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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 $<F,Ad_X(Y) >=<Ad^*_{X}F^*,Y >$ where $F \in \mathfrak{so}^*$ and $X,Y \in \mathfrak{so}$ where $<\cdot , \cdot>$ is the natural pairing between $\mathfrak{so}^*$ and $\mathfrak{so}$.

Problem 2: How is this natural pairing defined? Like $<L_i^*, L_j>= \delta_{ij}$ (with $\delta_{ij}$ the Kronecker delta)?

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.

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