Timeline for Surface in $\mathbb{P}^N$ covered by rational normal curves
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Jun 30, 2017 at 20:56 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| Jun 30, 2017 at 9:08 | vote | accept | Francesco Polizzi | ||
| Jun 29, 2017 at 19:14 | answer | added | Jason Starr | timeline score: 3 | |
| Jun 29, 2017 at 14:40 | comment | added | Francesco Polizzi | I see. I will have a look at Shen's thesis, thanks. | |
| Jun 29, 2017 at 14:38 | comment | added | Jason Starr | . . . In all of these examples, $N$ is no greater than $n^2$. Mingmin Shen's technique identifies a Zariski open neighborhood of a general $\mathcal{C}_p$ curve on a minimal desingularization of $\Sigma$ (abstractly) with a Zariski open neighborhood of a section curve, resp. line or conic, in a Hirzebruch surfaces, resp. projective plane. Then we can bound $N$ by the dimension of the corresponding linear system on this Hirzeburch surface / projective plane (blowing up at points and twisting down by exceptional divisors only decreases $N$). | |
| Jun 29, 2017 at 14:34 | comment | added | Jason Starr | Let $\Sigma$ be the image of the embedding of $\mathbb{P}^1\times \mathbb{P}^1$ into $\mathbb{P}^N$ by the complete linear system of $\mathcal{O}(a,b)$. Let $(a',b')$ be a pair of integers such that $1\leq a'\leq a$, $1\leq b'\leq b$ and $a'b$ equals $ab'$. Define $n$ to be $a+b+ab'=a+b+a'b$. Rational normal curves of degree $n$ in $\Sigma$ are images of curves of type $(1,1+b')$ or $(1+a',1)$. Any two points are connected by one of these. | |
| Jun 29, 2017 at 14:31 | comment | added | Francesco Polizzi | @Jason Starr. You mean the embedding of the quadric by the linear system $L+2M$, where $L$ and $M$ are the rulings? In that case, it seems to me that if we take two points in the same curve $M$ then the twisted cubic joining them will split and contain $M$, since $(L+2M)M=1$. Am I missing something? | |
| Jun 29, 2017 at 14:26 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| Jun 29, 2017 at 14:26 | comment | added | Jason Starr | The quadric surface example is fine if you take $n$ equal to $3$ (twisted cubics rather than plane conics). | |
| Jun 29, 2017 at 14:25 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
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| Jun 29, 2017 at 14:21 | comment | added | Francesco Polizzi | You are both right, thanks. I was thinking that "sufficiently general point" was enough for my pourposes, but I realized now that I really need any point. I edited the question accordingly. | |
| Jun 29, 2017 at 14:11 | comment | added | Jason Starr | Replacing "any" in property 2 by "sufficiently general" addresses the issue identified by @potentiallydense. My recollection is that there is a classification of these surfaces, and that they are roughly the rational surface scrolls together with the Veronese surfaces. I think that Proposition 2.3.3 of Mingmin Shen's thesis might be relevant (but I need to look it over more carefully): math.columbia.edu/~thaddeus/theses/2010/shen.pdf | |
| Jun 29, 2017 at 14:06 | comment | added | Lazzaro Campeotti | I think your first example for $n=2$ fails. Conics on the quadric surface are exactly the plane sections, but if $p$ and $q$ are on the same ruling of the quadric surface, then any plane containing both of them also contains the line of the ruling that joins them. So there is no smooth conic containing both. Am I making sense? | |
| Jun 29, 2017 at 13:53 | history | asked | Francesco Polizzi | CC BY-SA 3.0 |