Timeline for How many independent quadrics should one intersect to get the canonical curve.
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Nov 24, 2010 at 21:48 | comment | added | David Lehavi | @rita: thanks for the clarification | |
| Nov 23, 2010 at 21:28 | comment | added | rita | @David Lehavi: a canonical curve is the image via the canonical map of a curve of genus $g\ge 2$. So I can consider the canonical image $D$ of a smooth plane quintic $C$. Then $g(C)=6$ and $|K_C|$ is the restriction of $|O_{P^2}(2)|$. This means that the canonical map of $C$ is the restriction of the Veronese map $P^2\to P^5$ and $D$ lies on the Veronese surface. Let $Q(y_0,\dots y_5)$ be a quadric that vanishes on $D$: pulling $Q$ back to $P^2$ I get a quartic $F(x_0,x_1,x_2)$ that vanishes on $C$. By degree reasons, $F=0$, namely $Q$ vanishes on the Veronese surface. | |
| Nov 23, 2010 at 21:07 | comment | added | David Lehavi | @rita: I still don't undestand your remark on plane quintics: they are not canonical. | |
| Nov 23, 2010 at 20:45 | comment | added | rita | @David Speyer: I think that if one adds the equation $x_0x_{g-1}-x_1x_{g-2}=0$ to the ones you wrote, then you get the rational normal curve (set theoretically). | |
| Nov 23, 2010 at 19:35 | comment | added | David E Speyer | @rita Thanks, you are right. | |
| Nov 22, 2010 at 8:03 | comment | added | rita | @David: 1. I don't see how the statement can be correct. a) plane quintics: by adjunction, the bicanonical divisors are cut by quadrics of the plane. The canonical image $D$ of $C$ lies on the Veronese surface in $\pp^5$, which is the intersection of all the quadrics containing $D$. b) trigonal not hyperelliptic: if $A+B+C$ is in the $g^1_3$, then by geometric RR the points $A$, $B$,$C$ lie on a line in the canonical immage. Letting the divisor $A+B+C\in g^1_3$ vary one gets a ruled surface containing $D$ which is the intersection of all the quadrics containing $D$. | |
| Nov 22, 2010 at 7:44 | comment | added | David Lehavi | @rita 1. the statement in the question is correct for non-he curves as it is (and plane quintics are certainly not canonical) 2. Thanks, fixed it. 4. Maybe, but both answers are immaterial to the question. | |
| Nov 21, 2010 at 23:24 | comment | added | rita | @David Lehavi: 1) maybe I misunderstood. I thought that in your question you were claiming that the intersection of the quadrics through a canonical curve is the canonical curve itself, without excluding trigonal and plane quintic. 2) In your question you seemed to imply that the number of quadrics is the same for hyperelliptic, which is not true. 4)David Speyer equations don't seem to work: if you set all the variables equal to $0$ except the first and the last you get a line on which all the given quadrics vanish. | |
| Nov 21, 2010 at 21:26 | comment | added | David Lehavi | @rita: per your paragraphs: 1) you can do much better, see the reference I gave in the question - ACGH VI.4.1. 2) sure. 3) Obviously d > g-2, but can you give a sane upper bound on it ? e.g. d subquadratic in g would be nice - I have no idea if this is true. 4) see David Speyers comment to my question for g-2 quadrics that give the RNC set theoretically. | |
| Nov 21, 2010 at 20:30 | history | answered | rita | CC BY-SA 2.5 |