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Let $\mathcal X\to \mathrm{Spec}(\mathcal O_K)=C$ be an arithmetric projective variety over $C$ , where $\mathcal O_K$, ring of number filed $K$. Let $\omega$ be a Kähler current of $\mathcal X(\mathbb C)$. Assume that the arithmetic first Chern class $c_1(\mathcal X)$ vanishes.

Then what can we say about $\mathrm{Ric}(\omega)$? Is there any Calabi-Yau like theorem?

Have we such formula $$c_1(\mathcal X(\mathbb C))=[\mathrm{Ric}(\omega)]$$

In fact arithmetic first Chern class is in Chow group and '$\mathrm{Ric}$' is in $H^2$ and they have different cohomology.

More generally, If the anti-canonical arithmetic line bundle is negative, then can we say $\mathrm{Ric}(\omega)=-\omega$ ?

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  • $\begingroup$ To the best of my knowledge, there is a forgetful functor from arithmetic cohomology to singular cohomology that respects Chern classes. E.g., if you represent $c_1(\mathcal X)$ by a divisor of the canonical bundle, you would obtain the Poincaré dual of that divisor. Assuming that $\mathcal X(\mathbb C)$ is smooth, the answer would then be "yes" by the classical Calabi-Yau theorem, because (as far as I know), arithmetic geometry does not impose any further constraint on the Kähler form. $\endgroup$
    – Sebastian Goette
    Commented Feb 9, 2016 at 17:04
  • $\begingroup$ you mean mathoverflow.net/questions/195180/… ? $\endgroup$
    – user21574
    Commented Feb 9, 2016 at 18:22
  • $\begingroup$ I don't see the connection. You don't need specific properties properties of the divisor, just that it gives $0$ in cohomology. But you already assumed $c_1(\mathcal X)=0$ even in the arithmetic setting. $\endgroup$
    – Sebastian Goette
    Commented Feb 9, 2016 at 18:28
  • $\begingroup$ I have used "Kahler current" and not Kahler form $\endgroup$
    – user21574
    Commented Feb 9, 2016 at 18:35
  • $\begingroup$ Please specify the kind of Calabi-Yau theorem you hope for. I thought you mean $\mathrm{Ric}=0$ on all of $\mathcal X(\mathbb C)$. This means, forget the Kähler current and find a Ricci-flat Kähler form in the specified Kähler class. If $\mathcal X(\mathbb C)$ is smooth, I don't see an obstacle. Or do you simply ask for the displayed formula, which does not look like a Calabi-Yau theorem to me? $\endgroup$
    – Sebastian Goette
    Commented Feb 9, 2016 at 18:51

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