# Can we extend a logarithmic form to some appropriate compactification?

Given some simple normal crossings divisor $D$ on a complex manifold $X$, which is not assumed to be compact. Given a form $\omega\in H^0(X,\Omega_X^1(\log D))$, when is it possible to find a compactification $X\to \tilde X$ of $X$, where $\tilde D:=(\tilde X\setminus X)\cup D$ is a simple normal crossings divisor, such that $\omega$ extends to a global section of $\Omega_{\tilde X}^1(\log \tilde D)$?

A related question, is the extendibility of $\omega$ maybe independent of the compactification?

As an example of the kind of thing I'm looking for, we can note that a necessary condition is that the residues along the component of $D$ are constant, since otherwise they cannot extend to global functions on the compactifications of the components.

It is not clear to me what "when is it possible" means, but it is certainly not always possible. Take $X=\mathbb A^1_{\mathbb C}=Spec\ \mathbb C[x]$, $D=0$ and $\omega=dx$. Then $X$ has only one smooth compactification, $\mathbb P^1$, and $\omega$ does not extend to a form with logarithmic singularities.
• So "when is it possible" means whether there are some condition e.g. on the residues or the divisor $D$ that ensure that such a compactification exists Aug 29, 2018 at 20:14
To get a condition on extending these forms you could look at the standard restriction sequences by adding one (extra) component of $\widetilde D\setminus D$ at a time. They fit into a short exact sequence, so one possible condition is that the relevant cohomology group vanishes. (OK, actually, probably you also need the restriction to an open set as well, but that also follows from the vanishing of some cohomology. At the end this probably does not lead to a simple criterion, but if there was such you probably would have found that already. :)
Regarding independence, I think that it is true that if you can extend $\omega$ on one compactification then you can extend it on all and this follows from the theorems listed and linked in this answer.