# Is there a coordinate-free proof of the hamiltonian character of the geodesic flow?

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.

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Despite the apparent simplicity of the question, I think this is a very good question for this site. –  Deane Yang Feb 16 '12 at 13:57
I have had three interesting answers until now (not being ruled out that there are others in the future), meanwhile I have thought to accept the one giving more details and more supporting references. –  Giuseppe Tortorella Feb 18 '12 at 7:56

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: $$d_\theta H(\xi)=\frac{d}{dt}H(z(t))\big|_{t=0}$$

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

$$=g(K_\theta(\xi),v)$$

$$=g(d_\theta \pi(L_\theta (v)),K_\theta(\xi))$$

$$=g(d_\theta \pi(G(\theta)),K_\theta(\xi))$$

$$=\omega_\theta (G(\theta),\xi)\quad\square$$

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Thank you Renato, I like very much your answer. I need to find out my way through this material, but surely your reference will be helpful –  Giuseppe Tortorella Feb 16 '12 at 18:44

This is a very particular case of something more general: extremals for a non-degenerate Lagrangian correspond to solutions of the corresponding Hamiltonian system. There is no need to use connections at all. Check out Abraham and Marsden.

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Good point. But any chance you could provide a few details for those of us who don't have easy access to Abraham and Marsden? –  Deane Yang Feb 16 '12 at 18:30
@Deane: you might like this (even legal) way to take a look cds.caltech.edu/~marsden/books/Foundations_of_Mechanics.html –  Stefan Waldmann Feb 16 '12 at 20:17
With @Stefan, I highly appreciate the Marsden posts his books online, and I hope others do too. But I also think @alvarezpaiva should elaborate the answer, for two reasons: (1) it is good practice to write such elaborations, (2) the few mouse-clicks, the better. –  Theo Johnson-Freyd Feb 17 '12 at 4:43
Viewing things this way, one still needs to demonstrate that the Euler-Lagrange equations are the geodesic equations for the metric. Abraham and Marsden demonstrate this in Theorem 3.7.1 (page 224), but using coordinates, which Giuseppe was trying to avoid. It seems that showing this invariantly would still require the introduction of a connection, no? (Presumably the connection mentioned by A&M in the remarks on the bottom of page 227 is the same one introduced by the two posters above - though I would need to check to be sure.) –  user17945 Feb 17 '12 at 6:05
BTW apologies for the bump - accidentally posted an answer instead of a comment - doh! –  user17945 Feb 17 '12 at 6:11
Here is an outline of a proof. It relies on the splitting of the tangent bundle of the tangent bundle $$TTM=VTM\oplus HTM$$ into the vertical part $VTM=ker d\pi,$ where $\pi\colon TM\to M,$ and into the Horizontal part, which is induced by the Levi-Civita connection as follows: Let $v\in T_pM$ and $\gamma$ be a curve centered at $p.$ Take the parallel section $v$ along $\gamma$ through $v$ wrt Levi Civita: Then, by definition, $v'(0)\in H_vTM\subset TvTM,$ and this construction gives a well-defined bundle complement of the vertical space in $TTM.$ Note that both $VTM$ and $HTM$ are canonically isomorphic to $\pi^*TM.$ This enables one to define a almost complex structure $J$ on TTM by switching in the appropriate way from $HTM$ to $VTM$ and vice versa. (I don't want to give a formula, as this would cause sign problems..). Moreover one gets an induced metric from M, which makes $HTM$ orthogonal to $VTM,$ denoted by $G.$
The first thing is to show that (lets call the symplectic form $\omega$) $$\omega(A,B)=\pm G(A,JB)$$ for all $A,B\in T_vTM.$ This again uses the definition of $HTM$ as well as Cartan formula. This can be done without coordinates.
The derivative of $K$ along the vertical space is just given as $$d_vK=G(\hat v,.),$$ where $\hat v\in V_vTM\subset T_vTM$ is the vertical vector corresponding to $\in T_pM.$ Therefore, the hamiltonian vector field $\xi$ of $K$ is given by $\pm J\hat v$ which is (after taking care with the sign) exactly the horizontal lift of $v\in T_pM$ to $H_vTM.$ An integral curve $\hat\gamma$ of $\xi$ is automatically a parallel section of $TM$ along its projection $\gamma=\pi\circ \hat\gamma,$ and as $\gamma'(t)=\hat\gamma(t),$ $\gamma$ must be a geodesic.