I do not know if this question is appropriate for this site, but I posted [here](https://math.stackexchange.com/questions/107804/coordinate-free-proof-of-the-hamiltonian-character-of-the-geodesic-flow) without having answers, so now I make this attempt.


Let be $(M,g)$ a pseudoriemannian manifold. Let us identify the tangent and the cotangent bundles through the musical isomorphism $g^\flat:u\in TM\to g(u,\cdot)\in T^\ast M.$ 


It is well known that:



> 
> The geodesics of $(M,g),$ i.e. the solutions of
>  $\frac{D}{dt}\gamma=0,$ are integral curves for the hamiltonian vector
>  field of $K:u\in TM\to \tfrac{1}{2}g(u,u)\in\mathbb{R}$ w.r.t. the
>  canonical symplectic form.
> 
> 
> 


**Question** Knowing how to show it using coordinates and Christoffell symbols, I am wondering how to prove it in an intrinsic way.



Is there a coordinate-free proof of the hamiltonian character of the geodesic flow?
You may find an elegant proof of this fact on **Paternain's book "Geodesic Flows"** (Birkhauser), in the very first pages. For convenience, I will reproduce the main parts of the argument here:


The most important step is to understand the geometry of $TTM$, the tangent bundle to the tangent bundle of $M$. Henceforth denote $\pi:TM\to M$ the footpoint projection. Note that, along the zero section, there is a canonical identification of $T\_{(x,0\_x)}TM=T\_xM\oplus T\_xM$, nevertheless this is not the case for arbitrary points in $TM$. Existence of a canonical horizontal complement to the vertical space $\ker d\pi$ is equivalent to having a connection.


Connection map:
---------------


Fix a connection and consider the *connection map* $K:TTM\to TM$ defined as follows. Consider $\xi\in T\_\theta TM$ and a curve $z:(-\epsilon,\epsilon)\to TM$ s.t. $z(0)=\theta$, $\dot z(0)=\xi$. These give rise to a curve $\alpha=\pi\circ z$ (the projection of $z$ onto $M$)and a vector field $Z$ along $\alpha$, s.t. $z(t)=(\alpha(t),Z(t))$. Then $K$ is defined by $$K\_\theta (\xi)=(\nabla\_{\dot\alpha} Z)(0).$$


Horizontal lift:
----------------


Now, define the horizontal lift $L\_\theta:T\_xM\to T\_\theta TM$ as follows: given $v\in T\_xM$ and $\beta:(-\epsilon,\epsilon)\to M$ a curve s.t. $\beta(0)=x$ and $\dot\beta(0)=v$, let $W(t)$ be the parallel transport of $v$ along $\beta$ and $\sigma:(-\epsilon,\epsilon)\to TM$ be the curve $\sigma(t)=(\beta(t),W(t))$. Then set $$L\_\theta(v)=\dot\sigma(0)\in T\_\theta TM.$$


Finally, we recall that in the above language, the geodesic vector field $G:TM\to TTM$ is clearly given by $G(\theta)=L\_\theta(v)$.


Symplectic structure of $TM$:
-----------------------------


One can verify that, in the above notation, the canonical symplectic structure of $TM$ can be invariantly written as $$\omega\_\theta(\xi,\eta)=g(d\_\theta \pi (\xi),K\_\theta(\eta))-g(K\_\theta(\xi),d\_\theta\pi(\eta)).$$




---



> 
> **Prop.** The geodesic field $G$ is the symplectic gradient of the Hamiltonian $H(x,v)=\tfrac12 g\_x(v,v)$, i.e., for all $\theta \in TM$ and all $\xi \in T\_\theta TM$, $$d\_\theta H(\xi)=\omega\_\theta (G(\theta),\xi).$$
> 
> 
> 


**Pf.** With a curve $z:(-\epsilon,\epsilon)\to TM$ s.t. $z(0)=\theta$, $\dot z(0)=\xi$, we have: $$d\_\theta H(\xi)=\frac{d}{dt}H(z(t))\big|\_{t=0}$$


$$=\tfrac12\frac{d}{dt}g\_{\alpha(t)}(Z(t),Z(t))\big|\_{t=0}$$


$$=g(K\_\theta(\xi),v)$$


$$=g(d\_\theta \pi(L\_\theta (v)),K\_\theta(\xi))$$


$$=g(d\_\theta \pi(G(\theta)),K\_\theta(\xi))$$


$$=\omega\_\theta (G(\theta),\xi)\quad\square$$


