Let $(M, \Omega)$ be a symplectic manifold and $G$ be a compact Lie group acting on $M$, such that the 2-form $\Omega$ is $G$-equivariant. Let $m \in M$. Let $V_m := {T_m(G.m)}^\Omega/({T_m(G.m)}^\Omega \cap T_m (G.m) )$, where $G.m$ denotes the orbit of M through the point $m$ and ${T_m(G.m)}^\Omega$ is a subspace of $T_m M$ which is orthogonal to $T_m(G.m)$ with respect to the symplectic form $\Omega$.
The vector space $V_m$ is symplectic with the symplectic form $\Omega_{V_m}$ defined by $$\Omega_{V_m}([v],[w]):= \Omega_m(v,w)$$
For any $v, w \in T_m(G.m)$.
Let $H:= G_m$ be the isotropy group of $m$. The mapping $(h, [v]) \rightarrow [h.v]$, with $h \in H$ and $ [v] \in V_m$, defines an action of the Lie group $H$ on $V_m$, where $h.v$ denotes the tangent lift of the $G$-action on $TM$.
Why is the action $H$ on $V_m$ as described above is free ?