Let $M$ be a symplectic manifold with a symplectic action of a Lie algebra $\mathfrak{g}$. I am interested whether there exists a $\mathfrak{g}$-invariant symplectic connection on $M$. Where does the obstruction live and how to construct it?



Obstruction to the existence of an invariant symplectic connection
Let $\nabla$ be any symplectic connection. It is easy to check that the map $\mathfrak{g} \to \Omega^1 (M, \operatorname{End}\_{\omega} (TM))$ given by $X \to \mathcal{L}\_X (\nabla)$ is a Chevalley-Eilenberg 1-cocycle. Thus, we have an element in $H^1 (\mathfrak{g}, \Omega^1 (M, \operatorname{End}\_{\omega} (TM))$. The claim is that this cohomological class is the obstruction to the existence of an invariant symplectic connection.


Suppose that it vanishes. It means that there exists $A \in \Omega^1 (M, \operatorname{End}\_{\omega} (TM))$ s.t. for any $X \in \mathfrak{g}$ we have $\mathcal{L}\_X (\nabla) = \mathcal{L}\_X (A)$, then it is easy to see that $\nabla - A$ is the desirable connection.


Simiraly, if $\nabla$ is any symplectic invariant connection, then $\mathcal{L}\_X (\nabla) = 0$ for any $X \in \mathfrak{g}$, then the class is, of course, zero.


