0
$\begingroup$

Let $\pi:X \longrightarrow C$ be a smooth projective morphism onto a smooth projective curve. If the general fibers are of nonnegative Kodaira dimension, is $\pi_{\ast} \mathcal{O}(k K_{X/C})$ nonzero for sufficiently divisible $k$? If it is, is there an algebraic proof? That is, without using invariance of plurigenera.

$\endgroup$
6
  • $\begingroup$ This question is an exact duplicate of your previous question mathoverflow.net/questions/102165/… $\endgroup$
    – YangMills
    Jul 16, 2012 at 21:59
  • $\begingroup$ not exactly. Here I assume the general fibers are of nonnegative Kodaira dimension $\endgroup$
    – Hu Zhengyu
    Jul 16, 2012 at 22:56
  • $\begingroup$ I have to say I still don't understand what "central fiber" means in your other question. Moreover your new question is surely equivalent to the other one (a posteriori using invariance of plurigenera). $\endgroup$
    – YangMills
    Jul 17, 2012 at 0:43
  • $\begingroup$ I think Artie's other's proof (which he deleted) did this case (unless I'm misremembering/reading). $\endgroup$ Jul 17, 2012 at 1:43
  • $\begingroup$ Re YangMills: "central fiber" means a fibre over a fixed closed point on C $\endgroup$
    – Hu Zhengyu
    Jul 17, 2012 at 9:04

1 Answer 1

3
$\begingroup$

The stalk of $\pi_* (\omega_{X/C}^{\otimes k})$ at the generic point $\eta$ of $C$ equals $H^0(X_\eta, \omega_{X_\eta/\kappa(\eta)}^{\otimes k})$. If the generic fiber $X_\eta$ has nonnegative Kodaira dimension, then there exists an integer $k$ such that $H^0(X_\eta,\omega_{X_\eta/\kappa(\eta)}^{\otimes mk})$ is nonzero for every positive integer $m$. Since the stalk $\pi_*(\omega_{X/C}^{\otimes mk})_\eta$ is nonzero, in particular $\pi_*(\omega_{X/C}^{\otimes mk})$ is nonzero.

$\endgroup$
3
  • $\begingroup$ I wonder how you can ensure the generic fibre has nonnegative Kodaira dimension? $\endgroup$
    – Hu Zhengyu
    Jul 17, 2012 at 9:03
  • $\begingroup$ @Hu Zhengyu: I suppose this depends on what you mean by "general fibers". If you are working over $\overline{\mathbb{F}_p}$, then I agree that your definition of "general fibers" (whatever it is) might disagree with the standard definition. However, if you are working over an uncountable algebraically closed field, or over any "sufficiently large" algebraically closed field, then this follows by upper semicontinuity, cf. Theorem III.12.8 of Hartshorne. $\endgroup$ Jul 17, 2012 at 12:14
  • $\begingroup$ @HU Zhengyu: To spell it out, for every $k$, by Theorem III.12.8, there exists a proper Zariski closed subset $Z_k \subset C$ such that for every $p\in C\setminus Z_k$, '$h^0(X_p,\omega_{X_p}^{\otimes k})$' equals '$h^0(X_\eta, \omega_{X_\eta}^{\otimes k})$'. If your ground field is uncountable, then for every dense open subset of $C$, or even for every countable intersection $G$ of dense open subsets, there exists a point $p$ of $G$ which is no $Z_k$. So for this "general point", for every $k$, $h^0(X_p,\omega_{X_p}^{\otimes k})$ equals `$h^0(X_\eta, \omega_{X_\eta}^{\otimes k})$'. $\endgroup$ Jul 17, 2012 at 12:21

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.