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.](https://books.google.fr/books?id=Qh_yBwAAQBAJ&pg=PA70&lpg=PA70&dq=Infinitesimal%20version%20symplectic%20group%20action&source=bl&ots=Drj_t058vp&sig=jzyhQPUfn0_6znmaarJMCFBoluI&hl=de&sa=X&ved=0CEgQ6AEwBWoVChMI2cidwcT5xgIVRrMUCh3JGg-P#v=onepage&q=Infinitesimal%20version%20symplectic%20group%20action&f=false)


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."



Check symplectomorphism property on infinitesimal generators
I think it is sufficient to assume that $G$ is a connected Lie group.
It is not necessary to assume that the exponential map is surjective
(intuitively because one can follow broken one-parameter groups). 


In fact, the set $H$ consisting of all $g\in G$ such that $\Phi\_g^\*\omega=\omega$ is a closed subgroup of $G$. A closed subgroup of a Lie group is a Lie group 
with the induced topology. What is the Lie algebra of $H$? The assumption that 
implies that the Lie algebra of $H$ equals the Lie algebra of $G$. In case $G$ is connected, this implies that $H=G$. 


*Edit*: Another way to think, perhaps more elementary. $H$ is an (abstract) group. The assumption that $L\_{\phi\_\chi}(\omega)=0$ for all $\chi\in\mathfrak g$ implies that 
$H$ contains a neighborhood of the identity, and $G$ (if connected) is generated by any neighborhood of the identity. 


To see that $H$ contains a neighborhood of the identity, one uses that the exponential map is onto a neighborhood of the identity and computes that
(denote $\varphi\_t=\Phi\_{e^{t\chi}}$):


$$\frac{d}{dt}\Big|\_{t=s}\varphi\_t^\*\omega=\frac{d}{dt}\Big|\_{t=s}(\varphi\_{t-s}\varphi\_s)^\*\omega=\frac{d}{du}\Big|\_{u=0}(\varphi\_u\varphi\_s)^\*\omega=\varphi\_s^\*\frac{d}{dt}\Big|\_{u=0}\varphi^\*\_u\omega=0$$


for all $s\in\mathbb R$, so $\varphi\_t^\*\omega=\omega$ for all $t\in\mathbb R$. 


