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Let $E\to X$ be a holomorphic vector bundle on a projective complex manifold. If $Y$ is another projective manifold diffeomorphic to $X$ then we are free to consider $E$ a smooth bundle over $Y$. Assuming that all of its Chern classes are still Hodge, does it admit a holomorphic structure? The answer is probably no, but I cannot think of any example.

(By the way, the question may become more interesting if we make certain assumptions, e.g. that the bundle is ample. Because there is a possibility that with this restrictions the answer may be positive which would have interesting implications for the Hodge conjecture: an algebraic Hodge class remains algebraic unless some of the other Hodge classes vanish.)

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    $\begingroup$ Take a quintic surface in $X \subset \mathbb{P}^3$ containing a line $L$, and set $$E=\mathcal{O}_X(L) \oplus \mathcal{O}_X(-L).$$ Now deform $X$ to a quintic surface $Y$ with $\mathrm{Pic}(Y) = \mathbb{Z}[H]$, where $H$ is the hyperplane section (the general $Y$ would do this, by Noether-Lefschetz, and is diffeomorphic to $X$ by Ehresmann)...[continue] $\endgroup$
    – Francesco Polizzi
    Commented May 22, 2019 at 8:28
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    $\begingroup$ ...The first Chern class of $E$ is trivial, so it surely remains of Hodge type on $Y$. Does $E$ have a holomorphic structure on $Y$? My guess is no. Surely it cannot be a holomorphic structure in which is it decomposable, since otherwise $$E=\mathcal{O}_Y(aH) \oplus \mathcal{O}_Y(-aH),$$ so passing to the second Chern class we would have $$3=-(L, \, L)_X=c_2(E)=a^2H^2=5a^2,$$ a contradiction. Can $E$ be given a indecomposable holomorphic structure? $\endgroup$
    – Francesco Polizzi
    Commented May 22, 2019 at 8:38
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    $\begingroup$ From [Holomorphic 2-vector bundles on nonalgebraic 2-tori, Paul Flondor / Vasile Brinzanescu, J. reine angew. Math, 1985]: "It is known (see Schwarzenberger) that on a projective surface every topological 2-vector bundle has a holomorphic structure iff its first Chern class is of the form $c_1(L)$ with $L$ a holomorphic line bundle." $\endgroup$
    – Alex Gavrilov
    Commented May 22, 2019 at 13:28
  • $\begingroup$ So $E$ has a holomorphic structure on $Y$? Weird (meaning that I did not expect this...) $\endgroup$
    – Francesco Polizzi
    Commented May 22, 2019 at 13:32

1 Answer 1

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This is more of an extended comment with guesses and references rather than an actual answer.

1) If in your question you ask $Y$ to be merely Kahler and not projective, then I can propose potential counter-examples. Namely, one could try to get them from Theorem 10 in https://webusers.imj-prg.fr/~claire.voisin/Articlesweb/takagifinal.pdf which is originally proven in https://webusers.imj-prg.fr/~claire.voisin/Articlesweb/hodgeimrn.pdf

This theorem answers negatively to the following question: "Are Hodge classes on compact Kahler manifolds generated over $\mathbb Q$ by Chern classes of coherent sheaves?" The counter-examples to this question are $4$-dimensional complex tori. At the same time the answer to the question for projective manifolds is positive.

2) I wonder if some projective examples could be constructed using counter-examples to integral Hodge conjecture, constructed by Kollar. This is Theorem 14 in the first paper (above) of Voisin. It states, for example, that on a generic hypersurace of degree $125$ in $\mathbb CP^4$ the class of each curve is divisible by $5$. However there exist such hypersufaces that contain lines. I wonder if on such a hypersurface there could be a rank $2$ bundle with the class of $c_2$ equal to the class of line (or at least not divisible by 5).

Anyway, whether 1 or 2 works or not, it looks like the paper of Voisin is reasonable place to start looking for a counter-example.

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