4
$\begingroup$

Let $X$ be a smooth projective complex curve of genus 2 with canonical divisor $K$. $X$ of course is hyperelliptic and has an involution that I denote by $j$.

There exists 3 possibilities for divisors of degree 4 on $X$, namely $2K$, $K+p+q$ where $p$ and $q$ are distinct points not identified by $j$ and $K+2p$ with $j(p)\neq p$.

I'm looking for a nice geometric interpretation and construction of the graded rings associated to each of these divisors, the case $2K$ is easily described in terms of the geometry of the pluricanonical ring but I had difficulties with the other two cases.

Take for instance $D=K+p+q$. By Riemann-Roch $D$ defines a birational map $\varphi_D$ whose image in $\mathbb{P}^2$ is a quartic with a node at $\varphi_D(p)=\varphi_D(q)$. If I start giving a basis in degree one, say $x,y,z\in H^0(X,D)$, I need at least one extra generator $t$ in degree 2 because there are only 6 quadratic monomials in three variables. Going on to degree 3 we get 13 monomials and Riemann Roch says $\dim H^0(X,nD)=4n-1$ So there are 2 relations in degree 3. If I'm right, these should involve only $x,y,z$ ($t$ can not be a rational function of $x,y,z$, because $2D$ is very ample). This means that the nodal curve passes through the $9$ points of intersection of these 2 cubics in the projective plane? Is there some practical method of determining these relations so we can derive an explicit equation for the quartic nodal curve?

This is the first time that I play with these graded ring stuff, just trying to gain some geometric intuition with explicit examples, so please forgive me if I wrote something that makes no sense. Any helpful comment will be greatly appreciated!

$\endgroup$
2
$\begingroup$

John Voight and I just wrote a long monograph generalizing this question. See chapter 4, which works out all of the possible logarithmic cases. (We don't do the case of K + 2p, because it isn't "logarithmic", but it is similar.)

http://arxiv.org/pdf/1501.04657.pdf

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.