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(i) In 1960 Serre proved a famous analogue of the Weil conjectures for Kähler manifolds. This poses an obvious question: Does there exist an analogue of a Kähler structure for (non-singular) projective varieties over a finite field. That is, do there exist things like almost complex structures, Lefschetz operators, Kähler identities, etc?

(ii) Moreover, the study of generalised Hodge structures is a well-studied field. Does there exist a subfield of generalised Kähler structures?

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This may not be quite the answer you want, but from my point of view, smooth projective varieties are the analogues of compact Kähler manifolds, where the metric corresponds to a choice of ample line bundle. Over $\mathbb{C}$, the first Chern class of an ample line bundle is precisely the Kähler class for a Fubini-Study metric. This is, however, an imperfect analogy; some things translate and some thing don't. For example, Lefschetz operators make sense in any Weil cohomology, and Deligne even gave a proof of the hard Lefschetz theorem for projective varieties by reducing to finite fields. There is a definition of sorts for the $\Lambda$ and $*$ operators within algebraic geometry (see Kleiman, Algebraic cycles and the Weil conjectures). But these are all at the cohomology level, not at the differential form level. So you shouldn't expect Kähler identities in the conventional sense (as far as I can tell).

I'm not sure how to interpret (ii).

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So they did everything in terms of a Weil cohomology because there is no normal differential structure. But why didn't they just use the Kahler differential forms? –  Jean Delinez Oct 1 '11 at 18:42
    
It is a long story, and one that I'm not well qualified to answer. You can certainly define de Rham cohomology using Kähler differentials, but over a field of characteristic $p>0$, it yields a theory with values in that field, which is not so useful if one is interested in counting points using the trace formula. Also it is somewhat pathological: the Poincaré lemma can fail etc. Even in char 0, it is not quite a good enough substitute for the $C^\infty$ de Rham complex. –  Donu Arapura Oct 1 '11 at 19:07

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