I stumbled over the following question:

First, let me give the basic definition of a symplectic group action: Let $(M, \omega)$ be a symplectic manifold and $G$ a Lie group. A smooth action $\Phi:G \times M \rightarrow M$ is symplectic if each $\Phi_g$ is a symplectomorphism, i.e. $\Phi_g^* \omega = \omega.$

Now, it is clear that if $\Phi$ is symplectic, then the Lie derivative $L_{\phi_\chi} \omega =0$ for all $ \chi \in \mathfrak{g},$ where $\phi_{\chi}(x) = \frac{d}{dt}|_{t=0} \Phi_{e^{t \chi}}(x)$ are called the infinitesimal generators.

This is easy to see as $L_{\phi_{\chi}}(\omega) = \frac{d}{dt}|_{t=0} (\Phi_{e^{t \chi}})^* \omega = 0 $ by definition.

What is a priori unclear to me is whether this ($L_{\phi_\chi} \omega =0$ for all $ \chi \in \mathfrak{g}$) is already sufficient to conclude that $\Phi$ is a symplectic group action.

Does anybody know when (maybe under which additional conditions) this is the case?

I should say that I stumbled over this when reading this book, where it is called the infinitesimal version of this equation:

read the reference on google books.

But maybe the term "infinitesimal version" has a different meaning in symplectic geometry and it is not in general possible to go back to the "global version."