Let $U$ be a smooth variety, and $U\hookrightarrow X$ an smooth compactification with snc boundary $D=X\setminus U$. Suppose that $\omega\in H^0(U,\Omega^n\_U)$ is global algebraic $n$-form on $U$. It defines a class in $H^n(U,\mathbb{C})=\mathbb{H}^n(X,\Omega\_X^\bullet(\log D))$.


The form $\omega$ extends to a meromorphic form on $X$, denote it by $\tilde{\omega}$. This is not necessarily an element of $H^0(X,\Omega\_X^n(\log D))$, since $\tilde{\omega}$ can have poles of higher order. Is there an element $\omega'\in H^0(X,\Omega\_X^n(\log D))$ such that $\omega'|\_U$ defines the same cohomology class as $\omega$ in $H^n(U,\mathbb{C})$?


Here are my thoughts: the Hodge spectral sequence degenerates, and so we have
$$Gr\_F^i(H^n(U,\mathbb{C}))=H^{n-i}(X,\Omega^i(\log D)),$$
and so non-canonically (I believe)
$$H^n(U,\mathbb{C})=\bigoplus\_i H^{n-i}(X,\Omega^i(\log D)).$$
Now it seems that the class defined by $\omega$ should be contributed by the summand $H^{0}(X,\Omega^n(\log D))$, which would imply the claim. However, to prove this I think one would need something analogous to what Peters and Steenbrink call a "Hodge decomposition in the strong sense" (page 45). However, I do not know if this kind of result exists for non-compact $U$?



Is every $\omega\in H^0(U,\Omega_U^n)$ representable in $H^n(U,\mathbb{C})$ by an element from $H^0(X,\Omega_X^n(\log D))$?
This is not true. Take for $X$ an elliptic curve, for $D$ a point $p\in X$. The restriction map $H^1(X,\mathbb{C})\rightarrow H^1(U,\mathbb{C})$ is an isomorphism, and $H^0(X,\Omega ^1\_X(\log D))=H^0(X,\Omega ^1\_X)$. There is a form $\tilde{\omega } $ with a pole of order 2 at $p$; its restriction $\omega $ to $U$ is not cohomologous to the restriction of a holomorphic form on $X$.


