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?