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Let $X$ be a projective complex manifold and $E\to X$ a holomorphic vector bundle. When $E=X\times\mathbb{C}$ there is an injection (for any $n$) $$H^n(X, \mathcal{O}(E))=H^{0,n}(X)\to H^n(X,\mathbb{C}).$$ The question is about possible maps $$H^n(X, \mathcal{O}(E))\to \text{some ``topological'' group}$$ in cases when $E$ is not trivial and $X$ is simply connected. It seems unlikely that topological invariants of this kind exist in general, but I am curious if there is anything interesting in special cases.

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    $\begingroup$ Where did you get that isomorphism from? For $X$ an elliptic curve, $H^1(X, \mathcal{O}_X)$ is one-dimensional while $H^1(X, \mathbb{C})$ is two-dimensional. I do not understand the second paragraph. $\endgroup$ Commented Jun 10, 2020 at 7:12
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    $\begingroup$ P.S. Maybe the following is of interest to you: if $E$ carries a holomorphic integrable connection $\nabla$, then the horizontal sections of $E$ form a local system of $\mathbb{C}$-vector spaces $\mathcal{E}$ on $X$ and there is an isomorphism $H^*_{\rm dR}(X, E) \simeq H^*(X, \mathcal{E})$, where the first group is the de Rham cohomology of $E$ i.e. hypercohomology of the complex $(\Omega^\bullet_X \otimes E, \nabla)$. $\endgroup$ Commented Jun 10, 2020 at 7:15
  • $\begingroup$ I edited it so hopefully now it makes some sense. $\endgroup$ Commented Jun 10, 2020 at 10:00
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    $\begingroup$ @Piotr Achinger, If a connection is integrable and the base is simply connected, doesn't it mean that the bundle is trivial? $\endgroup$ Commented Jun 10, 2020 at 10:34
  • $\begingroup$ Even if $X$ is simply connected, a Zarsiki open $U\subset X$ need not be. Suppose additionally that $D=X-U$ is a divisor with normal crossings. If one starts with a unitary representation of $U$ forms the associated vector bundle, and extends to bundle $E$ $X$ a la Deligne. Then this gives something nontrivial example along the lines of your question. I have no idea, if this is the sort of thing you want however. $\endgroup$ Commented Jun 10, 2020 at 13:27

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