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For a smooth projective complex variety $X$ of dimension $n$, let $H^i(X)$ denote its integral Hodge structure of weight $i$. Define $\tilde{ H^0}(X) = \bigoplus H^{2i}(X)\otimes \Bbb Z(i)$ and $\tilde{ H^1} (X)= \bigoplus H^{2i+1}\otimes \Bbb Z(i)$, respectively. It is known that any derived equivalence $\Phi:D^b(X)\to D^b(Y)$ induces isomorphisms of rational Hodge structures $\tilde{H^0_\Bbb Q}(X)\cong \tilde{H^0_\Bbb Q}(Y)$ and $\tilde{H^1_\Bbb Q}(X)\cong \tilde{H^1_\Bbb Q}(Y)$, however these isomorphisms are defined by characteristic classes whose coefficients aren't necessarily integral.

Are there any known examples where this isomorphism does not descend to the integral Hodge structures and, furthermore, $\tilde H^*(X)$ and $\tilde H^*(Y)$ are nonisomorphic?

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  • $\begingroup$ Probably, the theory of non-commutative motives (as developed by Gonsalo Tabuada and his numerous co-authors) yields a positive answer to your question. $\endgroup$ Commented Dec 12, 2015 at 13:11
  • $\begingroup$ @MikhailBondarko: By "positive answer", do you mean that there are no such counterexamples? I'm a bit familiar with Tabuada's theory; could you expand on your comment? $\endgroup$
    – Dominik
    Commented Dec 12, 2015 at 14:52
  • $\begingroup$ The non-commutative motive functor does satisfy certain universality property (actuallya, there are several functors with distinct universality properties;)): in theory, this should describe which functors factor through this one. $\endgroup$ Commented Dec 12, 2015 at 16:02
  • $\begingroup$ @MikhailBondarko, it's not quite so easy, I'm afraid. You would need to be able to define Betti cohomology and Hodge structures at the level of dg-categories, something which is most likely impossible in my opinion. The best you have are noncommutative Hodge structures on Hochschild homology and variants, studied by Kaledin and others. $\endgroup$
    – AAK
    Commented Dec 13, 2015 at 15:23
  • $\begingroup$ Probablly you are right. Yet it seems that noncommutative motives may give an answer. $\endgroup$ Commented Dec 13, 2015 at 16:09

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