebraic topology. Meanwhile, since you use the word holomorphic in your title, you should specify that you are interested in complex vector bundles, since otherwise it's not clear what you are interested in. I think you might have stumbled into a very deep question on accident. $\endgroup$ – Andy Sanders Apr 6 '20 at 21:44
  • $\begingroup$ Complex line bundles on a manifold $X$ are classified by the first Chern class $c_1\in H^2(X,\mathbb{Z})$. If $X$ is a compact Kähler manifold, such a line bundle is holomorphic if and only if $c_1$ is of type $(1,1)$. $\endgroup$ – abx Apr 7 '20 at 5:41

I interpret the question as "Are there any complex bundles over the complex manifold which do not admit the holomorphic structure". The answer to this is "yes", but I'm not sure if it is known in full generality for what manifolds the answer is positive and for what it is negative.

Line bundles case. Suppose X is compact Kahler, L - complex line bundle over X. Then it has a holomorphic structure iff it's 1st Chern class is (1,1). See the comment by @abx

Proof: the "only if" part of the statement is the fact that holomorphic bundle admits so-called Chern connection, which has curvature of type (1,1). "If" part comes from the following consideration:

Complex line bundles are classified by $H^1(X, C^{\infty}(X, \mathbb{C}^*))$. This sheaf is subject to the following exponential exact sequence:

$0 \rightarrow \mathbb{Z} \rightarrow C^{\infty}(X, \mathbb{C}) \rightarrow C^{\infty}(X, \mathbb{C}^*) \rightarrow 0$

Similarly, holomorphic bundles are classified by $H^1(X, \mathcal{O}^*)$. There is also an analogous exact sequence (which is also embedded in the previous one by the inclusion of the holomorphic functions into the smooth ones):

$0 \rightarrow \mathbb{Z} \rightarrow \mathcal{O} \rightarrow \mathcal{O}^* \rightarrow 0$

Now, middle term in the former exact sequence is acyclic, so $H^1(X, C^{\infty}(X, \mathbb{C}^*)) \simeq H^2(X, \mathbb{Z})$ (this delta homomorphism is one of the equivalent formulations of the 1st Chern class). The latter exact sequence, however, gives (after standard identification $H^k(X, \mathcal{O}) \simeq H^{0,k}(X)$ the following description for holomorphic bundles:

$0 \rightarrow H^{0,1}(X)/H^1(X, \mathbb{Z}) \rightarrow H^1(X, \mathcal{O}^*) \rightarrow H^2(X, \mathbb{Z}) \cap (H^{1,1}(X) \oplus H^{2,0}(X)) \rightarrow 0$

The third term of this sequence is actually $H^2(X, \mathbb{Z}) \cap H^{1,1}(X)$ (because $H^2(X, \mathbb{Z})$ is real, and $H^{2,0}(X)$ is conjugate to $H^{0,2}(X)$.

This gives the following obstruction for line bundle - it's first Chern class should be (1,1), and provided this obstruction holds there is a complex torus $H^{0,1}(X)/H^1(X, \mathbb{Z})$ of different holomorphic structures.

General case. I think in the general case the similar classification is not known. The criterion that all Chern classes should be of (p,p) type clearly holds, but I think it is not enough. I'm not up to date with the current state of this field but you can see from here http://www.numdam.org/item/?id=SB_1978-1979__21__80_0 that even for projective spaces when this criterion is empty the question is far from trivial (and if I've understood correctly has a negative answer). Also see @Angelo 's comment which has a reference to the recent survey of this problem https://arxiv.org/pdf/1506.08111.pdf

Maybe it is also worth noting that over a manifold of complex dimension 1 any complex bundle admits holomorphic structure: indeed, one can choose an almost holomorphic structure (take any connection and take it's (0,1) part), and then it is automatically integrable by dimensional reasons).

  • $\begingroup$ Angelo's comment above is relevant here: he gives a reference to some results on obstructions. $\endgroup$ – Ben McKay Apr 7 '20 at 9:51
  • $\begingroup$ Thanks, I didn't see it, will add to answer. This survey also has a reference to the text I linked. $\endgroup$ – Lev Soukhanov Apr 7 '20 at 11:10

The bundles of (0, 1)-forms are not holomorphic. (the transition functions are anti-holomorphic) In fact the bundles of $(p, q)$-forms are all not holomorphic if $q>0$. (and if both $p$ and $q$ are not 0, then the transition functions are neither holomorphic nor anti-holomorphic)

  • $\begingroup$ But these bundles admit a holomorphic structure, namely the opposite complex complex. I suppose I am looking for a smooth bundle that doesn't admit any holomorphic structure. $\endgroup$ – Nadia SUSY Apr 7 '20 at 8:57

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