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.


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Oct 16, 2020 at 11:10
  • 1
    $\begingroup$ 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. $\endgroup$ Oct 16, 2020 at 12:01
  • $\begingroup$ Thanks for the explanation. That construction works indeed, so you just silently suppressed the fiber resealing in your answer. $\endgroup$ Oct 17, 2020 at 11:08

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