Moment map and Hamiltonian

Question

Take the manifold $M$ to be $M=\mathbb{R}^6=\mathbb{R}^3\times\mathbb{R}^3$ (hence $x\in M$ is given by $x=(p,q)$ with $p$ and $q$ three dimensional vectors) and take the possion bracket on $M$ given by $$\{f,g\}=\sum_{i=1}^3{(\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i}-\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i})}$$ Assume $G=\mathbb{R}^6$ acts on M by $(p',q')\cdot(p,q)=(p+p',q+q')$. The question is to show that this action is Hamiltonian with momentum map $J(p,q)=(q,-p)$. By definition we have to show that $\delta_T(f)=\{J_T,f\}$ for each $T\in\mathfrak{g}$ (the Lie algebra of $G$ (which is in this case $\mathbb{R}^6$)) and each $f\in C^{\infty}(M)$. Here $J$ is the momentum map $J:M\rightarrow\mathfrak{g}^*$ and $J_T(x)=J(x)(T)$. But how to show that claim? Can someone help me with this problem? Thanks a lot.

Answer 1

@Arnetta, The action by translation is symplectic, thus hamiltonian ; since $\mathbb{R}^6$ is contractible. The fundamental vector fields associated to the basis $\{e_1,...,e_6\}$ of $\mathrm{Lie}(\mathbb{R}^6,+)$ are : $\partial_{x_i}$ $(i=1,...,6)$. Since the symplectic form is $\omega=\sum_{i=1}^3dx_i\wedge dx_{i+3}$, then, for $i=1,2,3$, $$i_{\partial_{x_i}}\omega=dx_{i+3}\quad\text{and}\quad i_{\partial_{x_{i+3}}}\omega=-dx_i$$ The moment map is of the form $J(p,q)=(q+u,-p+v)$, for any constant vectors $u,v\in\mathbb{R}^3$. But you can check, these maps are not equivariant!