3
$\begingroup$

Let $A$ be an abelian variety over a field $k$ of dimension $g\geq 2$.

There exists a finite morphism $A\to \mathbf{P}^g_k$. Here's the question.

Does there exist a finite morphism $A\to \mathbf{P}^g_k$ of degree two?

Can we say something about the minimal degree of a finite morphism $A\to \mathbf{P}^g_k$?

Improve this question
$\endgroup$
1
  • 5
    $\begingroup$ A better analogue of the degree 2 map $E\to\mathbb{P}^1$ for an elliptic curve is the degree 2 map $A\to A/\pm1$ from $A$ to the associated Kummer variety. (Of course, you need to blow up some points if you want a smooth quotient, and then then map is only rational, it's not a morphism.) $\endgroup$
    – Joe Silverman
    Commented Aug 1, 2012 at 14:05

1 Answer 1

Reset to default
11
$\begingroup$

For a very general, principally polarized Abelian variety $(A,\Theta)$ of dimension $g$ over $\mathbb{C}$, every Cartier divisor $D$ on $A$ is numerically equivalent to $m\Theta$ for some integer $m$. In particular, the intersection number $D^g$ is $m^g \Theta^g$. So the minimal degree of an effective, nonzero divisor is $g!$, not $2$.

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ Your argument works for any abelian variety of dimension at least $3$: by Riemann-Roch, $g!$ divides $D^g$ for any ample divisor $D$. One can also eliminate the $g=2$ case using the fact that any double cover of $\mathbb{P}^2$ with trivial canonical bundle is a $K3$ surface. $\endgroup$
    – naf
    Commented Aug 1, 2012 at 13:33
  • $\begingroup$ @ulrich -- I agree. $\endgroup$
    – Jason Starr
    Commented Aug 1, 2012 at 14:35
  • $\begingroup$ How to prove that any $\mathbb{P}^n$ is finitely covered by an Abelian variety of the same dimension? Maybe it is elementary, but I can not see it. $\endgroup$
    – Tong
    Commented Jun 20, 2016 at 22:39

You must log in to answer this question.

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