Let's consider a complete Riemannian manifold $\mathcal{M}$. The geodesic flow of $\mathcal{M}$ is a first-order flow on the tangent bundle $T\mathcal{M}$.
My question: Is it conservative? By conservative, we mean that there exists a potential $\Psi : T\mathcal{M}\to \mathbb{R}$ whose gradient induces the geodesic flow.
Thanks in advance!
Consider a compact Riemannian manifold. Its geodesic flow preserves its unit sphere bundle, also compact. On the unit sphere bundle, any potential will have a minimum and a maximum, so there will be points where the gradient flow doesn't move. But all geodesics with unit vector initial condition move at unit speed.
Suppose that there is a construction, to each Riemannian manifold $(M,g)$, constructing a Riemannian metric $h$ on an open set $T^oM\subseteq TM$ containing the unit tangent bundle, and a function $f$ on that same $T^oM$, so that $\nabla^h f$ is the geodesic flow, invariant under isometry, so that when we restrict to an open subset $U\subset M$, the resulting $h$ and $f$ are the restrictions:
$$h_U=\left.h_M\right|_{T^oU}$$, $$f_U=\left.f_M\right|_{T^oU}$$.
Take any point $m_0\in M$ in a Riemannian manifold $M$. Cut and paste to make a new Riemannian metric, on a compact manifold $M'$, which is locally isometric to $M$ by some isometry of open sets $U\subset M\to U'\subset M'$ with $m_0\in U$, and we contradict the result above.
So there is no conservative vector field giving the geodesic flow on compact Riemannian manifolds, and no local construction of such a vector field on Riemannian manifolds, invariant under local isometry.
Note that all vector fields, where they are nonzero, are locally diffeomorphic to one another, by the flow box theorem, so are locally conservative. So we can't do better than saying that there is no isometry invariant local construction: there is always a local consruction, but not an isometry invariant local construction.
