Suppose I have a Hamiltonian with three first integrals $L_1,L_2,L_3$ satisfying \begin{equation} \{ L_1, L_2 \} = L_3, \; \; \{ L_2, L_3 \} = L_1, \; \; \{ L_3, L_1 \} = L_2, \end{equation} where $\{ \cdot, \cdot \}$ is the Poissonbracket. I want to find the maximal amount of degrees of freedom by which the Hamiltonian can be reduced.
Specfically I want to use a generalization on the Marsden-Weinstein-Meyer theorem as indicated by this post.
Observe that the Lie brackets induces a Lie algebra $ \simeq \mathfrak{so}(3)$.
Problem 1: I am not really sure how to define the momentum map $\mu: M \rightarrow \mathfrak{so}(3)^*$. I would think that it is defined by $\mu:= (L_1,L_2,L_3)$. But I am not really sure.
Choosing values for $L_i$'s, i.e. a point $P\in\mathfrak{so}^*$, you can reduce the number of variables by $3+m$ where $m$ is the dimension of the stabilizer of $P$ under the coadjoint action.
So I need to find the coadjoint action. To do this I want to find $Ad^*$ such that $
Problem 2: How is this natural pairing defined? Like $
I think the coadjoint action will correspond to
\begin{equation} Ad^*_{L_1} (L_2^*) = L_3^*, \; \; Ad^*_{L_2} (L_3^*) = L_1^*, \; \; Ad^*_{L_3} (L_1^*) = L_2^*, \end{equation}
Problem 3: It is really unclear to me how to compute the stabilizer group. Do I just take a constant vector in $P \in \mathfrak{so}^*(3) $ and compute the dimension of
\begin{equation} \{X : Ad^*_X P=P \}? \end{equation}
Any help is welcome.