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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.

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 Is the accepted answer below correct ? If not, please unaccept it. – jvp Dec 2 at 11:39
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 at 12:17

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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 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 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 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 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 at 9:34
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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.

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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 at 9:27
I wonder if we know that these graphs are reduced and irreducible? – Ilya Nikokoshev Oct 24 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. – unknown (google) Oct 24 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 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. – unknown (google) Oct 24 at 12:45
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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 at 3:35
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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 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 at 21:29

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