Associating the Hamiltonian vector field $X_f$ to a function $f$ is a Lie algebra homomorphism Let $(M, \omega)$ be a complex analytic symplectic variety, i.e. $\omega \in \Gamma(\Lambda^2 \Omega_M)$ is a closed holomorphic 2-form, which is everywhere non-degenerate. We can view $\omega$ as an isomorphism $\omega: \Omega_M \to T_M$, and let $\theta \in \Gamma(\Lambda^2 T_M)$ be its inverse. Then the Poisson bracket for two functions $f, g \in \mathcal{O}_M$ is defined by
$$\{f, g\} = \theta(d f, dg) = \langle \theta, df \wedge dg \rangle,$$
where $\langle -, - \rangle$ denotes the usual evaluation of a vector and a form. This is a Poisson bracket, i.e. a Lie bracket which satisfies the Leibniz rule for multiplication.
Using this, on can define a vector field $X_f$ associated to $f$ by declaring
$$X_f(g) = \{f, g\}.$$
In Systèmes Hamiltoniens Complètement Intégrables Associés aux Surfaces K3, Beauville then claims that the map $f \mapsto X_f$ is a homomorphism of Lie algebras. To check this, I tried to evaluate
$$X_{\{f, g\}}(h) = \theta(d\{f, g\}, d h) = \theta\big(d\theta(d f, d g), d h\big), \tag{$*$} $$
but I don't know how to go on from here.

I get that $\{f, g\} = X_f(g) = \mathcal{L}_{X_f}(g)$, and $[X_f, X_g] = \mathcal{L}_{X_f}(X_g)$, where $\mathcal{L}_{X_f}$ is the Lie derivative with respect to $X_f$. But to conclude $X_{\{f, g\}} = [X_f, X_g]$ we would have to know that $\mathcal{L}_{X_f}$ commutes with the mapping $g \mapsto X_g$.

Any help would be appreciated :)
Vector fields get the lie algebra structure with lie derivative.{g, f} =dg(Xf) =w(Xg, Xf).  The isomorphism of Xf and f is obsious then. Since lie bracket has the same form ( structure) as lie derivative.
Vector fields get the lie algebra structure with lie derivative.{g, f} =dg(Xf) =w(Xg, Xf).  The isomorphism of Xf and f is obsious then. Since lie bracket has the same form ( structure) as lie derivative.