Consider a connected symplectic manifold $(M, \omega)$ of dimension $m=2n$. A few preliminary reminders (mostly to fix the notation): A vector field $X$ is symplectic if its flow preserves the symplectic form, ie. $L_X \omega = 0$, where $L_X$ denotes Lie derivative with respect to $X$. The Cartan formula shows that this is equivalent to the 1-form $i_X\omega = \omega(X, -)$ being closed. The Hamiltonian vector field associated to a smooth function $f$ is the vector field determined by $\omega(X_f, -) = df$; any symplectic vector field is locally Hamiltonian. The questions I'm interested in are of local nature, so we don't have to worry about the distinction.

**Question 1:** Which differential forms are invariant under all Hamiltonian flows (meaning $L_{X_f}\alpha = 0$ for all smooth functions $f$)?

Clearly, the symplectic form itself generates a truncated polynomial algebra (isomorphic to $\mathbb{R}[x]/(x^{n+1})$) inside $\Omega^*(M)$ which is invariant under all Hamiltonian flows. But is it possible that there are other than those? I believe I have shown that there are no invariant 1-forms using a horrible calculation in local (Darboux) coordinates, but I'm not sure if this method is suitable for higher degrees. In the even degrees, we know that the answer is not 0, and I can't see how to prove that an invariant $2d$-form is necessarily a constant multiple of $\omega^d$.

**Question 2:** What can one say about more general tensor fields on $M$? I am especially interested in the sections of the symmetric powers of $TM$ (ie. symmetric multi vector fields).

The proof that no $1$-forms are invariant is easily adapted to proving that no vector fields are invariant, but again, I'm not sure if this generalizes.

**Question 3:** Suppose we have a subalgebra $A\subset C^\infty(M)$ with the property that for each $p\in M$, $\{df_p \mid f\in A\} = T^*_pM$ (in other words, the Hamiltonian vector fields associated to the functions in $A$ realize every tangent vector on $M$). Do the answers to Questions 1 and 2 change if we only insist that the forms/tensor fields should be invariant under the Hamiltionian vector fields associated to the elements of $A$?

It is certainly important that we still have a whole algebra of functions available; on $\mathbb{R}^{2n}$, any constant coefficient tensor field is invariant under the Hamiltonian vector fields associated to the coordinate functions $x_j, y_j$ (which, up to a sign, are just the corresponding coordinate vector fields $\partial/\partial y_j, \partial/\partial x_j$). The proof that no invariant vector fields exists also requires one to consider the Hamiltonians associated to $x_j^2$ and $y_j^2$.