Timeline for A question about the dimension of a relatively ample divisor
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| Sep 5, 2017 at 13:05 | comment | added | diverietti | My comment add almost nothing to Jason's one, but indeed more generally, take any vector bundle $E\to Z$ or rank greater than one such that its symmetric powers (including the vector bundle itself) have no global sections. Now take $Y=\mathbb P(E)$ its projectivization. Then, the Serre (anti)tautological line bundle $\mathcal O_E(1)\to Y$ is relatively ample by construction. But $H^0(\mathbb P(E),\mathcal O_E(m))\simeq H^0(Z,S^m E)=\{0\}$. | |
| Sep 5, 2017 at 3:21 | comment | added | Jason Starr | Unfortunately that is not true. Let $Z$ be a smooth, projective, genus $0$ curve. Let $f:Y\to Z$ be a Hirzebruch surface fibered over $Z$ whose "directrix" $H$ has negative self-intersection, i.e., any Hirzebruch surface other than $\mathbb{P}^1\times \mathbb{P}^1$. The divisor class of $H$ is $f$-relatively ample, yet the only global sections of $\mathcal{O}_Y(m\underline{H})$ are scalar multiples of the section whose zero scheme is $m\underline{H}$. | |
| Sep 5, 2017 at 3:16 | history | asked | Li Yutong | CC BY-SA 3.0 |