Timeline for Hypersurfaces without variable cohomology
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 29, 2016 at 4:43 | comment | added | Honglu | Let us continue this discussion in chat. | |
| Jan 29, 2016 at 0:53 | comment | added | Will Sawin | @Honglu check this answer : ) | |
| Jan 29, 2016 at 0:12 | comment | added | Honglu | Right. That multiplicity bound is wrong. I just spelled out that proof from a picture I drew... But my point was that the intersection number of $C$ with exceptional divisor is pretty much the number (probably minus one?) that contribute to the decreasing of genus. But I couldn't find a handy reference to write a precise formula. Anyway, I'm really looking forward to see you argument with blowing up just two points! | |
| Jan 29, 2016 at 0:01 | history | edited | Will Sawin | CC BY-SA 3.0 |
added 1157 characters in body
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| Jan 28, 2016 at 23:11 | comment | added | Will Sawin | @Honglu I don't understand your multiplicity bound - I think $y^n- x^{n+1}$ becomes smooth when you blow up - but I see a way to do it just with intersection theory. Think it works already for blowing up 2 points, which I guess would be toric. I'll write that up soon. | |
| Jan 28, 2016 at 22:41 | comment | added | Honglu | Notice the intersection of $\overline C$ and the line is at least $\sum e_i$. So the arithmetic genus of $\overline C$ is at least $(\sum e_i-1)(\sum e_i-2)/2$. But the geometric genus of it is $0$. However, the difference between arithmetic genus and geometric genus can be calculated at the singularities $\{p_i\}$. Each of them should contribute a multiple of $e_i$ (forgot the formula). The point is, the difference between arithmetic/geometric genus should be linear in terms of $\sum e_i$. If $n$ is sufficiently large, it may not be possible. @Will Sawin | |
| Jan 28, 2016 at 22:41 | comment | added | Honglu | I think it's not true for blow-ups of $\mathbb P^2$. Let's blow up n points $p_1,…,p_n$ that are collinear in $X:=\mathbb P^2$ to get $\tilde X$. Let $E_1,\dotsc,E_n$ be the corresponding exceptional curves. Let $C\subset \tilde X$ be a smooth very ample rational curve. Its image in $X$ (denoted by $\overline C$) is a rational curve passing through $\{pi\}$. Each component near $\{p_i\}$ has multiplicity at most $2$. Since $C$ is very ample, $C$ must intersect each $E_i$ (say their intersection number is $e_i$). | |
| Jan 28, 2016 at 20:22 | comment | added | Will Sawin | @Honglu My suspicion is yes because the equation $D \cdot D + D \cdot K = -2$ becomes easier to solve when the Picard rank gets larger. But I don't know how to check which solutions are very ample. | |
| Jan 28, 2016 at 20:15 | comment | added | Honglu | Thanks. So the condition in the question is really strong even for surfaces. I'm also curious about whether it's true for rational surface. But it looks subtle even for blow-ups of $\mathbb P^2$. | |
| Jan 28, 2016 at 16:56 | history | answered | Will Sawin | CC BY-SA 3.0 |