I'm trying to understand why a certain action of a Lie Group is hamiltonian.


Let $(M,g)$ be a geodesically complete Riemannian manifold.
Then there exists a canonical one-form on the cotangentbundle $T^\*M$ given by $\lambda\_0 : T( T^\*M) \to R, (v, \alpha) \mapsto \alpha(\pi\_\*v)$, with $\pi\_\*$ the derivative of the natural bundle projection $T^\*M \to M$.
It is now possible to pullback this one-form to the tangentbundle using our riemannian metric, so $\lambda = (g^b)^\*\lambda\_0$. Using this one-form we define a symplectic structure on $TM$ by $\omega = -d\lambda$.


Now let $\phi:R\times \text{TM} \to \text{TM}, V \mapsto \dot \gamma\_V(t) $ be the action generated by the geodesic flow.


My question is, if $\phi$ acts by symplectomorphisms, that is, if $\phi\_t^\*\omega = \omega$ for all $t\in R$ or is it even exact symplectic, that is, if $\phi\_t^\*\lambda = \lambda$ for all $t\in R$? Because Im trying to show that this is an hamiltonian action, so that there exists a moment map for this action.


And it would be clear if we have an exact symplectic action.


Edit: To have a well-defined action of $R$, we assume that the manifold is geodesically complete.



Action generated by geodesic flow is hamiltonian
$\phi$ is the Hamiltonian flow corresponding to the ``Hamiltonian'' $E:TM\rightarrow\mathbb{R}$ given by $E(V) = \frac{1}{2}g(V,V)$. This is pretty standard, and you can read about it in most references on symplectic geometry, for example [Foundations of Mechanics](http://authors.library.caltech.edu/25029/) (in particular, Theorem 3.7.1). The corresponding momentum map of this flow is simply the Hamiltonian $E$ itself (with the canonical identification of $\mathbb{R}^\*$ with $\mathbb{R}$).


The flow is not however exact symplectic, and in fact the generator of Hamiltonian flow $X\_E\in\mathfrak{X}(TM)$ satisfies $\mathcal{L}\_{X\_E}\lambda = dL$, where $L$ is the Lagrangian, which equals $E$ in this case. To see this in coordinates, note that
$$
i\_{X\_E}\lambda = p\_i dq^i \left(\dot{q}^i\frac{\partial}{\partial q^i} \right) = p\_i\dot{q}^i.
$$
So using Cartan's Magic Formula
\begin{align\*}
\mathcal{L}\_{X\_E} \lambda &= i\_{X\_E}d\lambda + d(i\_{X\_E}\lambda) \\
&= -i\_{X\_E}\omega + d(p\_i\dot{q}^i) \\
&= d(p\_i\dot{q}^i - E) \\
&= dL.
\end{align\*}
For a more geometric treatment, see Foundations of Mechanics, sections 3.5-3.7.


