The natural action of $Sp(V)$ on $V$ is Hamiltonian with the co-moment map given by: $\tilde{\mu} : sp(V) \to C^\infty(V); A \mapsto \tilde{\mu}_A$,

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Given a symplectic vector space $(V, \Omega)$, consider the Lie group $G := Sp(V)$, consisting of all symplectomorphisms $\phi: V → V$. Show that if $(·, ·)$ is an invariant inner product, then the natural action of $G$ on $V$ is Hamiltonian with the co-moment map given by:

$$\tilde{\mu} : sp(V) \to C^\infty(V);\quad A \mapsto \tilde{\mu}_A,$$

where $\tilde{\mu}_A: V \to V;\quad v \mapsto \frac{1}{2}\Omega(Av, v)$.

I already proved that the action is symplectic and that $\tilde{\mu}_A$ is a Hamiltonian function for a vector field $X$ but I'm having problem proving that this is a lie algebras morphism , i.e ,

I need still to prove that

$$\tilde{\mu}_{[A,B]}=\{ \tilde{\mu}_A,\tilde{\mu}_B \}.$$

I know that $$\tilde{\mu}_{[A,B]}(v)=\frac{1}{2}\Omega([A,B]v, v)=\frac{1}{2}\Omega(ABv-BAv, v),$$ but I don't know how to relate this with $$\{ \tilde{\mu}_A,\tilde{\mu}_B \}(v)=\Omega(d\tilde{\mu}_A, d\tilde{\mu}_B).$$