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:

1. A Lagrangian foliation over its base $(q,p) \mapsto q$.

2. 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 everywhere non degenerate, $\ker D(\Phi)_x = \{0\}$. 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.

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:

1. A Lagrangian foliation over its base $(q,p) \mapsto q$.

2. 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.

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, but this is the direction I would take if I had to "solve" the question.

A cotangent bundle $T^*Q$ has two main characteristics:

1. A Lagrangian foliation over its base $(q,p) \mapsto q$.

2. 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 everywhere non degenerate, $\ker D(\Phi)_x = \{0\}$. 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 get from an étale map to a diffeomorphism and a complete answer to your question. I may come back on that soon if I have some idea.

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:

1. A Lagrangian foliation over its base $(q,p) \mapsto q$.

2. 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 everywhere non degenerate, $\ker D(\Phi)_x = \{0\}$. 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.