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In this question I asked if it could happen that two complex manifolds are homeomorphic, and one of them is a Calabi-Yau manifold but the other isn't. It turns out that there are complex surfaces that are homeomorphic to a K3 surface, but are not K3 surfaces themselves.

On the other hand, in the first comment to that question, Mike Miller asserts that this cannot happen for diffeomorphic surfaces.

Is it possible in higher dimensions that a Calabi-Yau manifold (compact Kähler with $h^{k,0} = 0$ for $0 < k < n$, trivial canonical bundle) is diffeomorphic to another complex manifold that is not Calabi-Yau?

doetoe
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  • Diffeomorphic simply-connected surfaces, at least. :) –  Sep 22 '18 at 17:35
  • As an observation, it is an open question whether given any closed almost complex manifold $(M,J)$ of dimension $2n > 4$, you may deform $J$ to $J'$, an almost complex structure coming from a legitimate complex structure. So one might want to ask the more basic question, "can $M$ support $J_1. J_2$ so that $c_1(J_1) = 0 \neq c_1(J_2)$?" My gut feeling is "sure, why not?" without any serious thought behind it. –  Sep 22 '18 at 17:44
  • @MikeMiller Interesting. My gut feeling unfortunately is almost completely lacking in this, so I'd have to chew on this for a bit. As for your other comment, I thought that all K3 surfaces were simply connected, actually that all were diffeomorpic. Maybe it depends on the exact definition you use. – doetoe Sep 22 '18 at 18:01
  • When someone says "All K3 surfaces are diffeomorphic", they mean "Any two compact complex surfaces with $c_1(M, J) = 0$ are diffeomorphic." But we have already put the complex structure on and assumed a property about it. Now, it is true that any other complex structure on the same underlying smooth manifold is Calabi-Yau, but this is a nontrivial theorem of Seiberg-Witten theory. 2) The second sentence looks like one is asserting this property for all surfaces, not just K3s, so I was being careful.
    • –  Sep 22 '18 at 20:08
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    An example of a four-manifold supporting $J_1$ and $J_2$ with $c_1(J_1) = 0$, $c_1(J_2) \neq 0$ is provided by any parallelizable $M^4$ with a non-torsion class $\alpha \in H^2(M; Z)$ whose square is 0. (By Wu’s criterion there are then J’s with $c_1 = 0$ and $c_1 = 2\alpha$.) An example of such a manifold is the torus, or more generally any nilmanifold. In dimension 6, from the answer here https://mathoverflow.net/questions/63439/how-can-we-detect-the-existence-of-almost-complex-structures we can conclude that the six-torus admits such J’s as well. – Aleksandar Milivojević Nov 04 '18 at 22:39