Timeline for Complex curves covered by smooth plane curves
Current License: CC BY-SA 3.0
10 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jan 9, 2013 at 11:17 | history | edited | Michael Zieve | CC BY-SA 3.0 |
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| Jan 7, 2013 at 10:38 | comment | added | aglearner | Sorry in the last comment I meant that degree four curves in $\mathbb CP^2$ form a Zariski OPEN (not only dense) subset in $M_3$. | |
| Jan 7, 2013 at 8:46 | comment | added | aglearner | Michael, I don't see how you would use in your reasoning the fact that genus $g$ curve is generic. If you remove this assumption, it is very easy to construct a counterexample to that fact that you have assumed. Namely, the space of genus $3$ curves that admit a morphism to an elliptic curve is Zariski dense (in the moduli space $M_3$ of all genus $3$ curves), but it is by no means open. Since degree four curves in $\mathbb CP^3$ from a Zariski dense subset of $M_3$ we see a contradiction. | |
| Jan 7, 2013 at 7:13 | comment | added | Michael Zieve | Really? Then I take it back. I thought it would be straightforward to get that, if there were a morphism from a smooth plane curve of degree $d$ to a generic genus-$g$ curve, then there is a family of morphisms from a Zariski-open subset of degree-$d$ plane curves to genus-$g$ curves. If that's not true, then I was wrong. | |
| Jan 7, 2013 at 6:47 | comment | added | Serge Lvovski |
@Michael: In principle it is possible that each curve og genus $g$ can be covered by a plane curve, but these mappings do not "glue" into a dominant rational mapping from the space of plane curves of a given degree onto $M_g$.
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| Jan 7, 2013 at 4:22 | comment | added | aglearner | After some thinking I don't see any (non-superficial) relation between the question you cite and my question, so I would like to ask you give details on how you want to deform $C'$ with its morphism. | |
| Jan 6, 2013 at 23:45 | comment | added | aglearner | Dear Zieve I would like to know why you can deform $C'$ in the way you descibe. I don't see this. | |
| Jan 6, 2013 at 23:33 | history | edited | Michael Zieve | CC BY-SA 3.0 |
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| Jan 6, 2013 at 23:05 | history | answered | Michael Zieve | CC BY-SA 3.0 |