Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $X$ be a smooth variety over $\mathbb{C}$. Let $D \subset X$ be an effective Cartier divisor.

Question 1 What is the definition of the logarithmic differential sheaf $\Omega^1_X (\log D)$ ?

I saw the definition in the book by Esnault-Viehweg (meromorphic form $\alpha$ such that $\alpha, d \alpha$ has pole of order 1 along $D$ ?). If there is another definition, I would be happy.

And I have another question. If $D$ is normal crossing, there is an sequence

$0\rightarrow \Omega^1_X \rightarrow \Omega^1_X(\log D) \rightarrow \nu_* \mathcal{O}_{\tilde{D}} \rightarrow 0 $

where $\nu: \tilde{D} \rightarrow D$ is the normalization.

Question 2 If $D$ is not normal crossing, is there a similar exact sequence? Is there a suitable way to define a residue map $\Omega^1_X(\log D) \rightarrow \nu_* \mathcal{O}_D$?

share|improve this question
1  
For (at least) Q1, you may check arxiv.org/abs/1006.5870 –  shenghao Sep 10 '11 at 12:39
add comment

1 Answer

Here is a way to define this sheaf algebraically over any field of characteristic zero. Let $\mathrm{T}_{X}$ denote the tangent sheaf on $X$. Choose a local equation $\phi_U$ for $D$ on $U$. Consider the following submodule:

$\mathrm{T}_X(-\log\phi_U)=\ ${$\partial\in\mathrm{Der}(\mathcal{O}_X(U))\mid \partial\phi_U\in (\phi_U)$}$\ \subset \mathrm{T}_X(U)$.

It is an easy exercise to check that this does not depend on the choice of equation and glues into a sheaf $\mathrm{T}_X(-\log D)$. Now take the dual $\Omega^1_X(\log D)=\mathrm{T}_X(-\log D)^*$.

The good thing about having normal crossings is that in this case $\Omega^1_X(\log D)$ becomes a locally free sheaf.

For your second question, look at Definition 2.5 in this paper by Dolgachev, arXiv:math/0508044

share|improve this answer
3  
I think that the usual notation is $T_X(-\log D)$. –  Francesco Polizzi Sep 9 '11 at 7:57
    
Thanks, Francesco. It definitely makes sense! I corrected my answer. –  Anton Fonarev Sep 9 '11 at 9:53
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.