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Given a vector field $v$ on a $2n$-dimensional manifold, how many symplectic forms are there on $M$ that make $v$ a hamiltonian vector field? Alternatively, take the set of all $(H,\omega)$ pairs, mod adding constants to $H$ and doing $(H,\omega)\to (\alpha H,\omega/\alpha)$. Each element of this set may be associated to a vector field via the map $v^i=\omega^{ij}\partial_j H$ (call the map $F$ for later reference). The question is: given a vector field $v$, what does the preimage of $v$ under this map look like (is it empty/one element/large/etc)?

This question is inspired by this answer, where Qmechanic gives this local existence theorem:

Let there be given a $2n$-dimensional manifold $M$ with a non-vanishing vector field $X\in\Gamma(TM)$. Does there locally exists a symplectic two-form $\omega$, such that $X$ is a Hamiltonian vector field?

Answer: Yes!

My question differs in that I am interested in uniqueness/non-uniqueness in addition to existence, and I am interested in any obstructions to the existence of a global two-form satisfying the quoted result (aside from the manifold itself not supporting a symplectic structure).


For completeness, here are my (haphazard and muddled) attempts at answering this question:

The zeroth thing I did was make the following (probably very flawed) degree of freedom counting argument, forgetting about topological obstructions etc. It takes $2n$ functions on $M$ to specify $v$, $2n-1$ functions on $M$ to specify $\omega$ (since we can add gradients to the tautological one form without changing $\omega$), and $1$ function for the hamiltonian. Thus the very naive reasoning suggests that vectors $v$ should be in correspondence with pairs $(H,\omega)$. As we see below this fails to happen (even for well-behaved examples), so I would be interested in how to make this sort of argument properly.

The next thing I tried was to work out some specific cases. The hamiltonian flow for the harmonic oscillator, which on $\mathbb{R}^2$ with coordinates $(q,p)$ looks something like $(p,-q)$, seems to be hamiltonian with respect to any symplectic form which is a function of $p^2+q^2$ (which seems like a lot of freedom). In particular, let $d\equiv (p^2+q^2)/2$ and consider the hamiltonian $H=f(d)$ and the symplectic form $\omega=f'(d) dq\wedge dp$. Given any function $f'(d)$ which is nonzero everywhere, $\omega $ so defined will be a globally defined two-form, and $H$ will be a globally defined hamiltonian, such that the image of $(H,\omega)$ under $F$ is $v$.

Now let us consider a slightly different situation. Let's work on $\mathbb{R}^2\setminus \{0\}$, and define $v$ to be the flow $(-q,-p)$ (i.e. a radially inward flowing vector field). This is obviously not hamiltonian under the usual form on $\mathbb{R}^2$, but if I did the Mathematica right, any symplectic form of the form

$$\omega = \frac{f'(p/q)}{q^2} dq\wedge dp,$$

with a hamiltonian of the form $H=f(p/q)$, will reproduce this vector field. In this case, unlike before, not any positive $f'$ will give us a globally defined symplectic form, since for instance $f(x)=x$ would produce singularities at $q=0$. Thus, even though locally we have the same freedom to pick $\omega$, globally we do not. However, the choice:

$$f(x)=\frac{x^2}{1+x^2}$$

(which I basically found by guess-and-check) seems to produce a globally defined symplectic form and hamiltonian, so there still exists a global symplectic form on our space with respect to which this vector field is hamiltonian.

Evidently according to this comment, we can also find vector fields on $\mathbb{R}^2\setminus \{0\}$ that don't admit any $(H,\omega)$ pairs globally, although presumably there is still a lot of freedom in picking those pairs locally.

Is there any obvious geometric feature/property that we can use to distinguish these three cases just from looking at $v$?

Lastly, I wanted to briefly note the following. On some manifolds we can choose vector fields that are locally hamiltonian but not hamiltonian (like the last example before; see, for instance, Exercise 7 here). In this situation, we can find a globally defined symplectic form $\omega$, and a collection of locally defined hamiltonians, such that on each patch $v$ arises via hamilton's equations. However, one could imagine a situation where no one global symplectic form will do, no matter how "multivalued" we allow $H$ to be. Is there one? I couldn't think of how to come up with one.

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    $\begingroup$ Perhaps this question is better suited for Mathematics $\endgroup$ – Danu Sep 12 '15 at 18:10
  • $\begingroup$ Ah I asked it here because it seemed like a generalization of the linked physics.SE questions. Can I move the question myself? Should I delete it and re-ask it at math.SE? $\endgroup$ – commutatertot Sep 12 '15 at 20:10
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    $\begingroup$ No, simply flag it and describe your request in a custom flag. $\endgroup$ – Danu Sep 12 '15 at 20:17
  • $\begingroup$ Any progress on answering the original question or on a given symplectic manifold, when is a vector field Hamiltonian for some function? $\endgroup$ – Jim Stasheff Jan 27 '17 at 20:43
  • $\begingroup$ @JimStasheff I haven't been working on this lately; any suggestions you might have would be very welcome! $\endgroup$ – commutatertot Jan 27 '17 at 22:08

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