Classical physics on the phase space $T^* M$ (with $M$ a smooth manifold) is done mostly in the following way: one endows $T^*M$ with a Riemannian structure $g^*$ (that will give the kinetic term) and considers a potential $V : M \to \Bbb R$ (that will give the potential term). Then, in a local trivialization, the Hamiltonian will be $H(x, p) = \frac m 2 g^* _x (p, p) + V(x)$ (that will give the total energy), and it occupies a very special place among the observables of the system because the equations of evolution (of either the state, or the observables) can be written in terms of it (and the Poisson structure).
My question is: on a symplectic manifold that is not necessarily a cotangent bundle, does it make sense to speak of the Hamiltonian (note the definite article) in the absence of a Riemannian structure? Or, at least, about a Hamiltonian (note the indefinite article)? Can one produce it only from the symplectic structure alone?
Please consider answering the same question also in the setting of a Poisson (insetad of symplectic) manifold, and in the even more general case of a Poisson algebra (which is the case most interesting to me).
In a Poisson algebra you pick a Hamiltonian by picking any element $H$ in the algebra. The equations of evolution then say that the time derivative of time evolution acting on the algebra is $\{ H, - \}$ (maybe up to a sign). If your Poisson algebra has a *-structure then $H$ should be self-adjoint.
Note that already in the classical setting you need to make choices; the Hamiltonian you describe isn't determined by the symplectic structure of the cotangent bundle.