Why are Gromov-Witten invariants of K3 surfaces trivial?  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? 
 A: 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
A: 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.
A: Here are two answers to your first question:


*

*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.

*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.
