Is there an algebraic version of Darboux's theorem?

Question

Let $M$ denote a smooth manifold, and $\omega \in \Omega^2(M, \mathbb{R})$ a symplectic form. The classical version of Darboux's theorem states that for any $x \in M$, there exists an open neighborhood $U$ of $x$ together with local coordinates $p_1,\dots, p_n, q_1, \dots, q_n$ on $U$ such that \begin{equation} \omega \vert_U = dp_1 \wedge dq_1 + \dots + dp_n \wedge dq_n. \tag{1} \end{equation} I learned from this question that the theorem also holds in the complex analytic context, which leads to my question. All of the ingredients in Darboux's theorem make sense algebraically. Specifically, let $X$ denote a smooth variety over an algebraically closed field $k$ of characteristic $0$. A symplectic form on $X$ is a closed 2-form $\omega \in H^0(X, \Omega^2_{X/k})$ inducing a non-degenerate pairing on the tangent space to each closed point of $X$. For example, we have the 'standard' symplectic form on $\mathbb{A}^{2n}_k = \text{Spec} \, k[p_1,\dots, p_n, q_1, \dots, q_n]$ given by (1).

My question, vaguely stated, is whether or not there is a version of Darboux's theorem for such symplectic forms.

For example, since $X$ is smooth, there exists an open neighborhood $U$ of $x$, and an étale $k$-morphism $f: U \rightarrow \mathbb{A}^n_k$, where $n = \text{dim}(X)$. Perhaps such a result would state that $n = 2m$ is even, and that $f$ can be chosen so that $\omega$ is the pullback of the standard form on $\mathbb{A}^{2m}_k$ under $f$.

Answer 1

In fact, the opposite is true. For $X$ a smooth proper variety over $\mathbb C$ and $\omega$ a nonzero symplectic form on $X$. Then there is no nonempty Zariski open set on which $\omega$ is the pullback (under any map) of the standard symplectic form. The same should work for etale open sets, and even for smooth morphisms.

The reason is that the standard symplectic form is exact (i.e. is the derivative of a 1-form), so its cohomology class vanishes - we can just use ordinary de Rham cohomology for this. Thus its pullback under any map has vanishing cohomology class.

But it is not possible for a nonzero holomorphic $2$-form on a smooth projective variety $X$ to have zero cohomology class when restricted to a nonempty open set $U$. If it did, then the associated class in $H^2(X, \mathbb C)$ would lie in the image of the natural map from the $H^2$ of $X$ supported in $X \setminus U$. But this cohomology group is generated by the classes of the $\dim X-1$-dimensional irreducible components of $X \setminus U$, which are sent to their divisor classes. So the associated class in $H^2(X, \mathbb C)$ would have to be a linear combination of divisor classes.

But divisor classes are sent by Hodge theory to $(1,1)$-forms, while holomorphic $2$-forms are $(0,2)$-forms, so they can never be equal.