Let us say that a closed symplectic manifold $X$ is $GW_g$-connected if there is a nonvanishing Gromov-Witten invariant of the form $GW_{g,n}^{X,A}(\beta,point, point,\alpha_3,\ldots,\alpha_n)$ --in other words a nonvanishing invariant that formally counts (pseudo-)holomorphic curves of genus g passing through two generic points and satisfying some other constraints $\beta$ coming from $\bar{M}_{g,n}$ and $\alpha_i$ coming from other incidence conditions in $X$.

When $g=0$ this is something like saying that $X$ is rationally connected in the algebro-geometric sense, and there's been recent work (such as arXiv:1006.2486) relating to the question of whether the notions are the same. But in higher genus an analogous statement should fail--for instance in the product X of two elliptic curves there's a (reducible) genus two curve passing through any two points, but X is certainly not $GW_2$-connected.

Question: For which symplectic four-manifolds (or Kahler surfaces) $X$ does there exist g such that $X$ is $GW_g$-connected?

My personal motivation for this question comes from the fact that if $X$ is $GW_g$-connected for some g then by a result of Lu $X$ has finite Hofer-Zehnder capacity; however the question seems reasonably interesting aside from that. I restrict to dimension four here only because I expect doing so to make the question more tractable; insights into higher-dimensional cases would also be welcome.

Here are some preliminary observations in the direction of an answer:

It's an easy consequence of a result of McDuff that the only symplectic four-manifolds that are $GW_0$-connected are the rational ones (i.e. those related to $\mathbb{C}P^2$ by blowups and blowdowns).

For larger g, I've convinced myself that it's likely that any ruled surface over a curve of genus g ought to be $GW_g$-connected, though I haven't written down a careful proof--if someone knows where one can be found or knows that I'm wrong about this I'd be glad to hear about it.

I'd expect that symplectic four-manifolds with $b^+>1$ (for complex surfaces this means $p_g>0$) should rarely if ever have this property, since they typically don't have GW invariants counting curves with nontrivial incidence constraints. In fact for Kahler surfaces with $p_g>0$ this follows from a result of Lee and Parker.

For symplectic manifolds with $b^+=1$ which are not rational or ruled I'm not really sure what to expect. These usually have a decent supply of nontrivial Gromov-Witten invariants (as can be seen from Taubes-Seiberg-Witten theory), but it's not clear to me in general whether one should expect a nonvanishing invariant with two point constraints.

EDIT: Since originally posting this question I looked a little more carefully at the literature on four-manifolds with $b^+=1$, and found that work of Li, Liu, and others based on Taubes-Seiberg-Witten theory is enough to show that any closed symplectic four-manifold with $b^+=1$ is $GW_g$-connected for some $g$. I've provided details of the argument in the appendix of this preprint.

So it seems likely that the answer to the original question is that a closed symplectic four-manifold is $GW_g$-connected iff it has $b^+=1$: the backward implication is always true (by Li-Liu), and the forward implication is definitely true if one restricts to Kahler surfaces (by Lee-Parker), and there are also many non-Kahler examples for which it can be checked. It seems a good deal harder to say anything about higher-dimensional cases.


1 Answer 1


I am not familiar with symplectic geometry so let's assume everything here is at least K\'ahler.

If $g=1$, then the condition $\langle [pt], [pt], \ldots \rangle^X_{1, [C]}\neq 0$ implies that the variety is uniruled, which is equivalent to $\langle [pt], \ldots \rangle^X_{0, {C}}$.

I hope it is true that for a rationally connected fibration over a curve of any genus, your condition $\langle \beta, [pt], [pt], \ldots \rangle^X_{g, [C]}\neq 0$ is always true. And it is true when the fiber dimension is at most $2$. Basically as long as you know that there is a section which gives non-zero GW invariant, you can glue this section with curves in a general fiber which is minimal among all curves with non-vanishing GW invariant $\langle [pt], [pt], \ldots \rangle$.

For ruled surface, what you said is true. The methods used in the paper here certainly work.

  • $\begingroup$ Thanks, Zhiyu! I wasn't aware of the fact that the condition when g=1 implied uniruledness. Is there a simple explanation for why this is? Any reason to think it might be true for higher genera? $\endgroup$
    – Mike Usher
    Commented Dec 21, 2010 at 2:44
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    $\begingroup$ In case of $g=1$, the condition implies that there is a curve of arithmetic genus 1 passing through 2 general points. If this is a irreducible embedded curve, then you can deform it with 1 point fixed. Then bend and break tells you that there is a rational curve through the fixed point. If this is not an irreducible smooth embedded curve, then there are components of genus 0, passing through 1 of the two general points. In any case, there is a rational curve through a general point. In general, I think you need a genus g curve with g+1 points. $\endgroup$
    – Zhiyu
    Commented Dec 21, 2010 at 15:55
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    $\begingroup$ Sorry but you need to be a little bit more careful about the bend-and-break. It only works if you can deform the elliptic curve with a fixed complex structure. But if you can vary the complex structure, then it will specialize to a nodal rational curve. In this case you are still fine. $\endgroup$
    – Zhiyu
    Commented Dec 21, 2010 at 16:28
  • $\begingroup$ Dear Zhiyu, is $\langle [pt], \dots \rangle^{X}_{0, C}$ a statement? I, being a non-expert, think that it is a number so that '...equivalent to $\langle [pt], \dots \rangle^{X}_{0, C}$' does not make much sense. $\endgroup$
    – user74900
    Commented Feb 19, 2018 at 15:52

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