Timeline for The locus of rational/elliptic curves on a special surface in $\mathbb{P}^3$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 10, 2014 at 5:37 | comment | added | mehdi | Dear Will, in my case $P$ and $Q$ are as follows: $P(x)=(x-a_1)\cdots (x-a_n)$ and $Q(x)=(x-\overline{a}_1)\cdots (x-\overline{a}_n)$ where $a_i$ are general complex numbers. I think, I could derive contradiction if we could prove a certain proportion of the zeros of P(f) and Q(g) are single roots. I mainly apply complex analysis specially nevanlinna theory. | |
| Sep 9, 2014 at 7:29 | comment | added | mehdi | I mean that if $P(x)=(x-a_1)\cdots(x-a_n)$ then want to see that all roots of the equation, for example,$f(z)=a_1$ are of multiplicity 1. | |
| Sep 9, 2014 at 6:31 | comment | added | mehdi | thanks Will, but by applying Riemann-Hurwitz to the map $P(f):\mathbb{P}^1 \rightarrow \mathbb{P}^1$ we have: $2g_{\mathbb{P}^1} -2=(2g_{\mathbb{P}^1}-2)deg(P(f))+\sum_{p\in\mathbb{P}^1}(e_p -1)$ so we have $-2=-2d +\sum (e_p-1)$ where $d=deg(P(f))$. How we can deduce from the above that there are simple roots? | |
| Sep 8, 2014 at 20:34 | answer | added | Remke Kloosterman | timeline score: 1 | |
| Sep 8, 2014 at 19:21 | comment | added | Will Sawin | Using Riemann-Hurwitz you can show a certain proportion of the zeros of $P(f)$ and $Q(g)$ must be single roots, which means a certain proportion of the zeros of $P(f)$ must be zeros of $Q(g)$ and vice versa. Here I'm thinking of $f,g$ as just rational functions on a curve. One could try to derive a contradiction from that, but I don't immediately see how. | |
| Sep 8, 2014 at 12:33 | comment | added | mehdi | Yes, you are right, this is the trivial part. But do you have any idea about this problem? As I mentioned in above, how to prove that the above functional equation has no non-constant solution. Or maybe you want to think by using the algebrogeometric tools to solve the problem. | |
| Sep 8, 2014 at 10:40 | comment | added | Jason Starr | "So if we assume this conjecture it is enough to prove ..." Actually, you are only using the direction that is known (and trivial). Every nonconstant rational map from an Abelian variety induces (many) entire curves whose images sweep out the image of the Abelian variety. So the union of (nonconstant) images of Abelian varieties is contained in the union of (nonconstant) images of entire maps. | |
| Sep 8, 2014 at 4:42 | review | First posts | |||
| Sep 8, 2014 at 7:15 | |||||
| Sep 8, 2014 at 4:37 | history | asked | mehdi | CC BY-SA 3.0 |