Associating the Hamiltonian vector field $X_f$ to a function $f$ is a Lie algebra homomorphism

Question

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 :)

Answer 1

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.

Answer 2

For clarity, let's distinguish between $\omega\in\Gamma(\wedge^2T^*_M)$ and $\tilde{\omega}:T_M\to T_M^*$, given by $\tilde{\omega}(X) := \omega(X,\cdot) = i_X\omega$. From \begin{split} dg(X_f) &= X_f(g) = \lbrace f,g\rbrace = \theta(df,dg)=-\theta(dg,df)\\ &=-\omega(\tilde\omega^{-1}(dg),\tilde\omega^{-1}(df)) = -dg(\tilde\omega^{-1}(df)), \end{split} we see that $X_f = -\tilde\omega^{-1}(df)$, or equivalently $i_{X_f}\omega = \omega(X_f,\cdot) =-df$. Since $d\omega=0$, Cartan's magic formula implies $$\mathcal{L}_{X_f}\omega = (di_{X_f}+i_{X_f}d)\omega = d(-df) = 0.$$ Now using the identity $i_{[X,Y]} = [i_X,\mathcal{L}_Y]$, we get $$ i_{[X_f,X_g]}\omega = [i_{X_f},\mathcal{L}_{X_g}]\omega = -\mathcal{L}_{X_g} (i_{X_f}\omega) = \mathcal{L}_{X_g}(df) = d(\mathcal{L}_{X_g}f) = d\lbrace g,f\rbrace = -d\lbrace f,g\rbrace = i_{X_{\lbrace f,g\rbrace}}\omega. $$ By non-degeneracy of $\omega$, this implies $[X_f,X_g] = X_{\lbrace f,g\rbrace}$.