Fundamental vector field and free group action

Question

I am trying to understand this:

from http://www.math.harvard.edu/~auroux/253y18/2%20-%20Moment%20maps%20and%20symplectic%20reduction.pdf .

G is a compact Lie group acting freely on M such that the action is hamiltonian with hamiltonian mpa $\mu$. I see that the converse is true but why do we have this yellow implication?

Thank you very much

Answer 1

The orbit of $x$ by flow of the vector field $Y$ is the unique solution of the differential equation:

${d\over{dt}}_{t=t_0}\phi_t(x)=Y(\phi_{t_0}(x)$, with the initial condition $\phi_0(x)=x$.

Suppose that $Y(x)=0$, write $\phi_t(x)=x$ we have ${d\over{dt}}_{t=t_0}\phi_t(x)=Y(\phi_{t_0}(x))=Y(x)=0$, This implies that $\phi_t(x)=x$ is a solution of that equation and is the unique solution.

Take $Y=a_1X_1+...+a_nX_n$.