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.


1 Answer 1


$\phi$ 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 (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.


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