Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Why is GW invariants of K3 surfaces are trivial? My naive guess is that GW invariants are deformation invariant and you can always deform your K3 surface to non-projective one, which has no subcomplex manifold except points. Then GW invariants (or GV invariants) naively count the number of curves, so they must be trivial.

Are there any more rigorous proof of this fact? Or can we make the argument above rigorous? Since GW invariants are symplectic invariant, I wonder if there is a proof in symplectic geometry too.

Another question is that, are DT invariants of K3 surfaces also trivial?

share|improve this question
1  
The Donaldson invariants are non-trivial for K3 surfaces; however this is probably not related to Donaldson-Thomas invariants. –  Ian Agol Sep 10 '12 at 23:25
1  
I wonder what happens if we restrict ourselves to algebraic K3 surfaces, or lattice polarized ones. –  Atsushi Kanazawa Sep 11 '12 at 1:07
    
Thanks for the comments. –  Daniel Sep 12 '12 at 3:18
add comment

3 Answers

Here are two answers to your first question:

  1. Yes your argument works. To see that there are no curves for a generic K3 observe that their homology classes must be Poincare dual to an integral (1,1)-class and there are none of these for a generic K3.

  2. Another argument goes like this: K3 is hyperKaehler so it admits a sphere of symplectic structures/compatible complex structures. In particular, by moving in this sphere you can go from $(\omega,J)$ to $(-\omega,J')$. If there's a nonvanishing GW invariant then there's a curve class $A\in H_2(K3;\mathbf{Z})$ with positive $\omega$-area represented by a $J$-holomorphic curve which persists under the deformation and is also represented by a $J'$-holomorphic curve. Therefore it has to have positive $-\omega$-area as well, which is a contradiction!

I don't know about your DT question.

Edit: Your worry that GW invariants don't count curves is not an issue in this case because there are no curves to count. Let me explain. GW invariants count stable curves (weighted by signs coming from orientations or factors coming from automorphism groups) when all those curves satisfy some transversality condition.

  • When transversality fails there may be more curves than there should be and the GW invariant is counting something else (e.g. solutions of a perturbed Cauchy-Riemann equation). Certainly when there are no curves at all, all curves satisfy transversality!

  • The only other issue is that sometimes stable curves exist whereas smooth curves don't. For instance, there are no smooth genus 1 curves in $S^2$, but there are stable genus 1 curves which lead to a nonzero GW invariant. Consider the domain to be a torus union a sphere, connecting them at a single nodal point. Map the torus to a single point $p$ and map the sphere holomorphically to $S^2$, taking the node to $p$. That's a stable curve which you have to count. In the case when there are no nonconstant holomorphic curves at all, this is, again, not an issue.

share|improve this answer
2  
Argument 1. works for every counting theory deformation invariant which is the case for DT invariants. –  user25309 Sep 10 '12 at 11:47
    
Thank you for the answer. My concern is that GW invariants naively count the number of curves (with fixed data); they may not coincide with the actual number of curves and my deformation argument may fail in that case. Your second proof seems very nice to me. It seems to me that your argument works for arbitrary hyperKaehler manifolds. –  Daniel Sep 12 '12 at 3:23
    
@Daniel: I've edited my answer to address your concern. Certainly the argument applies to all closed hyperKaehler manifolds and more generally to any manifold where you can connect $\omega$ and $-\omega$ by a family of symplectic forms (e.g. Kodaira-Thurston). –  Jonny Evans Sep 14 '12 at 8:54
1  
As long as one uses the reduced theory, the GW invariants do count curves. In fact, if the curve class is primitive (not a multiple of another class), and the K3 surface is generic (in the family of K3 surfaces having the given curve class (1,1)), then the reduced GW invariants are enumerative in the strongest sense --- they count each exactly once without multiplicity. This follows from a strong genericity result of Xi Chen which says that all rational curves in a primitive class on a generic algebraic K3 surface are nodal. –  Jim Bryan Sep 16 '12 at 15:50
add comment

While it is true that the ordinary GW invariants of a K3 surface are trivial (by the deformation argument you cite), the reduced GW invariants are non-trivial and capture a lot of interesting enumerative information and structure. The reduced invariants are obtained by modifying the obstruction theory on the space of stable maps and result in curve counting invariants which are invariant under deformations of the K3 surface which preserve the (1,1)-type of the polarization. The generating functions for these GW invariants can be expressed in terms of quasi-modular forms. http://www.ams.org/journals/jams/2000-13-02/S0894-0347-00-00326-X/S0894-0347-00-00326-X.pdf

Donaldson-Thomas invariants are only defined for threefolds. In order to define them for a surface, one must create a "local surface" --- the threefold given by the total space of the canonical bundle over the surface. This introduces some non-compactness which then can be dealt with in various (sometimes equivalent) ways (for example, one can work equivariantly and use localization, or one can use Euler characteristics weighted by the Behrend function in place of virtual classes). After dealing with these issues, one finds that the DT invariants of K3 are indeed non-trivial and are related to the reduced GW invariants by a MNOP type relationship.

This is an extensive subject with lots of work. It began with the formula of Yau-Zaslow and the conjecture of Gottsche back in 1995. Leung and I defined the reduced GW invariants of K3 in our 2000 paper. Recently, there has been a resurgence of work on the invariants of K3 by Pandharipande, Maulik, Thomas, and others, for example:

http://arxiv.org/abs/1001.2719

http://arxiv.org/abs/0808.0253

http://arxiv.org/abs/0807.2477

share|improve this answer
    
Thank you for the answer. Your paper seems very interesting (I have heard of your paper, but totally forgot about it.) THe DT invariants of local CY via equivalent cohomology seems dependent of torus action , although there is a natural one in case of total space of the canonical bundle. Behrend's approach seems more natural. –  Daniel Sep 12 '12 at 3:32
    
This may not be directly relevant to my question, but is is possible for you to suggest good papers where I can start leaning computation of local CY GW or DT invariants? –  Daniel Sep 12 '12 at 19:07
2  
The section on Gromov-Witten theory in the big Mirror Symmetry book (Hori, et al) has a good introduction to localization in GW theory. The Katz and Cox book also has it. Localization in the toric context leads to the topological vertex machinery which is an effective computational tool for computing. In particular, one can compute the invariants for O(-1)+O(-1) in this way. You could look at section 2.3 of my paper with Kai Behrend for example. I nice general introduction to this area is the paper "13/2 ways of counting..." by Pandharipande and Thomas. –  Jim Bryan Sep 12 '12 at 20:21
1  
@Daniel: Indeed Jim Bryan's answer is related in an interesting way to the hyperKaehler answer I mentioned below, as is pointed out by Bryan-Leung in their paper. The reduced GW invariants have a symplectic interpretation in this instance as a "family GW invariant" associated to the family of K3s equipped with symplectic forms from the hyperKaehler sphere (also known as the twistor family). –  Jonny Evans Sep 14 '12 at 9:03
add comment

The most direct answer to the original question is provided by the DMJ paper of Junho Lee. A (2,0) form on a Kahler surface X determines an almost complex structure J on X so that all J-holomorphic curves lie in the zero set of the (2,0)-form. Since a K3 admits a nonwhere zero (2,0)-form, there are no J-holomorphic curves in K3 for the corresponding almost complex structure J.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.