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An HH type is the oriented homotopy type of a closed simply-connected Kähler manifold together with the Hodge structure on cohomology.

Two HH types are deformation equivalent if they are represented by closed Kähler manifolds that are deformation equivalent.

If two HH types are equivalent as oriented homotopy types and have the same Hodge diamonds are they deformation equivalent?

Closed simply-connected complex surfaces with $p_g\neq0$ (e.g. complete intersections) can supply counterexamples. I haven't yet verified if the statement holds for them.

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According to the two answers here, "Catanese and Manetti gave various examples of orientedly diffeomorphic but not deformation equivalent smooth projective algebraic surfaces of general type", and "For a Kaehler surface, the Hodge numbers are topological invariants". So, this would imply that the answer is no.

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  • $\begingroup$ My apologies, I am unable to follow the logic of this answer. $\endgroup$
    – user164740
    Oct 8 '20 at 11:23
  • $\begingroup$ Your question seems to be asking if two Kahler manifolds that are (oriented) homotopy equivalent and have the same Hodge numbers are deformation equivalent. Catanese and Manetti gave an example of two oriented diffeomorphic but not deformation equivalent smooth projective surfaces. Since Hodge numbers are topological invariants of surfaces, these two must have the same Hodge numbers. $\endgroup$ Oct 8 '20 at 15:07
  • $\begingroup$ We are asking if the oriented homotopy types together with the Hodge structures have Kähler representatives that are deformation equivalent. The absence of such representatives in your example is not clear to me. For instance there could be an oriented homeomorphism between the surfaces respecting the Hodge structures. Say, if the off-diagonal Hodge numbers vanish any homeomorphism must respect the Hodge structures. $\endgroup$
    – user164740
    Oct 8 '20 at 15:12
  • $\begingroup$ Ah, I see, so what you would need is that any pair of Kahler manifolds homotopic to the Catanese-Manetti example (where the homotopies respects the Hodge structures) are necessarily not deformation equivalent. Is that right? In any case, perhaps you should look at their paper and see if you can make that work, since it's close to what you want. $\endgroup$ Oct 8 '20 at 15:26

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