What is a symplectic form of the rotation group SO(n)

Question

I need to prove that the Hamiltonian system of the rigid body motion $$ \begin{cases} \dot{R}_t=P_tJ^{-1},\\ \dot{P}_t=2R_t\Lambda,\quad\text{$\Lambda$ is the Lagrange multiplier}\\ R_t^T R_t-I=0, \end{cases} $$ corresponds to the Lagrangian $$ \begin{array}{lccl} \mathbb{L}:&[a,b]\times SO(n)\times TSO(n)&\longrightarrow& \mathbb{R}\\ &(t,R_t,\dot{R}_t)&\longmapsto&\mathbb{L}(t,R_t,V)=\frac{1}{2}tr(\dot{R} J \dot{R}^T), \end{array} $$ is completely integrable system, so what is the usual symplectic form of the $SO(n)$-group?

Answer 1

If I understand well, I would deduce from the domain of definition of your Lagrangean that $SO(n)$ is the configuration space of your system and not the phase space. The phase space should be the cotangent bundle $T^*SO(n)$ and then the symplectic form is the standard form which is available on any cotangent bundle. (As noted in the comment of @JasonDeVito, $SO(n)$ itself does not carry a natural symplectic form, it may even have odd dimension ...)