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Let $C$ be a genus 2 curve over $\mathbb{C}$. Let $X=J(C)$. Consider the involution $i$ on $X$, $x\mapsto -x$. Let $Y=\frac{X}{(i)}$. This is a singular surface with 16 points of singularity - these are the images of the 16 2-torsion points of $X$. Let $f:X\longrightarrow Y$ be the quotient morphism, which is finite of degree 2.

Let $C$ be a curve on $X$, consider it's image $f(C)=C'$ in $Y$. Then what is the relation between $f^*\mathcal{O}_Y(C')$ and $\mathcal{O}_X(C)$? This is what I think they are:

1) if $C$ is not preserved by involution, $f^*\mathcal{O}_Y(C')=\mathcal{O}_X(f^{-1}(C'))=\mathcal{O}_X(C +i (C))=\mathcal{O}_X(C)\otimes\mathcal{O}_X(i(C))$.

2) if $C$ is preserved under involution, by the same argument, $f^*\mathcal{O}_Y(C')=\mathcal{O}_X(2C)$.

Are these right?

And if $C$ passed through the 16 2-torsion points, the $f|_C:C\longrightarrow C'$ is ramified over those 16 points. In that case also 1) and 2) hold?

Thanks in advance!

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1) is correct, 2) is not.

Indeed, if $i(C)=C$, then the map $C \rightarrow C'$ is a double cover, and $f^*{\mathcal O}_Y(C')={\mathcal O}_X(C)$ since in a neighbourhood of a general point of $C$ the map $X \rightarrow Y$ is biregular.

For a double cover $X\rightarrow Y$, in an analogous situation, you get $f^*{\mathcal O}_Y(C')={\mathcal O}_X(2C)$ when $C$ is contained in the branch locus. The point here is that $C$ is not just invariant, but a curve of fixed points. This is not your case, as your double cover has only finitely many branch points.

If $C$ passes through some of the nodes, of all of them, then ${\mathcal O}_Y(C')$ is no more necessarily a line bundle. Still, both statements still hold with the same proof.

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  • $\begingroup$ so in this situation, we will never get $f^*\mathcal{O}_Y(C')=\mathcal{O}_X(2C)$, since the branch locus is only finitely many points. Is that right? $\endgroup$ Nov 8, 2015 at 12:06
  • $\begingroup$ if $C$ passes through all the double points and preserved under involution, $C'$ is the quotient of $C$ by a finite group, hence $C'$ is smooth right? And therefore $\mathcal{O}_Y(C')$ is a line bundle. $\endgroup$ Nov 8, 2015 at 12:25
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    $\begingroup$ 1)Yes , you will never get $f^*{\mathcal O}_Y(C')={\mathcal O}_X(2C)$ if the map has finitely many branch points as in your case. $\endgroup$ Nov 8, 2015 at 12:39
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    $\begingroup$ 2) No. You are right that if $C$ is smooth then $C'$ is smooth. However smoothness has nothing to do with being Cartier. If $i(C)=C$, $C$ is smooth and passes through one of the $16$ points, then in a neighbourhood of the corresponding node of $Y$ $C'$ will NOT be befined by a single equation, and $C'$ will NEVER be Cartier. Think to a line $l$ through the vertex of a quadric cone $Q$; it is the same situation: $l$ is not a Cartier divisor of $Q$, just a Weil divisor (2-Cartier, to be precise). $\endgroup$ Nov 8, 2015 at 12:43

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