Let $(M,\omega)$ be a $4$-dimensional closed symplectic manifold. Assume there exists a Hamiltonian $S^1$-action on $M$, let $\mu:M \to \mathbb{R}^\*$ be its moment map and let $M\_{\text{red}}=\mu^{-1}(r)/S^1$ denote the symplectic reduction/symplectic quotient, where $r$ is a regular value of $\mu$ and where we assume that $S^1$ acts freely on $\mu^{-1}(r)$. For dimensional reasons, $M\_{\text{red}}$ is a closed manifold with $\dim M\_{\text{red}}=2$, i.e. a closed surface. 


Q: (How) can one determine the genus of $M\_{\text{red}}$ from the topology of $M$ and the $S^1$-action?



Symplectic reduction of 4-manifolds with circle actions
By a result of Hui Li <https://arxiv.org/abs/math/0605133>, the fundamental groups of $M\_{red}$ and $M$ are isomorphic, so one can deduce the genus of $M\_{red}$ from the fundamental group of $M$.


