1
$\begingroup$

The question is almost contained in the title.

I am looking for interesting examples of smooth, projective threefolds $X$ (preferably over a number field) such that $ H^0(X, \Omega^2_X) $ or equivalently $H^2(X, \mathcal{O}_X)$ has dimension one. Who can provide such examples?

$\endgroup$
2
  • $\begingroup$ By "interesting" do you mean threefolds which are not $\mathbb{P}^{1}$ bundles over a surface with $h^{2,0}=1$? $\endgroup$ Jun 27, 2013 at 15:21
  • $\begingroup$ Not necessarily $\endgroup$
    – 3fold
    Jun 27, 2013 at 15:49

2 Answers 2

4
$\begingroup$

You can prove that such threefolds are either P^1-fibration over a base, which is a surface with $h^{2,0}=1$, or have pseudoeffective canonical bundle: arXiv:1304.7891, Corollary 4.3. In the later case you can run the minimal model program, obtaining that your variety is either general type or is a (singular) fibration with Calabi-Yau fibers over a general type variety with canonical singularities. However, since $h^{2,0}=1$, your base has $h^{2,0}<2$, and this restriction is pretty strong: either you have an elliptic fibration over a surface, or a K3 or toric fibration over an elliptic curve.

$\endgroup$
1
  • 1
    $\begingroup$ One comment: a $\mathbb{P}^1$-fibration in this context may only be etale locally trivial (away from the discriminant locus). There are "conic bundles" that are not Zariski locally trivial over surfaces with $h^{2,0} =1$. $\endgroup$ Jun 28, 2013 at 17:25
2
$\begingroup$

Let $S$ be a K3 surface, and let $S^{[2]}$ be the Hilbert scheme parametrizing length-2 subschemes of $S.$ Then $S^{[2]}$ is a smooth projective 4-fold with $h^{2,0}=1.$ If $D \subset S^{[2]}$ is a smooth ample divisor on $S^{[2]},$ then Kodaira vanishing implies that $D$ is a smooth projective threefold with $h^{2,0}=1.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.