Let $(X,\omega)$ be a symplectic 4-manifold such that $\omega$ has a rational cohomology class. I am interested in Donaldson divisors (surfaces) $D$ in $(X,\omega)$ whose complement is a 1-handle body. Is there any obstruction for existence of such $D$? How about a transverse pair of them: Is there any obstruction for the existence of a positively intersecting transverse pair of such divisors such that $X-(D_1\cup D_2)$ is a $1$-handle body?
1 Answer
It seems to me that there is a kind of obstruction. Indeed, Donaldson divisors are usually surfaces of high genus, so they are of negative Euler characteristic. On the other hand the majority of known symplectic $4$-manifolds have positive Euler characteristics. There exists a folklore (or maybe Gompf's) conjecture stating that any simyplectic $4$-manifold of negative Euler characteristics is a blow up of an irrational ruled surface. Now
$$\chi(M\setminus \Sigma)=\chi(M)-\chi(\Sigma).$$
So this quantity will be positive in the majority of known cases. However the Euler characteristic of a $1$-handle body is non-positive (unless this is a ball).
So, it seems to me, that the number of examples of the type that you want will be very limited. Such as $\mathbb CP^2$ with a line in, or $\mathbb CP^1\times \mathbb CP^1$ with two lines.
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1$\begingroup$ I think you are correct. The boundary of a neighborhood of such high genus $\Sigma$ does not look like the boundary of a 1-handle body either. I should have thought more before asking here. $\endgroup$ Commented Dec 5, 2018 at 22:29