Curves with negative self intersection in the product of two curves

Question

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)

Answer 1

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 ?

Answer 2

He is a construction of such a product, I would like to thank Fedor Bogomolov, for providing the answer.

Construction. It is well known that the symmetric square $S^2(C)$ of a curve $C$ of genus $2$ is an Abelian surface blown up at one point (the canonical divisor of $C$). Consider the degree $16$ cover of $S^2(C)$ corresponding to the sub-lattice $2\mathbb Z^4$ in $H_1(C,\mathbb Z)$. The quotient of this surface by an involution is a Kummer surface, and hence it has infinite number of $-2$ rational curves. Let $C'$ be the degree $16$ cover of $C$ (again corresponding to the subgroup $2\mathbb Z^4$ in $H_1(C,\mathbb Z)$). Then one can check that there is a map $C'\times C'$ to the Kummer surface. $C'\times C'$ is the surface we are looking for.

Answer 3

It seems likely to me that the (graphs of) Hecke operators on the self-product of a modular curve have this property. This might be a little hard to verify because of the cusps, so it is better to work with suitable Shimura curves (quotients of the upper half plane by a torsion free arithmetic subgroup of an indefinite rational quaternion algebra).

In the case of Shimura curves one gets a curve C of genus > 1 (lots of them in fact) and infinitely many curves Gamma_i in C \times C such that both projection maps from Gamma_i to C are finite and etale. This shows that the self intersection of each Gamma_i is negative. The degrees of these maps go to infinity, hence so does the genus of the Gamma_i.

(In the case of the usual modular curves the projection maps are not etale which is what makes the computation of the self-intersection more difficult.)

THIS DOESN'T WORK! (Sorry.)

The problem is that even though we get curves Gamma_i with two (distinct) etale maps to C (a Shimura curve, say) the image in C x C might be singular, so the self-intersection could well be positive. For the case of modular curves this is in fact the case as may be seen by reducing the mod p. This suggests that the self-intersection numbers are also positive for Shimura curves.

Answer 4

Regarding JSE's idea: The appropriate vector space is H^{1,1}, not H^2, since we are dealing with classes of curves. And the bilinear form is not the Euclidean form, but has signature (1,k), by the Hodge index theorem.

As I understand it, JSE's idea is that it should be impossible to have infinitely many vectors v_i in (1,k) Minkowski space such that < v_i, v_i> < 0 but < v_i, v_j > > 0 for i \neq j. I disagree.

Consider the vectors (1-e_i, sin(pi/2^i), cos(pi/2^i)) where

0 < e_i < (1/2)(1-cos pi/2^{i+1}),

in the vector space with norm |(t,x,y)| = t^2 - x^2 - y^2. If I am not mistaken, the inner products between these vectors have the required signs.

Answer 5

An idea. Identify H^2(C_1 x C_2, R) with R^k. Now your curves E1, E2, .... are identified with an infinite sequence P1, P2, .... in R^k. You have Ei^2 < 0 and Ej^2 < 0, but (since all your curves are irreducible) Ei Ej >= 0. Is there such a sequence in H^2(C_1 x C_2, R)?

EDITED to reflect that David Speyer observes that yes, there are infinite sequences of points like this (and that the subspace H^{1,1} of H^2 is what one wants to consider.) David's comment below refers to the version prior to this edit.

Given the existence of such a sequence of cohomology classes, one then asks whether the cohomology classes are represented by irreducible curves, which is what Dmitri wants.

Answer 6

What does the self-intersection number calculation in homology say?

Update, corrected: Here's what I mean. Consider a curve that has class xC1 + yC2, then its square is

               x^2 (C1)^2 + y^2 (C2)^2 + 2xy(C1)(C2)

which is just 2xy. Perhaps this will shed some light on the subject.