22
$\begingroup$

We know that each genus 2 curve is embedded into its degree 1 Jacobian.

Under which conditions on $C$, $A$, $g$ and $n$ is it possible for a genus $g$ smooth curve $C$ to be embedded in an Abelian variety $A$ of dimension $n$?

What can be said in the case $n=2$?

And what if $A$ is a Jacobian (and e.g. $n=2$)?

Improve this question
$\endgroup$
8
  • 9
    $\begingroup$ If an abelian variety is simple, it cannot contain a curve of genus smaller than its dimension: this follows easily by considering the restriction of global differential forms. If an abelian variety of dimension g contains a curve of genus g, then the abelian variety is isogenous to the jacobian of the curve: this follows easily by using the fact that the Jacobian of the curve is also the Albanese variety of the curve. $\endgroup$
    – damiano
    Commented Apr 15, 2010 at 10:01
  • 1
    $\begingroup$ (even though this is not clear from what i wrote above, you need the abelian variety to be simple also in the second part.) $\endgroup$
    – damiano
    Commented Apr 15, 2010 at 10:09
  • 1
    $\begingroup$ @damiano: even if it doesn't completely answer to the question (I don't know if it's actually possible to have a complete answer), it's a valuable comment: you could have posted it as an answer. $\endgroup$
    – Qfwfq
    Commented Apr 15, 2010 at 10:57
  • 2
    $\begingroup$ If a curve C maps to an abelian variety A, then the Jacobian J(C) also maps to A. This is another way of realizing damiano's remark. It also suggests how one might try to find curves on A in general. $\endgroup$
    – mdeland
    Commented Apr 15, 2010 at 13:29
  • 1
    $\begingroup$ You might be interested in Ernst Kani's paper "Elliptic curves on abelian surfaces". $\endgroup$
    – Dan Petersen
    Commented Apr 15, 2010 at 21:27

4 Answers 4

Reset to default
15
$\begingroup$

For any $g >0$ there exists an abelian surface $A$ containing a smooth curve $C$ of genus $g$; the surface can be assumed to be simple if $g > 1$:

For $g=1$, one can just consider $C \times C$. For $g>1$, consider the generic abelian surface with a polarisation of type $(1, g-1)$. By the adjunction formula, a smooth curve in the linear system corresponding to the polarisation will have genus $g$. (Such abelian surfaces can be constructed by considering suitable quotients of principally polarised ones ($n=1$))

One gets abelian varieties of any dimension $>1$ by taking products. However, this does not produce a fixed abelian surface containing curves of all genera $g>1$ but it seems likely that such surfaces should exist. It would be interesting to know whether this is possible for the generic principally polarised surface, or more generally ppav of any dimension $n$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Regarding your last paragraph: if A is an abelian surface with End(A) = Z, then it contains smooth curves of very few genera. NS(A) is generated by the polarization L in this case, and the genus of a smooth curve in |nL| is n^2(L.L)/2 - 1. The abelian surfaces you ask about do exist however: certain (but not all) products of elliptic curves will do the trick. $\endgroup$
    – Ari Shnidman
    Commented Jun 26, 2016 at 2:30
8
$\begingroup$

Gian Pietro Pirola in [Curves on generic Kummer varieties, Duke Math. J. Volume 59, Number 3 (1989), 701-708] proves a rigidity theorem for curves of genus $g\le q-2$ in the Kummer variety of a $q$-dimensional abelian variety. As a consequence, he shows that a generic abelian variety of dimension $q\ge 3$ does not contain hyperelliptic curves of any genus.

Improve this answer
$\endgroup$
4
$\begingroup$

In his paper "Abelian surfaces with (1,2)-polarization" Barth proves that a smooth genus $3$ curve can be embedded into an Abelian surface if and only if it admits an elliptic involution, i.e. a map of degree $2$ onto an elliptic curve.

Improve this answer
$\endgroup$
1
$\begingroup$

If $C$ has a map onto a smooth curve $B$, there is a surjective map $J(C)\to J(B)$ hence $C$ maps to $J(B)$ and it is probably often an embedding.

Improve this answer
$\endgroup$
1
  • 9
    $\begingroup$ The map from $C$ to $J(B)$ factors through the map from $C$ to $B$ and hence is almost never an embedding. $\endgroup$
    – Torsten Ekedahl
    Commented Apr 15, 2010 at 15:38

You must log in to answer this question.

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