Timeline for Given a curve, under which condition is the set of gonal morphisms finite
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 31, 2011 at 19:08 | comment | added | roy smith | as an example generalizing rita's and confirming ulrich's comment, consider a smooth plane quintic, which has genus 6 and gonality 4, and project from points of the curve. | |
| Aug 22, 2011 at 13:13 | comment | added | naf | No. This will only be true for general curves, i.e. curves corresponding to some non-empty Zariski open subset of the moduli space $M_g$. | |
| Aug 20, 2011 at 17:08 | vote | accept | Ariyan Javanpeykar | ||
| Aug 20, 2011 at 13:14 | comment | added | Ariyan Javanpeykar | @ulrich: Did I understand correctly that your comment implies a positive answer to Question 3 if we stick to curves with even genus? | |
| Aug 20, 2011 at 13:02 | history | edited | Ariyan Javanpeykar | CC BY-SA 3.0 |
added 189 characters in body
|
| Aug 20, 2011 at 12:49 | comment | added | Ariyan Javanpeykar | @rita: i should have said i always mod out the set modulo the action of Aut$(\mathbf{P}^1)$. | |
| Aug 20, 2011 at 12:44 | comment | added | rita | A $g^1_3$ is a linear system on a curve of degree 3 and (projective) dimension 1. In general, a $g^r_d$ is a linear system of degree $d$ and dimension $r$, so the gonality $\gamma$ is the smallest $\gamma$ such that $C$ has a $g^1_{\gamma}$. | |
| Aug 20, 2011 at 12:41 | comment | added | naf | Generalising rita's example, a general curve of genus $g > 2$ with $g$ odd has a $1$-dimensional family of gonal morphisms; this can be seen by counting parameters. In contrast, for general curves of even genus there are only finitely many gonal morphisms. (See the book by Arbarello, Cornalba, Griffiths and Harris "Geometry of Algebraic Curves, Volume I" for more information.) | |
| Aug 20, 2011 at 12:39 | answer | added | Felipe Voloch | timeline score: 7 | |
| Aug 20, 2011 at 12:38 | comment | added | rita | I do not understand the difference between Q.1 and Q.4. If you do not divide by the action of $Aut({\mathbb P}^1$ then you always get infinitely many gonal morphisms. | |
| Aug 20, 2011 at 12:18 | comment | added | Ariyan Javanpeykar | thank you for this answer. What is $g_3^1$ stand for? I came across this notation a couple of times. Could you give a definition? I'm guessing $g_3$ means gonality three? What does the 1 on top mean? | |
| Aug 20, 2011 at 12:16 | comment | added | rita | About question 4: a smooth plane quartic $C$ has gonality 3 and has infinitely many $g^1_3$, corresponding to projecting to ${\mathbb P}^1$ from a point of $C$. | |
| Aug 20, 2011 at 10:38 | history | asked | Ariyan Javanpeykar | CC BY-SA 3.0 |