This is related to a question that I already asked [Curve through the 16 singular points of a Kummer surface](https://mathoverflow.net/questions/222057/curve-through-the-16-singular-points-of-a-kummer-surface/222068#222068). I am new to Abelian varieties and I am trying to understand more about them. 


Let $X=J(C)$ be the Jacobian of a genus 2 curve over $\mathbb{C}$, so $X$ is an abelian surface. We can choose an embedding of $C$ in $X$ such that the hyperelliptic involution on $C$ extends to the involution on $X$. Consider $L=\mathcal{O}(nC)$, where $n$ is an even positive integer. Then $L$ is a totally symmetric line bundle. So $L=f^\*L'$ where $f:J(C) \longrightarrow K(C)$ is the quotient map from the Jacobian to the associated Kummer surface, and $L'$ is an ample line bundle on $K(C)$. 


Consider a general smooth curve $C'\in |L|$ through all sixteen 2-torsion points. Let $C''$ be its image in $K(C)$, this passes through all sixteen singular points of $K(C)$.


a) For a general smooth $C'$ as above, will the restriction of $f$ from $C' \longrightarrow C''$ be a double cover, or an isomorphism or what?


b) For a general smooth $C'$ as above, is the image $C''$ smooth?


c) When is such a $C'$ isomorphic to its image $C''$?


Thanks in advance for the help!



A curve in an abelian surface and its image in the Kummer surface
Let $b:\hat{X}\rightarrow X$ the blowing up of the 16 2-torsion points, and $E\_1,\ldots ,E\_{16}$ the exceptional $(-1)$ curves on $\hat{X}$. The involution $x\mapsto -x$ lifts to an involution $\sigma $ of $\hat{X}$. 


You are looking at the linear system $|nb^\*C|-\sum E\_i$. The involution $\sigma $ acts on $H=H^0(\hat{X}, \mathcal{O}(nb^\*C-\sum E\_i))$, which splits as $H^+\oplus H^-$. The fixed curves of the system are those defined by elements of $H^+$ or $H^-$. Using for instance the holomorphic Lefschetz formula, one sees that these two subspaces are nonzero for $n\gg 0$. Thus a general $C'$ in the system is **not** fixed by $\sigma $, and therefore maps birationally to the curve $C''\subset K(C)$. However that map is **not** an isomorphism : there are pairs of points of $C'$ which map to the same point of $C''$, namely those which lie also in $\sigma (C')$. Since $\sigma $ acts trivially on $H^2(\hat{X})$, we have $C'\cdot \sigma (C')= (C')^2=2n^2-16$, so there are plenty of such points. From general principles it should be easy to show that for $C'$ general the intersection consists of $2n^2-16$ distinct points, so that $C''$ has $2n^2-16$ ordinary double points and no other singularities, and $C'\rightarrow C''$ is its normalization.


