I wonder if the following is known: Are there two compact curves C1 and C2 of genus>1 defined over complex numbers, such that their product contains infinite number of irreducible curves of negative self-intersection and arbitrary large genus? It we aks the same question replacing "negative" by "zero", the answer will be yes, moreover there will be examples of with C1 and C2 of any genus. These examples can be obtained as ramified covers ExE where E is an elliptic curve. 


(**PS.** The answers given to this question in 2009 did not solve it)



Curves with negative self intersection in the product of two curves
**Disclaimer.** The answer below is a variation of Bogomolov's argument and it would not come to be without Dmitri's answer. If you feel like upvoting this, please upvote his answer.


**Curves on products of isogeneous elliptic curves.** 
As already suggested in the body of the question, if we start with a pair of elliptic 
curves, say $E\_1$ and $E\_2$, admitting a non-constant morphism $f : E\_1 \to E\_2$ then
given any point $p \in X=E\_1 \times E\_2$ we have infinitely many elliptic curves with self-intersection zero on $X$ passing through $p$. It suffices to consider translates of the graphs of endomorphisms of $E\_2$ (there are at least $\mathbb Z$ of them) composed with $f$.


If we blow-up $p$ then we get a surface $S$ containing infinitely many (elliptic) curves
with negative self-intersection.


**Jacobians of genus $2$ curves.** As pointed out in Dmitri's answer the 
natural morphism
$$
\mathrm{Sym}^2 C \to \rm{Pic}^2(C) \cong \rm{Jac}(C)
$$
identifies $\mathrm{Sym}^2 C$ with the blow-up of $\rm{Jac}(C)$ at a point. 
Thus if 
we have a genus $2$ curve with Jacobian isogenous to the square of an elliptic curve
then the discussion in the previous paragraph shows that $C^2$ has infinitely many
curves of negative self-intersection since we can pull-back the 
negative curves on $\mathrm{Sym}^2 C$ through the natural morphism $C^2 \to \rm{Sym}^2 C$. 
Notice also that the negative curves have unbounded intersection with the diagonal $\Delta \subset C^2$. It is not hard to verify that the pull-backs of the negative elliptic curves to $C^2$ will have unbounded genus. 


**Explicit example.**
If $C$ is a genus $2$ curve admitting a morphism
$\pi : C \to E$ to an elliptic curve $E$ then $\rm{Jac}(C)$ is isogeneous to the product of $E$ with another
elliptic curve $E'$ ( the connected component of the kernel of $\pi$ through zero). 
Automorphisms of $C$ act naturally on $\rm{Jac}(C)$. If there is an element  

$\varphi \in \mathrm{Aut}(C)$ with induced action on $\rm{Jac}(C)$ not preserving $E'$ then $E$ is isogeneous to $E'$ since $$\pi\_\* \circ \varphi\_\* : \rm{Jac}(C) \to \rm{Jac}(E)\cong E$$ restricted to $E'$ is an isogeny. Therefore $\rm{Jac}(C)$ is isogeneous to the square of $E$.


To have a concrete example we can take $C = \lbrace y^2 = x^6 - 1\rbrace$ which maps to $E =\lbrace y^2 = x^3 -1\rbrace$ and has automorphism group isomorphic to $\mathbb Z\_3 \rtimes D\_8$ (which is not the automorphism group of any elliptic curve). From the discussion above
it follows that $C^2$ has infinitely many curves of negative self-intersection and unbounded genus. 


**Question.** Suppose $C$ is genus $2$ curve such that $C^2$ contains infinitely many
curves of negative self-intersection. Is the Jacobian of $C$ isogeneous 
to the square of an elliptic curve ?


