# Problem:Gromov-Witten;Moduli space

Let us consider a map from a $\Sigma_g \longrightarrow N$, where $N$ is a symplectic manifold.

Then we define the moduli space as

$M= \{ f | f \mbox{ is a pseudoholomorphic map } \Sigma_g \to N, f_* ([\Sigma]) = [A]\}$, where $A \in N$.

We assume the dimension of $N$ is $2n$. We can cauculate the dimension of the moduli space $M$ is $n(2-2g)+2c_1(A)$

Questions:

1. Is the moduli space non-empty for any Riemman surface and any $N$ and any $[A] \in H_2(N)$ ?

2. Is there any simple way to judge whether the moduli space is not empty? What about this problem when N is a four dimension manifold?

The case I want to know most is when $N$ is a four-dimensional symplectic manifold with infinite fundamental group.

-
This is not an answer, just a comment. Even when $N$ is a complex manifold these are difficult questions. Most (positive) existence results I know of are either by explicit construction (not too hard, for instance, if $N$ is a low degree hypersurface in $\mathbb{C}P^n$), or by Mori's Bend-and-Break result. Mori's technique works by passing to algebraic varieties over positive characteristic fields, so it only applies in the algebraic case. –  Jason Starr Jul 15 '12 at 11:08
Two comments: For question 1, the answer is trivially yes if you take $A = 0$ and constant holomorphic curves... is that a reasonable interpretation of the question? There is also a typo in the index formula. If you want the real dimension of $M$, the formula is $n(2-2g) + 2c_1(A)$. –  Sam Lisi Jul 15 '12 at 14:28
The simplest way to get existence for all choices of $\Sigma_g$, $g$ positive, is to first prove existence for a genus $0$ Riemann surface. Since $\Sigma_g$ has positive degree maps to the Riemann sphere, the composition will give maps of $\Sigma_g$ to $N$. –  Jason Starr Jul 15 '12 at 16:26
I would say that without more information you can't say whether there is a curve or not. Concerning your question in a comment, even assuming that N is Kaehler isn't much use: a generic K3 surface has no curves because curves represent integral (1,1)-classes in cohomology and you can generically assume that the subspace $H^{1,1}$ is irrational (though I guess by Lefschetz's theorem on (1,1)-classes this is the only obstruction for Kaehler surfaces). Do you have a specific example in mind where you want to find a holomorphic curve? –  Jonny Evans Jul 15 '12 at 17:32
Is this a new question, or have I misunderstood the original questions? If you want a sequence of manifolds $N_i$ for which there exist non-constant pseudoholomorphic curves of genus $g$, then take your favourite sequence of symplectic manifolds $X_i$ and let $N_i=X_i \times \Sigma_g$. On this, you can take a split almost complex structure, $A= [pt]\otimes [\Sigma_g]$ and the pseudoholomorphic curve will be $\{pt\} \times \Sigma_g$. –  Sam Lisi Jul 16 '12 at 10:43

• If we allow ourselves to find a $J$-holomorphic curve for a very special (not generic!) almost complex structure $J$, it suffices to find an embedded symplectic surface and then construct $J$ to make this surface $J$-holomorphic. In dimension 4, we can find a symplectic surface by finding a Donaldson divisor.
• If there exists a $J$-holomorphic sphere in $N$, then there exist $J$-holomorphic maps from domains of all genus, by composing with a branched cover.
• Another family of examples come from 4-manifolds $(N, \omega)$ for which the Gromov-Taubes invariant is non-vanishing. For instance, if $c_1(TN) \ne 0$.