1
$\begingroup$

Let $X$ be a smooth projective variety over a field $k$ of characteristic zero and let $D$ be a simple normal crossing divisor on $X$, with irreducible components $D_i$.

Does there exist a nonzero global section $\omega \in H^0(X, \Omega^1_X(\log D))$ such that $\mathrm{Res}_{D_i}\omega \neq 0$ for every component $D_i$?

I think the case of curves should follow somehow from Riemann-Roch but I don't see how.

Thanks for your help

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Take $X$ a curve and $D$ a single point... probably you should assume that $\mathrm{gr}^W_2 H^1(X \setminus D) \neq 0$. $\endgroup$
    – Dan Petersen
    Commented Nov 8, 2013 at 9:51
  • 1
    $\begingroup$ Since abx and Dan have shown that your initial guess is wrong, I recommend that you try some simple examples on your own before asking further questions about this. $\endgroup$
    – Jason Starr
    Commented Nov 8, 2013 at 14:38

1 Answer 1

Reset to default
2
$\begingroup$

With no further hypothesis, the answer is no. Take $X$ with $H^0(X,\Omega ^1_X)=0$, and $D$ a smooth divisor. Then from the exact sequence $$0\rightarrow \Omega ^1_X\rightarrow \Omega ^1_X(\log D) {\buildrel {\mathrm{Res}}\over {\longrightarrow}} \mathcal{O}_D\rightarrow 0$$you get $H^0(X,\Omega ^1_X(\log D) )=0$ (the section $1$ of $\mathcal{O}_D$ goes to the class of $D$ in $H^1(X,\Omega ^1_X)$, which is nonzero).

Improve this answer
$\endgroup$
6
  • $\begingroup$ Oh, thank you for this observation. What happens if we assume $H^0(X, \Omega^1_X) \neq 0$? $\endgroup$
    – resid
    Commented Nov 8, 2013 at 12:27
  • $\begingroup$ That doesn't change the problem : the map $\mathrm{Res}: H^0(X,\Omega ^1_X(\log D))\rightarrow H^0(D,\mathcal{O}_D)$ is still the zero map, for the same reason. $\endgroup$
    – abx
    Commented Nov 8, 2013 at 12:51
  • $\begingroup$ Are you saying that every global section of $\Omega^1_X(\log D)$ has zero residues at all $D_i$? $\endgroup$
    – resid
    Commented Nov 8, 2013 at 13:16
  • $\begingroup$ In the case there is just one $D_i$, yes -- and therefore every such section comes from $H^0(X,\Omega ^1_X)$. $\endgroup$
    – abx
    Commented Nov 8, 2013 at 13:29
  • $\begingroup$ What happens if there are several $D_i$? $\endgroup$
    – resid
    Commented Nov 8, 2013 at 13:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .