Timeline for Movable Divisors
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 5, 2015 at 5:00 | comment | added | user47305 | If you only blow up one, it's linearly equivalent to any other plane through the point, which may or may not contain the line... | |
| Jan 5, 2015 at 4:12 | comment | added | Allen Knutson | Oh indeed. The thing I was missing was that $D$ moves, but its deformations all intersect in that curve. (Not sure why you blew up two points instead of one.) | |
| Jan 5, 2015 at 1:10 | answer | added | Sándor Kovács | timeline score: 4 | |
| Jan 4, 2015 at 23:21 | comment | added | user47305 | I'm not sure I understand. What if $X$ is $\mathbb P^3$ blown up at two points, and $D$ the strict transform of a plane through the points? Its deformations cover $X$, but there's a curve in the base locus. | |
| Jan 4, 2015 at 23:12 | comment | added | Allen Knutson | I don't see the complete argument, but I'd like to say that since $X$ is irreducible and $D$ moves, that its deformations cover $X$, and hence the linear system has no basepoints. Then use Bertini on the image of the map to projective space, to show that the general deformation of $D$ is reducible, absent the case $\dim X = 1$. | |
| Jan 4, 2015 at 23:10 | comment | added | Allen Knutson | If you relax "divisor", you can take $D$ to be a union of two planes in $\mathbb P^4$. | |
| Jan 4, 2015 at 22:05 | answer | added | Jérémy Blanc | timeline score: 4 | |
| Jan 4, 2015 at 22:04 | answer | added | user47305 | timeline score: 2 | |
| Jan 4, 2015 at 21:41 | history | asked | user56259 | CC BY-SA 3.0 |