# Embeddings of magnetic cotangent bundles over surfaces into closed symplectic 4-manifolds

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.

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.
• Why does $t \omega + \sigma'$ give you the magnetic form $\omega_{\textrm{can}} + \sigma$ on $U$ (seen as a subset of $T^* \Sigma$)? Don't you need $t = 1$ for this? Oct 16, 2020 at 11:10
• Tobias, you don't really need $t=1$ for this. The point is that if you consider a map from the cotangent bundle to itself that sends $\alpha\to t\alpha$, the canonical form $\omega$ pulls back to $t\omega$. So, any cotangent bundle with $\omega_{can}$ is simplectomorphic to one with $t\omega_{can}$. At the same time, this operation of scaling the fibers by $t$ doesn't affect the pullback of $\sigma$. Hopefully this answers your question. Oct 16, 2020 at 12:01