Let $\pi:X \longrightarrow C$ be a smooth projective morphism onto a smooth projective curve, and $F$ be a central fiber. If the Kodaira dimension $\kappa (F)$ is nonnegative, is $\pi_{\ast} \mathcal O_X (k K_{X/C})$ nonzero for sufficiently divisible $k$? If it is, can anyone arrange an algebraic proof?

$\begingroup$ I think the question is equivalent to say that, if some central fiber is of nonnegative Kodaira dimension, is $X$ of nonnegative Kodaira dimension? $\endgroup$– Hu ZhengyuJul 13, 2012 at 20:07

1$\begingroup$ No, consider the case X=P^1 x C, where C is a curve of genus at least 1, and the map is projection onto P^1. $\endgroup$– user5117Jul 13, 2012 at 20:30

$\begingroup$ I think he means that $X/C$ is of nonnegative relative Kodaira dimension. $\endgroup$– Karl SchwedeJul 13, 2012 at 23:54
1 Answer
I think one can argue as follows. (Let me know if I made a mistake!)
Choose $k$ large enough so that $kK_F$ has a nonzero global section for some fibre $F$. For any fibre $F$, we have $K_F = K_{X/C}~_{F}$. So deformation invariance of plurigenera says that the function
$$h^0(X_p, k K_{X/C}~_{X_p})$$ is constant on $C$ (where now $X_p$ denotes the fibre over the point $p \in C$). Therefore by Grauert's theorem (Hartshorne III.12.9) any global section of kK_F must come from the sheaf $\pi_* O_X(kK_{X/C})$. Since we have chosen $k$ large enough that $kK_F$ has a nonzero global section, the sheaf $\pi_* O_X(kK_{X/C})$ must be nonzero.
Presumably this doesn't satisfy your requirement of an algebraic proof, because it use deformation invariance of plurigenera, which currently only has an analytic proof in general.

$\begingroup$ Is it possible to give an algebraic proof? I guess this problem is easier than invariance of plurigenera. $\endgroup$ Jul 13, 2012 at 20:49

$\begingroup$ You're right; I edited to make the proof simpler. $\endgroup$– user5117Jul 13, 2012 at 21:10

$\begingroup$ Artie, I'm confused about the new proof. Suppose the section $F$ lies over $0 \in C$. How do you know that $0 \in U$? $\endgroup$ Jul 13, 2012 at 23:53

$\begingroup$ Karl: I'm not sure I understand your comment. If a divisor lies over a point in C, then its restriction to the generic fibre will be trivial. Let me know if I'm missing your point. $\endgroup$– user5117Jul 14, 2012 at 20:46

$\begingroup$ Artie, maybe I'm being dumb with terminology, you are saying that $F$ is the generic fiber? I somehow thought that $F$ was supposed to be a closed fiber (the original questioner called it a central fiber, which I interpreted as a fiber over a special closed point). $\endgroup$ Jul 14, 2012 at 23:50