Skip to main content

Timeline for Double dual of ample sheaf

Current License: CC BY-SA 3.0

2 events
when toggle format what by license comment
Jul 1, 2016 at 7:43 comment added Chieh LIU Finally, I find that this may be not true in general. If $\mathcal F$ is of rank $1$, the double dual of $\mathcal F$ is just the determinant bundle $\det(\mathcal F)$. Moreover, there is a closed subscheme $Z\subset X$ of codimension $\geq 2$ such that $\mathcal F= \mathbb P(\det(\mathcal F))\otimes \mathcal I_Z$. Since $\mathbb P(\mathcal F)$ is just the blowing-up of $\pi\colon Bl_Z X\to X$, if $Z$ is smooth, ampleness of $\mathcal F$ just means $\pi^*\det(\mathcal F)-E$ is ample, where $E$ is the exceptional divisor. In general, this doesn't imply that $\det(\mathcal F)$ is ample.
Jun 29, 2016 at 10:48 history asked Chieh LIU CC BY-SA 3.0