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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)

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jvp thanks for this comment! I accepted the answer, since it looked very promissing, meanwile I hoped to learn more about Shimura curves to see if it is really correct. But now since the auther of the post tells it does not work I unaccept it. Sorry... – Dmitri Dec 2 '09 at 12:17
can you give an example with even one curve of negative self intersection? – Vivek Shende Nov 24 '10 at 23:06
Digaonal in CxC, where C has genus at least 2 – Dmitri Nov 24 '10 at 23:09
up vote 5 down vote accepted

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 ?

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Thank you very much for writing this answer! It is really cute so I'am accepting it :) . Also, the question that you ask in the end is interesting. I am curious as well if one could describe the Ample cone of these products C×C, that you suggest to consider. – Dmitri Jun 6 '11 at 10:55

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.

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Very nice! It is perhaps simpler to start with a genus 2 curve with Jacobian isogeneous to the square of an elliptic curve. Since these abelian varieties have infinitely many elliptic curves with self-intersection zero through 0, theirs blow-ups have infinitely many negative elliptic curves. – Jorge Vitório Pereira Jun 3 '11 at 23:41
No. $C'\times C'$ clearly maps to that Kummer. I was confused by something else, but I see the mistake I made. Sorry. – Sándor Kovács Jun 3 '11 at 23:43

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.)


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.

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Thanks a lot for this comment! This seems very promissing, indeed all curves will etale projections will have negative self intersection. Unfortunatelly I don't know anything about Shimura curves. Are they compact quotinets of H^2? Where can I learn a bit on this? I also wonder what will happen if we let the curve get larger and larger --- will it converge to some lamination? – Dmitri Oct 24 '09 at 9:27
I wonder if we know that these graphs are reduced and irreducible? – Ilya Nikokoshev Oct 24 '09 at 9:43
@Dmitri. Yes, Shimura curves are compact quotients of H^2 but the main point is that these correspond to very specific discrete subgroups of PSL(2,R). By a search on google I found the book "Quaternion Orders, Quadratic forms and Shimura curves" by Alsina and Bayer; it looks like a good reference. I am not sure about what happens when the curve gets larger; I will add a comment later if I have anything sensible to say about it. – ulrich Oct 24 '09 at 10:53
Thanks for the refference, I'll try to study it. I do belive that this can work but I just wonder, can you estimate how many pages would take an honest proof of this fact? At least what will be the steps? Also I am curious if these curves will be geodesic with respect to the product metric? Thanks again. – Dmitri Oct 24 '09 at 11:19
The only thing which is not elementary is the fact that Shimura curves are compact; the Hecke cycles can be defined very explcitly using some simple group theory and the properties I mentioned are immediate from the definitions. It might be easier to first think about the case of classical modular curves (associated to congruence subgroups of SL(2,Z)) and the Hecke cycles on self products of these; the case of Shimura curves works in essentially the same way but because they are compact one can deduce the negativity of the self-intersection without any calculations at all. – ulrich Oct 24 '09 at 12:45

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.

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Say, I can't delete my own posts! Assuming this is right there's no reason for either mine or yours to stay up. Is a sequence of cohomology classes like yours going to be in the effective cone? If so, I suppose you'd find in the classes the curves that Dmitri wants? – JSE Oct 24 '09 at 4:30
I have no idea whether this is in the effective cone. Do you know what the effective cone looks like here? – David Speyer Oct 24 '09 at 4:37
By the way, I disagree that there is not point in our posts staying up. It is worthwhile to know what doesn't work. If you like, I'll edit mine to stand alone. – David Speyer Oct 24 '09 at 4:43
OK, I edited mine to remove the stuff that was wrong but preserve the suggestion to ask first whether the phenomenon Dmitri asks about can happen "numerically." – JSE Oct 24 '09 at 5:07
The problem here is that for a generic product C1 x C2 its space H^(1,1)(Z) has dimension 2, and generated by curves C1 and C2. The effective cone is the positive octant aC1+bC2 with a>=0, b>=0. So for a generic product C1xC2 all curves inside have non-negative self intersections... – Dmitri Oct 24 '09 at 9:34

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.

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I don't think this works. See my comment. – David Speyer Oct 24 '09 at 3:35

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.

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Sorry I don't quite get your question:) If we consider C1xC1 as a topological 4-fold, there will many 2-surfaces inside with negative self intersection, for example the diagonal. But overwise a curve with negative self intersection has negative expected dimension, so on a generic product C1xC2 all complex curves have postive self-intersection – Dmitri Oct 23 '09 at 21:02
If x>0, y>0, then x^2 (C1)^2+y^2(C2)^2+2xy= 0 + 0 + 2xy >0, so positive. If you take a generic product C1xC2, then all curves on it will be numerically equivalent to xC1+yC2 with x, y>0, so the square is always positive. Of course I forgot to say, that I am looking for irreducibe and reduced curves --- i.e. of multiplicity one. If a product C1xC2 satisfying conditions that I want exists, that will be something extremely rigid... – Dmitri Oct 23 '09 at 21:29

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