Inclusion of logarithmic de-Rham complex into differentials - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T20:59:33Z http://mathoverflow.net/feeds/question/108201 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/108201/inclusion-of-logarithmic-de-rham-complex-into-differentials Inclusion of logarithmic de-Rham complex into differentials LMN 2012-09-27T00:59:14Z 2012-09-27T03:13:24Z <p>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:</p> <p>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"?)</p> <p>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 $j^{-1}, j_*$?</p> http://mathoverflow.net/questions/108201/inclusion-of-logarithmic-de-rham-complex-into-differentials/108205#108205 Answer by Sándor Kovács for Inclusion of logarithmic de-Rham complex into differentials Sándor Kovács 2012-09-27T03:13:24Z 2012-09-27T03:13:24Z <p>Answers to the numbered questions:</p> <ol> <li>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).</li> <li>I would actually say that the adjointness you are using <strong>is</strong> 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 $U$, $U\cap V\neq\emptyset$): $$\Gamma(V,\Omega_X^k(\log D))\to \Gamma(U\cap V, \Omega_X^k(\log D))\simeq \Gamma (U\cap V, \Omega_U^k)\simeq \Gamma (V, j_*\Omega_U^k).$$</li> </ol>