# is the geodesic flow on Hyperbolic Plane completely integrable?

I'm looking for examples of completely integrable systems and specifically geodesic flows. We remember that when we have a symplectic manifold $(M,\omega)$ (with $M$ of dimension $2n$) and $H:M\rightarrow\mathbb{R}$ a smooth function, its symplectic gradient is the unique field $X_H$ over $M$ satisfying

$$\textrm{d}H=\omega(X_H,\cdot)$$

and we say that the system $(M,\omega,H)$ is completely integrable is there exists $f_1,\ldots,f_{n-1}:M\rightarrow\mathbb{R}$ smooth functions Poisson commuting: $\{f_i,f_j\}=\{f_k,H\}=0$, where $\{f,g\}=\omega(X_f,X_g)$, and with $\textrm{d}f_1,\ldots,\textrm{d}f_{n-1},\textrm{d}H$ linearly independent in a dense set of $M$.

In the cotangent bundle $T^*M$ of a manifold $M$, there exists a canonical symplectic form,

$$\omega_\textrm{can}=\sum{\textrm{d}x_i}\wedge\textrm{d}\xi_i$$

$(x_1,\ldots,x_n,\xi_1,\ldots,\xi_n)$ local coordinates of $T^\star M$. Then, if we consider a riemannian manifold $(M,g)$, we can canonically define a symplectic form on $TM$ with the bundle isomorphism $\Phi:TM\rightarrow T^\star M$ given by

$$\Phi(p,v)=(p,v^\star)$$

where $v^*(w)=g(v,w)$ is the Riesz representation of a functional. Hence we can define $\omega=\Phi^{\star}\omega_\textrm{can}$. In this way, the geodesic flow can be viewed as the flow of the symplectic gradient of the metric hamiltonian $H(p,v)=\frac{1}{2}g_p(v,v)$. Then my question is if the geodesic flow on the tangent bundle of the hyperbolic plane is completely integrable and, if yes, what is the another function beside the metric hamiltonian $H$. Any help will be appreciated.

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Another example of a completely integrable geodesic flow is provided by peakon solutions of the Camassa–Holm shallow water equation; they give the geodesics for the manifold $\{ x_1 < x_2 < \dots < x_n \} \subset \mathbb{R}^n$ with inverse metric tensor $g^{ij}=\exp(-|x_i-x_j|)$. – Hans Lundmark Apr 22 '12 at 22:21

A complete system of independent and Poisson-commuting first integrals is given by: $$H_1=\frac{x'^2+y'^2}{y^2},$$ $$H_2=\frac{x'}{y^2}.$$ Obviously $H_1$ is the kinetical energy.
Instead the constant of motion $H_2$ is geometrically interpreted by means of the following fact:
the constant of motion $-H_1^{-1/2}H_2$ describes the euclidean curvature of the geodesics for the hyperbolic plane.