Let $\Sigma$ be a closed, orientable surface.


Then the cotangent bundle $T^\*\Sigma$ has a canonical symplectic form $\omega$, given as the derivative of the tautological Liouville one-form. We can modify it to a "magnetic" form by adding some two-form $\sigma$ on the base to the symplectic form.


The notation $T^\*\_\sigma \Sigma$ will denote the "magnetic cotangent bundle", i.e. the cotangent bundle equipped with a symplectic form $\omega + \sigma$.


Given this, my (rather broad) question is the following: in what cases (i.e. varying $\sigma$ or the genus of $\Sigma$) is it known that a small neighborhood of the zero section in $T^\*\_\sigma \Sigma$ symplectically embeds into a closed symplectic $4$-manifold?


For the purposes of this question, we will suppose that $\sigma \neq 0$. Otherwise, there are many examples with $\sigma = 0$, as by the Weinstein neighborhood theorem, we can just take a neighborhood of an embedded Lagrangian $\Sigma$ in a closed symplectic $4$-manifold.



Embeddings of magnetic cotangent bundles over surfaces into closed symplectic 4-manifolds
This can always be done.


Let's first treat the case when $\Sigma$ is not a torus. Then take any symplectic $4$-manifold $(M,\omega)$ where $\Sigma$ can be embedded as a Lagrangian surface. Now, take a small neighbourhood $U$ of $\Sigma\subset M$ that is symplectomorphic to a neighbourhood the zero section in $T^\*\Sigma$. Let $\pi: U\to \Sigma$ be the corresponding projection. Now take $\pi^\* \sigma$ on $U$ and extend it to a closed two form $\sigma'$ on $M$. This is always possible, since $\Sigma^2=-\chi(\Sigma)\ne 0$, so we don't have cohomological obstructions. Finally for some large $t$ the form $t\omega+\sigma'$ will be symplectic and it will induce on $U$ the desired magnetic form.


In case $\Sigma=T^2$ one can do everything directly on $T^4=T^2\times T^2$. We take standard symplectic $T^4=\mathbb R^4/\mathbb Z^4$ (with the form $\omega=dx\_1\wedge dy\_1 +dx\_2\wedge dy\_2$) and choose Lagrangian $T^2$ that is given by the $(x\_1,x\_2)$-plane. Project $T^4$ to this $T^2$ and pullback $\mathbb \sigma$ to $T^4$ from this $ T^2$. Then $\omega+\pi^\*\sigma$ does the job.


