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In defining Gromov-Witten invariants using symplectic geometry, most of the trouble is to achieve transversality for moduli spaces of pseudo-holomorphic curves which are multiple covers of simple ones. One technique which was invented by Ruan and Tian allows us to define the invariants for semi-positive symplectic manifolds. Namely they deform the equation $\overline{\partial}(u)=0$ to

$$ \overline{\partial}(u)= \nu $$

Solutions to this equation are no longer multiply covered and we can achieve transversality on stable $\mathbf{domains}$ using these ideas. The maps from unstable bubble domains which are multiple covers can be reduced to their simple image and can be seen to occur in codimension 2 . Therefore we can ignore them.

In their famous Annals paper, Ionel and Parker define relative Gromov-Witten invariants for pairs $(X,V)$ where $V$ is a symplectic submanifold of $X$ a compact symplectic manifold. They seem to indicate that they can achieve transversality under the assumption that both $X$ and $V$ are semi-positive symplectic manifolds using the same idea as Ruan and Tian.

What I want to know is: what allows us to rule out sphere bubbles from unstable domains in the relative setting?

My difficulty is that I don't see why semi-positivity is enough to ensure that when we reduce an unstable sphere with possible tangency of intersection conditions along the divisor $V$, the dimension of the reduced spheres is lower than the virtual dimension of the moduli space of $V$-regular maps.

In the case that our intersection multiplicity with $V$ is say $d$ at each marked point, then the condition might be elegantly expressed as "the root DM stack on $(X,V,d)$" be semi-positive and $V$ is semi-positive. I haven't thought too carefully about this though... In any case, it's not too hard to write down the condition that you need explicitly and it seems to be different from what I mentioned above, so I have the feeling I'm missing a key point in the definition.

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    $\begingroup$ Send me an email and I will share some notes with you which explains things. $\endgroup$ Feb 7, 2014 at 19:15
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    $\begingroup$ Have you sorted this question out? I'm also interested in the answer, though I haven't thought about it very much yet. $\endgroup$
    – Sam Lisi
    Feb 19, 2014 at 14:53

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