While symplectic vector fields are closed in relation to the symplectic form, i.e.
if we impose a stronger condition - namely that be an exact differential form, we get Hamiltonian vector fields.
Note that the exactness condition is stronger because if is exact, then it is necessarily closed since and .
This is often applied using a smooth function
(calling it for suggestive reasons), we want to learn something about the manifold and its symplectic structure from the function.
For physical systems, the smooth function has an intrinsic important meaning about the physical system.
For example, could represent “total energy” of the components of the system.
Since we are working with a smooth manifold, for a smooth function .
Thus,
Since is a symplectic form we know that for a point ,
is an isomorphism.
Therefore, we can look for the vector such that
Thus, collecting all these tangent vectors to get a vector field , we get
Making Hamiltonian vector fields
If are symplectic vector fields then (denoting the Lie bracket on vector fields) is a Hamiltonian vector field with Hamiltonian function .
Proof
It suffices to prove that is exact.
Which makes sense since is a smooth function.
Thus, is a Hamiltonian vector field.
Equivalence on Poisson and symplectic manifolds
The notions of Hamiltonian vector fields on Poisson manifolds and symplectic manifolds are equivalent when a Poisson manifold is symplectic.
Proof
Let be a Poisson manifold where is a nondegenerate Poisson bivector.
Let be the Hamiltonian vector field of in the notion of a Hamiltonian vector field of a symplectic manifold, that is
Taking the Poisson bracket of with any arbitrary function gives
The last line follows from the fact that is an isomorphism the construction of the symplectic form.
Note that looking for is equivalent to asking what is the vector field such that
Therefore, it is the Hamiltonian vector field .
This vector field acting on gives
Next let be a Hamiltonian vector field in the Poisson manifold sense, i.e.