Integrability of the tautological 1-form

[MathNeophyte]

TL;DR Summary

Trying to understand under what conditions the tautological/Liouville one form is integrable in the sense of having $n$ Poisson commuting integrals.

Apologies for potentially being imprecise and clunky, but I'm trying understand integrability of the following Hamiltonian
$$H(x,p)=\langle p,f(x) \rangle$$
on a 2n dimensional vector space
$$T^{\ast}\mathcal{M} =\mathbb{R}^{2n}.$$
Clearly this is just the 1-form $$\theta_{(x,p)} = \sum_{i}{p_i{}dq^{i}}.$$
I assume that anywhere along the flow of x given some x₀
$$x(\tau) = \Phi(x_{0},\tau)$$ the Hamiltonian is non-degenerate i.e. the Hessian of the Hamiltonian is full rank.

My questions are:

1. Is H(x,p) Liouville integrable along the orbits of x given the above non-degeneracy assumption? I.e. Does this Hamiltonian have n, non-constant integrals that are commuting with respect to the cannonical Poisson bracket?
2. How does this relate to the differential form version of the Frobenius theorem?

I would appreciate any and all pointers, this clearly isn't my domain of expertise but it is a question that arose in some work on dynamical systems.

 

[member38644]

The tautological 1-form is the canonical 1-form on the cotangent bundle $T^*\mathcal{M}$, given by $\theta_{(x,p)} = \sum_i p_i dq^i$. This is a special case of a symplectic form, which is a closed, non-degenerate 2-form on a symplectic manifold. In this case, the symplectic manifold is $T^*\mathcal{M}$, which has dimension $2n$.

Now, to answer your first question, the Hamiltonian $H(x,p) = \langle p,f(x) \rangle$ is not Liouville integrable in general. Liouville integrability requires the existence of $n$ independent, globally defined integrals of motion that commute with each other. In this case, we only have one integral of motion, given by $H(x,p)$, and it is not guaranteed to be globally defined. Furthermore, even if we have $n$ independent integrals of motion, they may not necessarily commute with each other.

To understand this better, let's look at the differential form version of the Frobenius theorem. The Frobenius theorem states that a distribution on a manifold is integrable if and only if it is involutive. In the case of a symplectic manifold, the distribution is given by the kernel of the symplectic form, which is the set of all vectors that are orthogonal to the symplectic form. This distribution is involutive, which means that it is closed under the Lie bracket operation. In other words, if we have two vectors $X$ and $Y$ in the distribution, then their Lie bracket $[X,Y]$ is also in the distribution.

Now, let's relate this to our Hamiltonian. The Hamiltonian vector field $X_H$ associated with $H(x,p)$ is given by $X_H = \frac{\partial H}{\partial p_i}\frac{\partial}{\partial q^i} - \frac{\partial H}{\partial q^i}\frac{\partial}{\partial p_i}$. This vector field is tangent to the distribution, since it is in the kernel of the symplectic form. Furthermore, it is closed under the Lie bracket operation, which means that it generates a flow that preserves the symplectic structure. However, this does not guarantee that the Hamiltonian is Liouville