Let $G$ be a compact Lie group, $M$ be a non-compact symplectic manifold, $\psi_g:M\to M$ is a symplectic action, and $\omega$ be a symplectic exact form i.e. $\omega=-d\lambda$ for some $1$-form $\lambda$. We say that the symplectic action $\psi_g:M\to M$ is exact if $\psi_g^*\lambda=\lambda$.
I want to show that every exact symplectic action is Hamiltonian which is given by the map $$\mathfrak{g}\to C^{\infty}M:\xi\longmapsto H_{\xi}$$ where $H_{\xi}=\iota(X_{\xi})\lambda$.
So, I think I need to show two things:
- $H_{\xi}$ is Hamiltonian vector field.
- $\xi\longmapsto H_{\xi}$ is a Lie algebra homomorphism i.e. $\{H_{\xi},H_{\nu}\}=H_{[\xi,\nu]}$ for all $\xi,\nu\in\mathfrak{g}$ where $\{\cdot,\cdot\}$ is the Poisson bracket.
Can you please tell me if I am in the right direction?
To show the $(1)$, it's enough to observe, using excactness, that $$0=\mathcal{L}_{X_{\xi}}(\lambda)=i_{X_{\xi}}d\lambda+d\lambda(i_{X_{\xi}})$$ which gives $i_{X_{\xi}}\omega=dH_{\xi}$.
To show (2), I need to check $\{H_{\xi},H_{\nu}\}=H_{[\xi,\nu]}$ $(*)$ using local coordinates. I choose Darboux chart $(U,q_1,\dots,q_n)$ of $M$ where $\lambda|_U=p_1dq_1+\dots+p_ndq_n$ and did computation when $\dim(M)=1$ to convince myself that $(*)$ holds for $n=1$. If I want to show $(*)$ for any $n$, is there a coordiante free proof?