Let $\Phi: G \times \mathbb{R}^4 \rightarrow \mathbb{R}^4$ be a group action where $G = \mathbb{R}/(2 \pi \mathbb{Z}).$

Then $$\Phi(\theta,(x_1,x_2,p_1,p_2)) = ( R(\theta) (x_1,x_2)^T, R(\theta) (p_1,p_2)^T)$$

with $R(\theta):=\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \\ \end{bmatrix}$ is the standard rotation matrix.

Then I know that the corresponding Hamiltonian vector field is

$$\phi_{\zeta} = \zeta(-x_2,x_1,-p_2,p_1).$$

Afaik the corresponding vector field is defined by $\phi_{zeta}(x_1,x_2,p_1,p_2)=\frac{d}{dt}|_{t=0} \Phi(e^{t \zeta},(x_1,x_2,p_1,p_2))$ but this does correspond to the end result at all. Does anybody know how we can derive this vector field?


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