Thank you for your kind answer. Actually, I study about a closed non-hamiltonian symplectic $S^1$-manifold $(M,\omega,J)$ with non-empty fixed point set.

I have one another question. Let X be a fundamental vector field of the action and let $\gamma(t) : [0, \infty) \to M$ be an integral curve for $JX$ with $\gamma(0) = z$, where $z$ is a fixed point for the given action. I wonder whether $\gamma(t)$ has the end point or not. (I mean whether $JX$-flows converges to another fixed point or not)

If $\gamma(t)$ has no end point, then $S^1 \cdot \gamma([0,t])$ is a $J$-holomorphic disk for any $t \in (0,\infty)$. Hence I think it should have infinte symplectic volume.

In Hamiltonian action case, the integral curve $\gamma(t)$ of course converges to some fixed point by compactness of $M$

Thank you for your kind answer. Actually, I study about a closed non-hamiltonian symplectic $S^1$-manifold $(M,\omega,J)$ with non-empty fixed point set.

I have one another question. Let X be a fundamental vector field of the action and let $\gamma(t) : [0, \infty) \to M$ be an integral curve for $JX$ with $\gamma(0) = z$, where $z$ is a fixed point for the given action. I wonder whether $\gamma(t)$ has the end point or not. (I mean whether $JX$-flows converges to another fixed point or not)

*** If $\gamma(t)$ has no end point, then $S^1 \cdot \gamma([0,t])$ is a $J$-holomorphic disk for any $t \in (0,\infty)$. Hence I think it should have infinte symplectic volume. But as you said, this question seems not to be related to the symplectic volume of disk.

*** In Hamiltonian action case, the integral curve $\gamma(t)$ of course converges to some fixed point by compactness of $M$