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 $y\_j^2$.



Which tensor fields on a symplectic manifold are invariant under all Hamiltonian vector fields?
Any symplectic linear transformations in $T\_xM$ is locally realizable as a Hamiltonian vector field, thus for questions 1 and 2, one can profitably use representation theory of the symplectic group.


**FACT** (*Lefschetz decomposition*) Let $W$ be a $2n$-dimensional symplectic vector space, $\bigwedge^\ast W$ its exterior algebra, and $\omega\in\bigwedge^2 W$ the invariant two-form. Exterior multiplication by $\omega$ and the contraction with $\omega$ define a pair of $Sp(W)$-equivariant graded linear transformations $L, \Lambda$ of $\bigwedge^\ast W$ into itself of degrees $2$ and $-2,$ and let $H=\deg-n$ be the graded degree $0$ map acting on $\bigwedge^k$ as multiplication by $k-n.$ Then $L,H,\Lambda$ form the standard basis of the Lie algebra $\mathfrak{sl\_2}$ acting on $\bigwedge^\ast W$ and the actions of $Sp(W)$ and $\mathfrak{sl\_2}$ are the commutants of each other.


See, for example, Roger Howe, *Remarks on classical invariant theory*.


**Corollary** Every homogeneous $Sp(W)$-invariant element of $\bigwedge^\ast W$ is a multiple of $\omega^k$ for some $0\leq k\leq n.$


Since, conversely, every polynomial in $\omega$ is invariant under the Hamiltonian vector fields, this gives a full description of the invariant differential forms.


For question 2, locally every invariant tensor must reduce to an $Sp(W)$-invariant element of the tensor algebra. For the special case of symmetric tensors, the answer is trivial.


**FACT** Under the same assumptions, the $k$th symmetric power $S^k W$ is a simple $Sp(W)$-module (non-trivial for $k>0$).


General case can be handled using similar considerations from classical invariant theory. A more involved question of describing the invariant local tensor *operations* on symplectic manifolds (an analogue of the well-known problem of invariant local operations on smooth manifolds, such as the exterior differential or Schoutens bracket) was considered in an old article by A.A.Kirillov. 


