# Which tensor fields on a symplectic manifold are invariant under all Hamiltonian vector fields?

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 $y_j^2$.

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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.

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Just some preliminary thoughts (which I think should work), via a Darboux basis.

Firstly, under the local coordinate, the coordinate functions generate the coordinate derivatives as their Hamiltonian vector fields. Now for the coordinate derivatives, their Lie action on tensors is simply partial derivation on the tensor coefficients in the Darboux coordinate, so this show that any invariant tensors, when written in local coordinates, must be constant coefficient.

Thus it suffices to show at one point that the coefficient is trivial.

Now observe that the Hamiltonian vector fields associated with the function $x_j^2 + y_j^2$ is the rotation vector field in the $x_j$-$y_j$ plane. There are no rotationally invariant 1-forms in $T_0\mathbb{R}^2$, and the only rotationally invariant two-form is $dx\wedge dy$.

By a counting argument, for forms of odd degree, there will be at least "an odd coordinate out" that doesn't pair into $dx_j\wedge dy_j$ pairs. By the rotational symmetry any such term must have vanishing coefficient. So there are no odd invariant forms.

Now, let us write $w_i = dx_i\wedge dy_i$. For forms of even degree, the argument above shows that it must be a linear combination of terms of the form $w_i \wedge w_j \wedge ... \wedge w_k$.

Now, the function $x_i y_j - x_j y_i$ generates the vector field that simultaneously rotates in the $x_i$-$x_j$ plane and $y_i$-$y_j$ plane. This implies that point-wise an invariant form must be obtained through a symmetric polynomial in $w_i$. Using that $w_i^2 = 0$ this here should imply that the only invariant even forms are given by powers of $\omega$.

So I think this allows me to answer Question 1 in the affirmative (in the sense that the only invariant forms are the trivial ones) and Question 3 in a manner differently from posed (that locally it suffices you consider the functions $x_i, y_i, x_i^2 + y_i^2, x_iy_j - x_j y_i$ to settle the problem for forms).

For arbitrary tensor fields the question is a bit more delicate. Since there are larger classes of invariant objects under rotation. For example, the tensor field $\sum dx_i\otimes dx_i + dy_i \otimes dy_i$ in local coordinates is invariant under all of the operations I've considered above. (Of course, as Deane noted below in the comments, the infinitesimal symmetries of this tensor is finite dimensional, so by a dimensional counting argument it cannot be invariant under all Hamiltonian vector fields.) It is, however, not clear to me how to rule out general tensor fields using a fixed, finite dimensional set of vector fields as I've done above for forms.

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This is a nice discussion, where you have reduced to looking at the standard symplectic structure on $R^n$. So can't you rule out the standard flat metric (mentioned in the last paragraph) by simply using a Hamiltonian vector field that is not an infinitesimal isometry? More generally, isn't it the case that if the group of local diffeomorphisms preserving a tensor is finite-dimensional, then it can't be invariant under all Hamiltonian vector fields? –  Deane Yang Jun 29 '10 at 17:17
@Deane: you are absolutely correct. I wrote that last paragraph with half-a-mind on whether it suffices to consider only a finite dimensional set of vector fields for ALL (say symmetric) tensors. I shall update the answer to reflect your observation. –  Willie Wong Jun 29 '10 at 18:45