2
$\begingroup$

I have the following question. It's a long shot, but worth the try.

Let X be a compact connected Riemann surface of genus $g\geq 2$. Does there exist an effective divisor $D$ on $X$ of degree $g$ such that, if $D=\sum_{i=1}^g D_i$ is the prime decomposition of $D$, the image of the line bundle $\mathcal{O}_X(D_i-D_j)$ is torsion in the Picard group of $X$ for all $i,j = 1,\ldots,g$?

(Note that the $D_i$ are not necessarily distinct.)

The answer is yes for modular curves by Manin-Drinfeld. Namely, take $D$ to be a degree $g$ divisor supported on the cusps. The answer is also yes for Fermat curves.

The answer would be no if there would exist a Jacobian variety without torsion. Fortunately, Jacobians have a lot of torsion.

If it helps assume that $X$ can be defined over some number field (as an algebraic curve).

If the answer to my question is not completely trivial, there should be some general theory about such divisors. If yes, does there exist a good reference?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ What is to stop $D = gP$ for some point $P\in X$? Also, is $\mathcal O_X(D_i-D_j)$ supposed to be torsion for every $i,j$? $\endgroup$
    – Jack Huizenga
    Commented Dec 14, 2011 at 17:38
  • $\begingroup$ You're right. I wanted to exclude this example. And yes, I want $\mathcal{O}_X(D_i-D_j)$ to be a torsion element in the Picard group for all $i,j=1,\ldots,g$. Moreover, the $D_i$ are points (with multiplicity 1). $\endgroup$
    – Ariyan Javanpeykar
    Commented Dec 14, 2011 at 18:18
  • 2
    $\begingroup$ It is not true that the answer is always yes for modular curves : for example, there could be only one cusp. In this direction Matthew Baker has proved that if $N>479$ is prime then there is no torsion packet at all on the modular curve $X_0^+(N) = X_0(N)/W_N$ (see his article Torsion points on modular curves). $\endgroup$
    – François Brunault
    Commented Dec 14, 2011 at 22:42

1 Answer 1

Reset to default
4
$\begingroup$

Are the $D_i$ supposed to be points? This is not made clear in the question. If the $D_i$ are points and you exclude Jack's example where they are all the same, then the answer is no for the general curve. That is because, for the general curve (of genus at least three) $P-Q$ is not torsion for any pair of distinct points of the curve. I don't have a reference for this but it's not very hard.

Improve this answer
$\endgroup$
9
  • $\begingroup$ To answer your question: yes the $D_i$ are points. Sorry for being unclear about thas. Say we fix $g$. Do I understand correctly that the answer to my question is no for all but finitely many curves of genus $g$? $\endgroup$
    – Ariyan Javanpeykar
    Commented Dec 14, 2011 at 18:22
  • $\begingroup$ The answer is no for curves on the complement of a countable union of proper subvarieties of the moduli space of curves for each genus. There could be infinitely many curves of some genus for which the answer is yes. $\endgroup$
    – Felipe Voloch
    Commented Dec 14, 2011 at 18:32
  • $\begingroup$ I understand. But I don't see how to prove this easily. Fixing a point Q on X, the claim is that if X is general of genus at least three, we have that the image of C in the Jacobian via the map $P\mapsto P-Q$ contains no torsion elements...What kind of argument should I look for? $\endgroup$
    – Ariyan Javanpeykar
    Commented Dec 14, 2011 at 19:39
  • $\begingroup$ For any given $n$, the set of curves that has distinct points $P,Q$ with $P-Q$ of order $n$ is closed in moduli, that's more or less clear. Now you have to find, for each $n$, a curve not in this closed subset and conclude that this closed subset is not the whole moduli space. Off the top of my head, I don't remember all the details. $\endgroup$
    – Felipe Voloch
    Commented Dec 14, 2011 at 19:59
  • 1
    $\begingroup$ If you take X hyperelliptic then any divisor that is the sum of distinct Weierstrass points has the property you ask for. Letting X vary, you get a family of dimension 2g−1 of curves with this property. $\endgroup$
    – rita
    Commented Dec 15, 2011 at 14:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .