Constructing projective Calabi-Yau varieties

A nice way of constructing a non-projective Calabi-Yau threefold is to take the total space $$Y:= \mathrm{Tot}(\omega_S) = \mathbf{Spec}(\mathrm{Sym}^\bullet(\omega_S^\vee))$$ of the canonical bundle of a surface $$S$$. Here $$\mathbf{Spec}$$ is relative Spec. This is Calabi-Yau because it can be shown that $$\omega_Y \cong \mathcal{O}_Y$$.

The projective version of the total space construction is $$\mathbb{P}(\mathcal{F}) = \mathbf{Proj}(\mathrm{Sym}^\bullet(\mathcal{F}))$$ where $$\mathcal{F}$$ is a (finite rank) locally free coherent sheaf. My question is: can such a sheaf $$\mathcal{F}$$ (on some variety $$X$$) be picked such that $$\omega_{\mathbb{P}(\mathcal{F})} \cong \mathcal{O}_{\mathbb{P}(\mathcal{F})}$$, i.e. such that $$\mathbb{P}(\mathcal{F})$$ is Calabi-Yau?

No, this construction will never give a Calabi--Yau variety.

For simplicity assume $$X$$ is smooth. By construction, the variety $$V=\mathbb P_X(F)$$ is a projective bundle, so it is uniruled (except possibly in the trivial case when $$F$$ has rank 1). It is a basic property of smooth projective uniruled varieties $$V$$ that

$$H^0(V,K_V^m)=0 \quad \text{ for all } m>0$$

On the other hand, if $$Y$$ is Calabi--Yau, then $$K_Y$$ is the trivial bundle, so $$H^0(Y,K_Y)=1$$.

A reference for the above is Kollár, Rational Curves on Algebraic Varieties, Corollary IV.1.11.

• It is easier to just say that $K_Y$ restricts to a fiber $\mathbb{P}^m$ of the projective bundle as $\mathcal{O}(-m-1)$, hence is non-trivial. Commented Oct 3, 2019 at 5:27