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3 Corrected some oversight as pointed out by Deane Young

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, and 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 obvious clear to me whether it can be ruled how to rule out under other Hamiltonian general tensor fields using a fixed, finite dimensional set of vector fields as I've done above for forms.

2 Corrected some notation.

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, this the argument above shows that it must be a linear combination of terms of the form $(dx_i\wedge dy_i)\wedge (dx_j\wedge dy_j)$. Let us write $w_i = dx_i\wedge dy_i$\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$in local coordinates is invariant under all of the operations I've considered above, and it is not obvious to me whether it can be ruled out under other Hamiltonian vector fields. 1 Just some preliminary thoughts (which I think should work), via a Darboux basis. Firstly, under the local coordinate, the coordinate functions generate the coordinate derivatives as their Hamiltonian vector fields. Now for the coordinate derivatives, their Lie action on tensors is simply partial derivation on the tensor coefficients in the Darboux coordinate, so this show that any invariant tensors, when written in local coordinates, must be constant coefficient. Thus it suffices to show at one point that the coefficient is trivial. Now observe that the Hamiltonian vector fields associated with the function$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$pairs. By the rotational symmetry any such term must have vanishing coefficient. So there are no odd invariant forms. For forms of even degree, this shows that it must be a linear combination of terms of the form$(dx_i\wedge dy_i)\wedge (dx_j\wedge dy_j)$. Let us write$w_i = dx_i\wedge dy_i$. Now, the function$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, and it is not obvious to me whether it can be ruled out under other Hamiltonian vector fields.