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This is a kind of a follow up to this question, which actually already had an answer here, in which it is asserted that Hodge numbers in general are not topological invariants. Could it be so extreme that a Calabi-Yau $n$-manifold (compact Kähler with $h^{k,0} = 0$ for $0 < k < n$, trivial canonical bundle) is homeomorphic to a non-Calabi-Yau complex manifold?

In other words, could there be a topological manifold that admits one complex structure for which it is Calabi-Yau, and another for which it isn't?

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    $\begingroup$ Among simply connected 4-manifolds, the only Calabi-Yau manifold is the $K3$ surface, so we need to know whether or not it can support complex structures with nonzero $c_1$. Maybe there is a low-tech answer, but in fact a Kahler structure on $K3$ always has $c_1 = 0$ by techniques of Seiberg-Witten theory from the 90s. I would guess there are counterexamples in higher dimensions. $\endgroup$
    – user98602
    Sep 19, 2018 at 23:27
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    $\begingroup$ Note that Hodge numbers $h^{k,0}$ are birational invariant. In particular, the blowup of a CY variety has the same $h^{k,0}$ as the original one. So, I don't think it is a good idea to define the CY property in terms of Hodge numbers. $\endgroup$
    – Sasha
    Sep 20, 2018 at 8:44
  • $\begingroup$ @Sasha In fact the condition on the Hodge numbers is equivalent to the holonomy being exactly $SU(n)$ rather than contained in it, but has the advantage that it is defined in purely algebraic terms. $\endgroup$
    – doetoe
    Sep 20, 2018 at 12:34
  • $\begingroup$ @Sasha Ah, I meant to add triviality of the canonical bundle but didn't write that. Was that your objection? I'll edit. $\endgroup$
    – doetoe
    Sep 20, 2018 at 21:27
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    $\begingroup$ @doetoe That was definitely his objection, as it is crucial. That's what gives the reduction to holonomy inside $SU$ (after applying Calabi's conjecture / Yau's theorem). $\endgroup$
    – user98602
    Sep 20, 2018 at 21:29

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Yes, it is!

There exist the homotopy type K3-surfaces which are homeomorphic but not diffeomorphic to K3-surfaces.

You can consult D. Huybrechts (2016) Lectures on K3-surfaces, Cambridge University Press; chapter 1, remark 3.6.ii.

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  • $\begingroup$ Great, thanks! I guess a follow-up question would be if a complex manifold that is not a Calabi-Yau manifold can be diffeomorphic to a Calabi-Yau manifold? $\endgroup$
    – doetoe
    Sep 21, 2018 at 10:37
  • $\begingroup$ Do you happen to know if the K3 surface it is homeomorphic to is algebraic? I'm wondering if it could be that dimension 2 is somehow exceptional. $\endgroup$
    – doetoe
    Sep 22, 2018 at 17:30
  • $\begingroup$ If I'm not wrong: if a $K3$-surface $X$ is homeomorphic to an algebraic one $Y$ then $X$ is algebraic as well. See Barth W., Peters C., Van de Ven A. (1984) Compact complex surfaces, Springer-Verlag; Chapter VIII. $\endgroup$
    – Armando j18eos
    Sep 24, 2018 at 10:01
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    $\begingroup$ I doubt that can be true, as all K3 surfaces are diffeomorphic (chapter VIII.2) $\endgroup$
    – doetoe
    Sep 24, 2018 at 10:43

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