Simple examples of Gromov-Witten invariants not being enumerative

Question

I understand why Gromov-Witten invariants in general are not enumerative, so it's not necessary to explain this. However to test something I'm working on, I'm looking for examples of concrete enumerative problems where both the actual answer and associated Gromov-Witten invariants are known. I'm particularly interested in algebraic G-W invariants here.

Answer 1

The simplest example I know is that of a general quintic threefold $ Q $. Indeed, such a threefold has $ n_2 = 609250 $ rational curves of degree $ 2 $ (this is the actual answer) while the corresponding Gromov-Witten invariant $ \deg [ \mathcal{M}_{0,0} ( Q , 2[L] ) ]^{vir} $ is a fraction, equal to $ 4876875/8 = 609250 + 2875/8 $. It should be no surprise that the number of lines $ n_1 = 2875 $ appears here: because the moduli stack $\mathcal{M}_{0,0} ( Q , 2[L] ) $ has, apart from the $ n_2 $ many 0-dimensional components corresponding to actual degree two rational curves, also $ n_1 $ many 2-dimensional components corresponding to double covers of lines in $ Q $.

Another example comes from Taubes GW-invariant. For a surface $ S $ of general type and target homology class the canonical divisor class, one can compute the Taubes GW-invariant for genus $ g(K_S) = 1 + K_S^2 $ without marked points as the number $ (-1)^{\chi (\mathcal{O}_S)} $ but clearly it is not enumerative. For instance, for a quintic surface of $ \mathbb{P}^3 $, it is just 'counting' hyperplane sections.

Answer 2

I'm not sure if this is quite what you're looking for, but I always found it a useful example to keep in mind. Take the genus 1 invariant for degree 1 maps into $P^1$. There are no degree 1 holomorphic maps from a torus to $P^1$, but there is a stable map: the domain is a torus union a sphere at a point; the map sends the torus to a single point and is degree 1 on the sphere. The GW invariant is nonzero because of this guy, so if you perturbed the holomorphic map equation using a generic Hamiltonian term then you would indeed find a solution whose domain is irreducible.