Lift of symplectic form to the cotangent bundle Consider a symplectic manifold $(M,\omega)$ on which a Hamiltonian group action $G\curvearrowright M$ is defined. Then we have the usual canonical induced action of $G$ on $X:=T^*M$, that also preserves the fibration $\pi:X\to M$. My question is:

Is there any symplectic form on $X:=T^*M$ that coincides with $\omega$ on the zero-section of $X$, that is also preserved by the induced action $G\curvearrowright X$ ?

I realize that the question may seem too broad or vague, but I would very much appreciate any reference to articles or books addressing the question. Feel free to assume additional regularity on the manifold, like compactness or being Kähler.
 A: Such a symplectic form exists and it is quite canonical.
Given any smooth manifold $M$, the (total space of the) cotangent bundle $X = T^{\ast}M$ has a canonical symplectic form $\omega_0$, the so-called tautological symplectic form. Any diffeomorphism $\psi : M \to M$ lifts to a symplectomorphism $\Psi = (\psi^{-1})^{\ast} : (X, \omega_0) \to (X, \omega_0)$.
Now suppose $M$ is itself symplectic, with symplectic form $\omega$. Its pullback under the bundle projection $\pi : X \to M$ gives a degenerate (as it vanishes on the fibers of $\pi$) 2-form $\pi^{\ast}\omega$ on $X$ which 'coincides' with $\omega$ on the zero section of $\pi$.
We can then add the two 2-forms on $X$ in order to get the simplest instance of Thurston's construction: set $\Omega := \omega_0 + \pi^{\ast}\omega$. $\Omega$ is still symplectic: If $0 \neq v \in T_pX$ lies in the kernel of $d\pi_p$, then $v \lrcorner \, \Omega = v \lrcorner \, \omega_0$ is non-zero by the nondegeneracy of $\omega_0$, whereas if it doesn't lie in $ker \, d\pi_p$ then there exists $w \in ker \, d\pi_p$ such that $\omega_0(v,w) \neq 0$ (exercise), but as $w \lrcorner \, \pi^{\ast}\omega = 0$ we have $\Omega(v,w) = \omega_0(v,w) \neq 0$.
Given a symplectomorphism $\psi : (M, \omega) \to (M, \omega)$, we have that $$\Psi^{\ast}(\pi^{\ast}\omega) = (\pi \circ \Psi)^{\ast}\omega = (\pi \circ (\psi^{-1})^{\ast})^{\ast}\pi = (\psi \circ \pi)^{\ast}\omega = \pi^{\ast}(\psi^{\ast}\omega) = \pi^{\ast}\omega, $$
whence $\Psi^{\ast}\Omega = \Omega$ by the 'naturality' of $\omega_0$. It proves that symplectomorphisms of $(M, \omega)$ lifts to symplectomorphisms of $(X, \Omega)$ (and group actions thereof).
Observe that we did not use the nondegeneracy of $\omega$ in any crucial way: everything works for a closed 2-form $\omega$ and for diffeomorphisms which preserve this form.
The form $\pi^{\ast}\omega$ is sometimes called the 'coupling form'. This terminology comes from the fact that when $M$ is space(time), $\omega$ can be thought of as an electromagnetic 2-form (acting on some implicitly given charged particle). It happens that the motion of this charge particle is described by the usual 'free' hamiltonian function on $X$ and the use of the symplectic form $\Omega$ (instead of the more customary scheme where one considers the hamiltonian function resulting from the 'minimal coupling' $p \mapsto p - eA$ and uses the canonical symplectic form $\omega_0$).
