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Here is my precise question. Let $M, \omega$ be a symplectic manifold and let $H: M \to \mathbb{R}$ be any smooth function. The symplectic form gives rise to an isomorphism between the tangent bundle and cotangent bundle of $M$, and in this way we can associate to the 1-form $dH$ a vector field $X_H$ which is characterized by the property that $\omega(X_H, Y) = Y(H)$ for any vector field $Y$. The one parameter group of diffeomorphisms associated to $X_H$ is the "Hamiltonian flow" associated to $H$.

An interesting special case of this construction is furnished by Riemannian geometry. For any manifold $M$, there is a canonical symplectic structure on $T^*M$ (regarded as a manifold in its own right) defined as follows. Given a tangent vector $X \in T(T^*M)$ sitting over a covector $p \in T^*M$, define $\eta_p(X) = p(d\pi_p(X))$ where $\pi: T^*M \to M$ is the natural bundle projection. Then $\eta$ is a 1-form on $T^*M$, and one checks that $d\eta$ is a symplectic form. If $M$ is equipped with a Riemannian metric $g$ then the metric yields an isomorphism between $TM$ and $T^*M$, and the construction of the previous paragraph produces a Hamiltonian flow associated to any smooth function on $TM$. If we consider the smooth function $H: TM \to \mathbb{R}$ given by $H(V) = g(V, V)$, then it is a fact that the resulting Hamiltonial flow $F_t$ is precisely the geodesic flow for $M$. In other words, given a tangent vector $W \in T_p M$, $F_t(W)$ is the velocity vector at time $t$ of the unique geodesic $\gamma$ with $\gamma(0) = p$, $\gamma'(0) = W$.

So I am wondering if there are interesting invariants - dynamical, geometric, topological, or otherwise - which help to determine whether or not a given Hamiltonian system is secretly the geodesic flow on some Riemannian manifold. This is kind of a screwy question from a geometric point of view, because it essentially asks if given a smooth function $H: M \to \mathbb{R}$ on a symplectic manifold there is a submanifold $N$ of $M$ such that there is a diffeomorphism $M \to TN$ which carries $H$ to a positive definite quadratic form on each fiber. But dynamically it boils down to a fairly natural question: how can one characterize geodesic flows among all Hamiltonian dynamical systems?

If this question has any sort of reasonable answer, I can think of half a dozen follow-up questions. Is there a natural notion of equivalence up to which $N$ is unique? To what extent does $H$ constrain the geometry and topology of $N$? If a Lie group acts on the pair $M, H$ can we choose $N$ which is invariant under the group action? For example, one idea along these lines that comes to mind immediately is the assertion that if the Hamiltonian flow for $H$ is not ergodic relative to a prescribed smooth invariant measure then $N$ cannot have nonpositive curvature. If you have an answer to this question and you can elaborate on the relationship between $H$ and the geometry of $N$, please do so.

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I am pretty sure this is somewhat of an open question. You can start from and go from there. – Willie Wong Oct 1 '10 at 23:28
Oh, and just remembered that Terry also wrote about this a short while ago on his blog: – Willie Wong Oct 1 '10 at 23:30
Reading Tao's blog, it seems that I could have been even more ambitious and asked when the Hamiltonian system is equivalent to the geodesic flow on a Lie group, not just any old Riemannian manifold. Thanks for the link! Poking around a little further, I am getting the impression that the phenomenon that I'm asking about is well understood in some specific examples but that there might not be a lot known in general. It's all very interesting anyway. – Paul Siegel Oct 2 '10 at 12:52
up vote 5 down vote accepted

My reading of the question is this: we're given $H\in C^\infty(M)$ with $M$ symplectic, and we want to know whether there's a submanifold $L\subset M$, a Riemannian metric $g$ on $L$, and a symplectomorphism $T^\ast L \cong M$ under which $H$ pulls back to the norm-square function. And we want to know if $(L,g)$ is unique.

Uniqueness is easy: we recover $L$ as $H^{-1}(0)$, and $g$ as the Hessian form of $H$ on the vertical tangent bundle (determined by the symplectomorphism) along $L$.

Basic necessary conditions:

(1) $L:=H^{-1}(0)$ is a Lagrangian submanifold of $M$.

(2) $L$ is a non-degenerate critical manifold of $H$ of normal Morse index 0.

These conditions imply that a neighbourhood of $L$ embeds symplectically in $T^\ast L$, and also (by the Morse-Bott lemma) that $H$ is quadratic in suitable coordinates near $L$. These two sets of coordinates needn't be compatible, so let's replace (2) by something much stronger (but still intrinsic):

(3) There's a complete, conformally symplectic vector field $X$ (i.e., $\mathcal{L}_X\omega=\omega$), whose zero-set is exactly $L$, along which $H$ increases quadratically (i.e., $dH(X)=2H$).

I claim that (1) and (3) are sufficient. With these data, you can locate a point $x\in M$ in $T^\ast L$. Flow $X$ backwards in time starting at $x$ to obtain the projection to $L$; pay attention to the direction of approach to $L$ to get a tangent ray, and use the metric (i.e., the Hessian of $H$ on the fibres of projection to $L$) to convert it to a cotangent ray. Pick out a cotangent vector in this ray by examining $H(x)$. If I'm not mistaken, this will single out a symplectomorphism with the desired properties.

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I want to thank everyone for your extremely helpful answers - I upvoted everyone. This particular answer works from the point of view that is most useful to me. – Paul Siegel Oct 9 '10 at 14:46

I wish to add an $\epsilon$ to the previous answer. Assume that your symplectic manifold is $T^*M$, where $M$ is a closed manifold, and consider a so called ``Tonelli Hamiltonian´´ $H:T^*M\rightarrow\mathbb R$, which is simply a function which is fiberwise (differentiably) convex and superlinear. Then consider the value $$ c(H)=\min_{u\in C^\infty(M;\mathbb R)} \max_{q\in M} H(q,du_q). $$

This special number in the literature is called ``Mañé critical value´´. Now, if you fix an energy value $h > C(H)$, you can easily build a new Hamiltonian $G$ such that

1) $H^{-1}(h)=G^{-1}(h')$ for some $h'\in\mathbb R$

2) $G(q,\lambda p)=\lambda^2 G(q,p)$ for each $\lambda>0$

3) $G$ is fiberwise convex

Now, the Legendre dual of the Hamiltonian $G$ will be a Finsler metric (not Riemannian in general).

The interesting thing here is that the critical value $c(H)$ can be defined in many other equivalent ways (in terms of Lagrangian action of closed loops, in terms of minimizing invariant measures, etc.). You can find more on this in the book by Contreras and Iturriaga:

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This is a kinda stupid answer but it is the simplest most useful test I know. Does your Hamiltonian vector field have a zero? If `yes': sorry, not a geodesic flow!

(Could be the reduction of a geodesic flow though.)

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Here is a (partial) answer to your question for the so-called natural Hamiltonian systems. The latter are defined as follows. Let $(M,g)$ be a (pseudo-)Riemannian manifold, $q^i$ be local coordinates on $M$ and $(q^i, p_i)$ be the adapted coordinates on $T^*M$ which is a symplectic manifold with a canonical symplectic structure as described in the question. In these coordinates the natural Hamiltonian is one of the form $$ H=\frac12 \sum\limits_{i,j=1}^n g^{ij}(q^1,\dots,q^n) p_i p_j + V(q^1,\dots,q^n), $$ where $n=\dim M$. Using the Legendre transformation we can pass to an equivalent Lagrangian $$ L=\frac12 \sum\limits_{i,j=1}^n g_{ij}(q^1,\dots,q^n) \displaystyle \frac{dq^i}{dt} \frac{dq^j}{dt}-V(q^1,\dots,q^n). $$

For such Lagrangians the answer to your question is in the affirmative if we restrict ourselves to the motion with a fixed energy $E$, i.e., on a hypersurface $H=E$. The respective solutions of the Euler--Lagrange equations associated with $L$ indeed can be viewed (see e.g. the paper Geometry of spaces with the Jacobi metric by Szydlowski, Heller and Sasin and references therein for details) as geodesics of the so-called Jacobi metric $$ \tilde g_{ij}=(E-V)g_{ij} $$ which arises from the Maupertuis principle.

Note that here one considers the curves being the solutions of the Euler--Lagrange equations and the geodesics of $\tilde g$ as (roughly speaking) one-dimensional submanifolds in $M$ ignoring their parametrization. The metric $\tilde g$ obviously vanishes in the points where $E=V$, so it is in fact a degenerate metric, see e.g. the above paper by Szydlowski et al. for details.

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