TITLE: Is there a coordinate-free proof of the hamiltonian character of the geodesic flow?
QUESTION [17 upvotes]: I do not know if this question is appropriate for this site, but I posted here 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.

REPLY [3 votes]: Suppose that $G$ and $G'$ be the vector fields on $\mathrm{T}M$ that describe the first and the second definition of the geodesic flow; we need to show that $G=G'$.
If the metric tensor $g$ is flat, then $G=G'$ (it should be obvious).
Now assume that two metric tenors $g_0$ and $g_1$ on $M$ coincide at a point $p$ up to first order.
Observe that $G_0=G_1$ and $G'_0=G'_1$ on $\mathrm{T}_p$.
Finally, observe that given a Riemannian metric $g_1$ there is a flat metric $g_0$  that coincides with $g_1$ up to first at a given point $p$.