Timeline for Hilbert scheme of curves in a degree $d$ surface in $\mathbb{P}^3$
Current License: CC BY-SA 3.0
12 events
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| Sep 21, 2012 at 9:07 | comment | added | Naga Venkata | @quim: As far as I understand Bertini's theorem is applicable to hyperplane sections. But $C$ does not need to be a complete intersection in $X$. Could you elaborate/give a reference on how to use negative self-intersection number of irreducible components to prove that there is no moving part of $|C|$? | |
| Sep 19, 2012 at 9:34 | comment | added | quim | By Bertini (at least if the surface is nonsingular) the moving part of |C| is irreducible, which combined with your statement that the selfintersection of every component is negative would give that there is no moving part. However, there are a couple of things to check: 1) that the general surface containing C is nonsingular, 2) that the components of every element D∈|C| also have negative selfintersection. I don't know how you argue this selfintersection, and how the argument interacts with the choices of general C -> general S -> general D | |
| Sep 19, 2012 at 3:44 | comment | added | Naga Venkata | @Huizanga: I apologize for my inaccuracy in my question. As it is clear from my last sentence in the question that the curves I am concerned with have self-intersection number negative which would imply trivial linear series. This happens when $d \ge 5$ and $d \ge e+2$. | |
| Sep 19, 2012 at 3:42 | history | edited | Naga Venkata | CC BY-SA 3.0 |
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| Sep 19, 2012 at 1:02 | comment | added | Jack Huizenga | If I'm not mistaken then looking at twisted cubic curves on cubic surfaces gives another counterexample. Thinking of such a surface as a blowup of $\mathbb{P}^2$ at six points, the image of any line missing the points is a twisted cubic, and varying the line shows it moves in a linear family. All that's left to do is show that if we fix a twisted cubic then the general cubic surface containing it contains it as a curve of class, but that doesn't seem unreasonable to me (I haven't checked, though). | |
| Sep 18, 2012 at 20:53 | comment | added | Naga Venkata | @quim: I have edited the question to non-plane curves | |
| Sep 18, 2012 at 20:52 | history | edited | Naga Venkata | CC BY-SA 3.0 |
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| Sep 18, 2012 at 20:51 | comment | added | user5117 | The hilbert-space tag didn't fit here, so I removed it. | |
| Sep 18, 2012 at 20:50 | history | edited | user5117 |
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| Sep 18, 2012 at 20:29 | comment | added | quim | Maybe you want to assume the curves are nondegenerate, ie, not contained in planes? | |
| Sep 18, 2012 at 20:27 | comment | added | quim | e=1, d=2 gives a counterexample. | |
| Sep 18, 2012 at 17:00 | history | asked | Naga Venkata | CC BY-SA 3.0 |