Timeline for Transversality of quadrics containing a projective curve
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Sep 23, 2016 at 15:06 | comment | added | Irfan Kadikoylu | Thanks a lot, I might need to talk to you later if I cannot figure out a solution to my problem. The Green-Lazarsfeld result that I mentioned is Proposition 2.4.2 in this paper: researchgate.net/publication/… | |
| Sep 21, 2016 at 16:29 | comment | added | meh | Roughly speaking a Clifford Index can be defined for any line bundle L and $Cliff(L) >1 $ is necessary for a curve to be defined by quadrics. When one projects from a point, that is going from $L \to L(-p) $ the Clifford index can go down by one. We can discuss in more detail off-line or you can look at my thesis (on the arxiv). | |
| Sep 21, 2016 at 16:26 | comment | added | meh | Further I think you are out of luck in the sense that the result you mention is sharp. Allow me to do 'approximate math' with fractions. Suppose C is a general curve of Clifford index $\frac{g-1}{2}$ computed by a divisor D of degree $\frac{g+3}{2} $. Pick $\frac{g-3}{2}$ points in D and project from them. The line bundle is of degree $\frac{3g-1}{2} $ and has a trisecant line (the other 3 points of D). If you demand your result be true for every line bundle of a given degree that is the best one can do. | |
| Sep 21, 2016 at 16:19 | comment | added | meh | I was referring to what you call the result of Saint-Donat, Fujita, and Green-Lazarsfeld. It was just meant to be just a comment since you asked 'is anything known' :). I don't have a reference for the Green-Lazarsfeld result. Can you provide one. I am familiar with the analagous statement for normal generation. | |
| Sep 21, 2016 at 9:40 | comment | added | Irfan Kadikoylu | Ok, actually you probably meant the result of Green, Lazarsfeld, which says that if the curve has no trisecant line and $d\geq 2g+2-2h^1(L)-Cliff(C)$ then it is cut by quadrics. The thing is, I need this result for generic curves of degree $=3g/2$ with $h^1=1$ and for generic curves, the bound in that theorem becomes (strangely enough) $d\geq (3g+1)/2$. It is not my lucky day, I guess! | |
| Sep 21, 2016 at 8:27 | comment | added | Irfan Kadikoylu | Thanks for the comment! You probably mean the result by Saint-Donat, Fujita, Green (and Lazarsfeld too I guess), which states that $\varphi_L(C)$ is cut out by quadrics if $d\geq 2g+2$. If you mean something different I would be happy to hear it! Moreover, this is an "absolute" statement in the sense that it is true for every curve and every line bundle. What I am curious about is if one can compromise on that by considering "only" generic curves and line bundles to improve the statement in the way that I described in my question (and possibly for lower degree range). | |
| Sep 20, 2016 at 19:39 | comment | added | meh | I don't recall off-hand who proved this, but if $2d \geq 2g+2$ , then for any line bundle L of degree d, one has that the base locus of $I_2(C,L)$ is exactly C. Frequently, the non-existence of a 3 pointed secant line is a sufficient (it's certainly necessary!) condition for C to be defined by the quadrics through it. | |
| Sep 20, 2016 at 14:56 | history | asked | Irfan Kadikoylu | CC BY-SA 3.0 |