TITLE: Why can we define the moment map in this way (i.e. why is this form exact)?
QUESTION [8 upvotes]: Given a symplectic manifold $(X, \omega)$ and a group $G$ acting on $X$ preserving the symplectic form, we define the moment map $\mu : X \to \mathfrak{g}^*$ so that
$$
\langle d\mu(v), \xi\rangle = \omega\big(\xi^*(x), v\big)
$$
where $\xi \in \mathfrak{g}$, $\xi^*$ is the vector field generated by $\xi$, and $v \in T_x(X)$.
Why can we do this? Why is $\mu$ defined?
I read the above definition as follows. Define the $\mathfrak{g}^*$-valued 1-form $\theta$ via
$$
\theta(v) = \omega(\xi^*,v).
$$
Then this form is exact, and so we define $\mu$ so that $d\mu = \theta$.
Why can we do this? Having done some simple computations, the fact that $G$ preserves the symplectic form seems to be imply that exactness holds (and from the above definition it is necessary), but I don't know why this should be the case.
My question is:

Why does the fact that $G$ preserves the symplectic form imply that the 1-form $\theta$ defined above is exact?

Also:

Is the preservation of the symplectic form equivalent to the exactness of $\theta$?

REPLY [6 votes]: There is yet another interpretation of the already mentioned obstructions for the existence and uniqueness of momentum maps (moment mappings, moment maps, momentum mappings....) in terms of Poisson geometry. This even generalizes then to Poisson manifolds and makes things perhaps a little bit more transparent:
For the existence of a (non-equivariant) momentum map you ask Poisson vector fields to be Hamiltonian, the obstruction therefore lies in the first Poisson cohomology, which is precisely the quotient of Poisson vector field modulo Hamiltonian vector fields.
In the symplectic case this quotient becomes canonically isomorphic to the first deRham cohomology as symplecticl vector fields correspond to closed one-forms while Hamiltonian ones correspond to exact one-forms via the musical isomorphism induced by $\omega$.
The uniqueness (and also the equivariance) is then controlled by the zeroth Poisson cohomology which are th functions with trivial Hamiltonian vector field, also called the Poisson center. In the symplectic and connected case, these are just the constant functions, but in the general Poisson case this might be a much more interesting cohomology. 
The advantage of this point of view is, perhaps, the fact that now all obstructions are somehow arising from the same complex: the canonical complex associated to a Poisson structure $\pi$. As a vector space this is just the sections of the exterior algebra of the tangent bundle (i.e. the multivector fields) and the differential is the Schouten bracet with $\pi$. In the symplectic case, this boils down to the deRham complex via the musical isomorphim.
You can find a detailed discussion of this in the (by the way very: nice) book of Ana Cannas da Silva and Alan Weinstein on "Geometric Models for Noncommutative Algebras".