From the definition of symplectic birationality given here (https://arxiv.org/pdf/0906.3265.pdf, Definition 2.1), two compact symplectic $2n$-manifolds $(M_{1},\omega_{1}),(M_{2},\omega_{2})$ are called symplectically birational cobordant if there is a connected, $2n+2$ dimensional symplectic manifold $Y$, with a semi-free, Hamiltonian $S^{1}$-action, generated by Hamiltonian $H : Y \rightarrow \mathbb{R}$ such that for some levels $c_{1},c_{2} \in H(Y)$ the symplectic reduction at level $c_{i}$ is symplectomorphic to $(M_{i},\omega_{i})$.
My question is about what happens if we drop the assumption of semi-freeness:
Question: Suppose that there is a connected, $2n+2$ dimensional, symplectic manifold $Y'$ along with Hamiltonian $S^{1}$-action, generated by Hamiltonian $H : Y' \rightarrow \mathbb{R}$. Suppose further that at levels $c_{1},c_{2} \in H(Y')$, the $S^{1}$-action is free on the level sets $H^{-1}(c_{i})$ (for $i=1,2$), so that the symplectic reductions at these levels are symplectic manifolds. But, we don't required the $S^{1}$-action on $Y'$ to be semi-free. Then are the symplectic reductions $H^{-1}(c_{i})/S^{1}$ symplectically birational cobordant in the above sense? are there any easy counter-examples?
(semi-free means the stabiliser of every point is $S^{1}$ or the identity).
