# Calabi-Yau theorem on arithmetic variety

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 $$\mathrm{Ric}(\omega)=-\omega$$ ?

• 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. – Sebastian Goette Feb 9 '16 at 17:04
• you mean mathoverflow.net/questions/195180/… ? – user21574 Feb 9 '16 at 18:22
• 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. – Sebastian Goette Feb 9 '16 at 18:28
• I have used "Kahler current" and not Kahler form – user21574 Feb 9 '16 at 18:35
• 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? – Sebastian Goette Feb 9 '16 at 18:51