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Given two smooth algebraic varieties (proper or not), if the two derived categories of the bounded complexes of coherent sheaves over them are equivalent (if necessary we assume there is a fully faithful equivalence between the two categories) , do they have the same hodge numbers, or even the same hodge decomposition in the hochschild homologies of the two categories?

Baranovsky in arXiv:math/0206256 discussed this problem and gave a affirmative answer, but I think he only proved that the two varieties have the same odd and even part of homologies, but not the individual degrees of homologies, i.e. I don't know why they have the same betti numbers from his argument, let alone the hodge numbers. The point I feel is that, we can define hochschild homology from a dg-category (e.g. a dg-enhancement of the derived category), but can't define hodge decomposition of the hochschild homology from only the information of the derive category.

Orlov in arXiv:math/0512620v1 discussed this problem as a consequence of his theorem on motives and gave a sufficient condition which demands the support of the corresponding Fourier-Mukai transform to be equal to the dimension of the concerned varieties. I think this condition is too strong and difficult to verify.

A related problem is, does the homological mirror symmetry conjecture imply the coincedence of hodge numbers of two mirror duals?

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Dunno if it's a misprint or not, but just in case: Volodia's last name is spelled with a single "n". – Anton Fonarev Nov 9 '11 at 0:14
I added in the links to the arXiv papers. – David Roberts Nov 11 '11 at 3:36
I corrected the spelling of Baranovsky's name. – S. Carnahan Nov 11 '11 at 8:42

For varieties of dimension 1, 2 and 3 it is known that derived equivalent varieties have the same Hodge numbers. As far as I know, this is open (though believed to be true) for higher dimensional varieties.

For surfaces this follows from the classification of Bridgeland and Maciocia in their paper "Complex surfaces with equivalent derived categories" whereas for threefolds the result is contained in this preprint of Popa and Schnell.

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Let me mention one situation in which this is known in all dimensions: varieties of general type. The key point in this setting is that derived equivalence implies K-equivalance, and K-equivalence implies equality of Hodge numbers. – user5117 Nov 8 '11 at 16:05
In general, what is known is that the sum of the vertical Hodge numbers is always preserved (at the very least, this is what homological mirror symmetry would suggest). This fact is used for derived invariance of Hodge numbers in dimensions 1,2,3. in dimensions When you know more about the Hodge diamond of your variety (say for Calabi-Yau varieties), then you can conclude equalities of some Hodge numbers. About non-dg derived equivalences: would you drink non-mammalian milk? – Chris Brav Nov 8 '11 at 19:36

This isn't quite responsive, but you can show for nice dg-categories that the Hodge-de Rham spectral sequence degenerates in many situations. See Kaledin's notes, 0708.1574. This is a special case of Kontsevich's degeneration conjecture. For a lot of discussion along these lines, see Katzarkov, Kontsevich and Pantev on nc-Hodge structures.

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Thanks to all of your answers! – Xiaowen Hu Nov 10 '11 at 17:06

I don't know the state of the art. In physics if two smooth Calabi-Yau varieties, $X$ and $X^\vee$, of dimension $n$ are mirror to each other then $h^{p,q}(X)=h^{n-p,q}(X^\vee)$. In general they will not smooth but the physics suggests that we take crepant resolutions $Y\to X$ and $Y^\vee\to X^\vee$ and hence we should have $h^{p,q}(Y)=h^{n-p,q}(Y^\vee)$. But, if a crepant resolution exists it is not necessarily unique. This motivated Maxim Kontsevich to introduce motivic integration and prove that if $X_1$ and $X_2$ are two crepant resolutions of $X$ then $h^{p,q}(X_1)=h^{p,q}(X_2)$.

As for the decomposition of Hochschild cohomology holding for derived equivalences, this is not true. In a paper by Oren Ben-Bassat, Jon Block, and Tony Pantev, they show that for a torus $X$ and its dual $X^\vee$ the Fourier-Mukai transfrom induces an isomorphism $H^0(X,\wedge^2T_X)\simeq H^2(X^\vee,\mathcal{O}_{X^\vee})$. In other words, a noncommutative deformation of $X$ goes to a gerby deformation of $X^\vee$.

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Thank you for your mentioning this work. – Xiaowen Hu Nov 15 '11 at 15:53

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