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Let $X$ be a complex manifold and $D$ a normal crossing divisor. Let $U = X - D$ and $j: U \rightarrow X$ the natural map. Voisen observes that there is a natural inclusion $$\Omega^k_X(\log D) \subset j_*\Omega_U^k$$ Why is this so? Certainly, by adjointness of $j^{-1}, j_*$ we get a natural map $\Omega^k_X(\log D) \rightarrow j_*\Omega_U^k$. It's not obvious to me that this map is injective at the level of stalks. Basically I have two questions:

1.) Is it infact obvious that the natural map produced by adjointness is injective at the level of stalks? (Does this follow from more general "sheaf theory" theorems"?)

2.) Are you able to see in a more obvious way that a map $\Omega^k_X(\log D) \rightarrow j_*\Omega_U^k$ exists, and is an inclusion of sheaves "geometrically", that is, without using adjointness of $j^{-1}, j_*$?

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    $\begingroup$ $j_*\Omega^k_U$ is the sheaf of differential forms which are holomorphic on $U$. You have an inclusion $\Omega^k_X(*D)\subset j_*\Omega^k_U$ of forms meromorphic along $D$. Furthermore, $\Omega^k_X(log D)\subset \Omega^k_X(*D)$ as forms having a first-order pole along $D$. $\endgroup$
    – Pavel Safronov
    Commented Sep 27, 2012 at 1:11
  • $\begingroup$ Why do you know that $j_*\Omega_U^k$ is the sheaf of $C^\infty$ differential $k$-forms on $X$ that are holomorphic on $U$? $\endgroup$
    – LMN
    Commented Sep 27, 2012 at 1:17
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    $\begingroup$ Sorry, I didn't understand your question at first. Forms in $j_*\Omega^k_U$ are definitely not $C^\infty$ on $X$, since they are not even defined at $D$. Let me try to be more precise. The kernel of $\Omega^k_X(*D)\rightarrow j_*\Omega^k_U$ consists of meromorphic forms on $X$ which vanish on $U$. Since they are zero on an open set, they are zero on the whole $X$. $\endgroup$
    – Pavel Safronov
    Commented Sep 27, 2012 at 1:40
  • $\begingroup$ Great, thanks! If you reply below, I'll be happy to accept your response as an answer. $\endgroup$
    – LMN
    Commented Sep 27, 2012 at 2:07
  • $\begingroup$ Sorry to be late to the party, but do the answers change if one replaces X by a smooth rigid analytic space? $\endgroup$
    – lemiller
    Commented Jan 25, 2018 at 0:46

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Answers to the numbered questions:

  1. Yes, it is in fact obvious that the natural map you describe is injective, because it is injective, in fact an isomorphism, on $U$ which is dense in $X$ and $\Omega_X^k(\log D)$ is locally free (it would be enough that it is torsion-free).
  2. I would actually say that the adjointness you are using is both geometric and obvious. In other words, your map is simply the restriction of logarithmic differentials from an open set $V\subseteq X$ to $U\cap V$. (Note that by the definition of $U$, $U\cap V\neq\emptyset$): $$ \Gamma(V,\Omega_X^k(\log D))\to \Gamma(U\cap V, \Omega_X^k(\log D))\simeq \Gamma (U\cap V, \Omega_U^k)\simeq \Gamma (V, j_*\Omega_U^k). $$
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