TITLE: When is a symplectic manifold equivalent to a cotangent bundle?
QUESTION [33 upvotes]: Let $X$ be a differentiable manifold. Its cotangent bundle $T^*X$ carries a canonical 1-form $
\alpha$ whose exterior differential $\omega = d\alpha$ endows $T^*X$ with the structure of a symplectic manifold.
But what about the converse question? Which symplectic manifolds are cotangent bundles? 
Clearly a necessary condition is that $\omega$ must be exact, so cohomological obstructions are relevant. Is that all? Compact symplectic manifolds have non-trivial de Rham cohomology in grade two, so the cohomological test passes muster for that important class of examples.
I'm also interested in examples of manifolds where different symplectic forms (modulo exact 2-forms) give qualitatively different dynamics with the same Hamiltonian. As symplectic structures on the same manifold are all locally equivalent by Darboux's theorem, one expects such phenomena would occur only on a global scale.

REPLY [5 votes]: I don't know if this question : "when a symplectic manifold is isomorphic to a cotangent bundle" has a complete and simple answer in the literature, in the way you want, but this is some comments that come in mind about this question. 
A cotangent bundle $T^*Q$ has two main characteristics: 

A Lagrangian foliation over its base $(q,p) \mapsto q$. 
A one parameter dilatation group $t\mapsto [(q,p) \mapsto (q,e^t
    p)]$.

On a symplectic manifold $(M,\omega)$, we usually call polarization a Lagrangian foliation when the space of leaves is a manifold. And we call Liouville 1-parameter group a 1-parameter group of diffeomorphisms $\varphi_t$ such that $\varphi_t^*(\omega) = e^t\omega$. The infinitesimal action of the Liouville group is a (complete) Liouville vector field, it satisfies ${\cal L}_\xi(\omega) = \omega$. Every Liouville vector field on a symplectic manifold gives a primitive of the symplectic form: $\alpha(\cdot) = \omega(\xi,\cdot)$, that is, $d\alpha = \omega$. We have then two necessary conditions for the symplectic manifold $(M,\omega)$ to be a cotangent space: 

There exists a polarization $\pi : x \mapsto q$ onto some manifold $Q$.
There exists a complete Liouville vector field $\xi$, tangent
to the polarization $\pi$.

Let's assume now that these two conditions are satisfied, we can define a natural map $\Phi : M \to T^*Q$ by 
$$
\Phi(x) = (q = \pi(x), p = [\delta q \mapsto \omega_x(\xi(x),\delta x)]) 
\quad \mbox{with} \quad \pi_*(\delta x) = \delta q.
$$
Here $\delta q \in T_qQ$, $\delta x \in T_xM$. Because the Liouville field is tangent to the polarization, $\omega_x(\xi(x),\delta x)$ depends only on $\delta q = \pi_*(\delta x)$, and therefore $p$ belongs to $T^*_qQ$. Now this map $\Phi$ satisfies:
$$
\Phi(\lambda) = \alpha \quad \mbox{and then} \quad \phi^*(d\lambda) = \omega,
$$
where $\lambda$ is the canonical Liouville 1-form $pdq$ on $T^*Q$. Now, since $d\lambda$ and $\omega$ are symplectic the tangent linear map $D(\Phi)_x$ is non degenerate, $\ker D(\Phi)_x = \{0\}$ for all $x$. Thus, $\Phi$ is an étale map, that is, a local diffeomorphism everywhere. Hence, $(M,\omega)$ is not far to be a cotangent bundle, we can already say that $\omega$ is the pullback of the standard symplectic form $d\lambda$ on a cotangent $T^*Q$ by an étale map which is already a bit of interesting information. It remains to give some conditions on the polarization to move from an étale map to a diffeomorphism. I don't know if it is exactly the sense you gave to your question but it may help to apprehend the situation.