2
$\begingroup$

Let $X=Jac(C)$ be an abelian surface over $\mathbb{C}$, the Jacobian of a genus 2 curve. Let $L$ be a symmetric line bundle. Let $Y$ be the Kummer surface, quotient of $X$ by the action of involution. Then $L^2$ is totally symmetric, hence there is a line bundle $L'$ on $Y$ which pulls back to $L^2$. Further since $L^2$ embeds $Y$ in $\mathbb{P}^3$, $L'$ is very ample.

1) Is $L'$ the unique line bundle which pulls back to $L^2$?

2) Consider a curve $C\in |L^2|$, which is smooth, preserved under involution and avoiding the 16 double points of $X$. Then the image of $C'$ is smooth and avoids the 16 singular points of $Y$. Can we say that $C'\in |L'|$?

Thanks!

$\endgroup$
1
$\begingroup$

1) Yes. If there is another one, it differs from $L'$ by a line bundle $M$ with $M^{2}\cong \mathcal{O}_Y$. Consider the resolution $\pi :\hat{Y}\rightarrow Y$ obtained by blowing up the double points $p_1,\ldots ,p_{16}$. Since $\hat{Y}$ is simply connected, we have $\pi^* M\cong \mathcal{O}_{\hat{Y}}\ $.

Thus $M_{|Y\smallsetminus \{p_i\}}$ is trivial, and this implies that $M$ is trivial.

2) Yes by 1), since the pull-back of $\mathcal{O}_Y(C')$ to $X$ is $\mathcal{O}_X(C)=L^2$.

| cite | improve this answer | |
$\endgroup$

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.