-2
$\begingroup$

Let $C$ be a smooth projective curve of genus $2$ and $J$ denotes the Jacobian of $C$. Let $\theta$ be the image of $C$ under the abel Jacobi map. Is there exist a divisor $D$ in $J$ such that $D.\theta=1$?

Is it true that the condition $D.\theta=1$ implies that $J $ biregular to $ D\times \theta$?

$\endgroup$
  • 2
    $\begingroup$ Assuming $g(C)>1$, you can not have $J=D\times \theta$, because an abelian variety (such as $J$) does not dominate any curve of genus at least two. $\endgroup$ – Ariyan Javanpeykar Aug 11 at 11:50
  • $\begingroup$ @AriyanJavanpeykar I have modified question now for genus $2$ case for simplicity. $\endgroup$ – PSUN Aug 11 at 12:39
  • 1
    $\begingroup$ There can be no isomorphism $J = D\times \theta$. If there is, then there would be a surjective morphism $J\to \theta$, but $\theta \cong C$, so $J$ would dominate the genus two curve $C$. That is impossible. $\endgroup$ – Ariyan Javanpeykar Aug 11 at 15:38
  • $\begingroup$ Is it true that the condition $D.\theta=1$ implies that $J $ biregular to $ D\times \theta$? $\endgroup$ – PSUN Aug 12 at 5:51
  • 2
    $\begingroup$ Please read Ariyan's answer. If an abelian surface is isomorphic to a product, both factors are elliptic curves. $\endgroup$ – abx Aug 12 at 7:30
3
$\begingroup$

I think I understand your question now: you want your $D$ to be effective, hence irreducible since $\theta $ is ample. Now the index theorem $(D\cdot \theta )^2\geq D^2\cdot\theta ^2\ $ implies $D^2=0$, hence $D$ must be a smooth elliptic curve. Consider the quotient map $p:J\rightarrow J/D$. The condition $D\cdot \theta=1 $ means that the restriction of $p$ to $\theta $ is one-to-one, which is impossible since $g(\theta )=2$.

$\endgroup$
  • $\begingroup$ Thanks. Is there exist any map from $J\to D\times \theta$? Just by using the fact $D.\theta=1$, where $D$ is effective divisor. I am very sorry if I am asking very silly question. $\endgroup$ – PSUN Aug 12 at 11:30
  • 1
    $\begingroup$ No, there is no nontrivial map from an abelian variety to a curve of genus $\geq 2$ (as Aryan said already). $\endgroup$ – abx Aug 12 at 11:54

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.