Let $C$ be a smooth projective curve over a field $k$ of characteristic zero and $S$ a reduced divisor on $C$ (so just a collection of points). Consider the sheaf of logarithmic differentials $\Omega^1_C(\log S)$.

Why is it possible to find a rational section $\omega$ of $\Omega^1_C(\log S)$ such that the residue of $\omega$ at all points of $S$ is non-zero?

Can one give an explicit formula in the case of $\mathbb{P}^1$?


1 Answer 1


I think what you mean is a regular section of $\Omega ^1_C(\log S)$, that is, a rational 1-form $\omega $ which is regular outside $S$ and has a simple pole at each point of $S$. Consider the exact sequence of sheaves $$0\rightarrow \Omega ^1_C\rightarrow \Omega ^1_C(\log S)\rightarrow \oplus_{s\in S}\ k(s)\rightarrow 0$$ and the associated cohomology sequence $$H^0(C,\Omega ^1_C(\log S))\stackrel{\mathrm{Res}}{\longrightarrow} k^S\stackrel{\partial}{\longrightarrow} H^1(C,\Omega ^1_C)\cong k\rightarrow 0\ .$$ It is easy to see that $\partial$ is just the sum map, so given $(r_s)\in k^S$, there exists a form $\omega \in H^0(C,\Omega ^1_C(\log S))$ with residue $r_s$ at $s$ if and only if $\sum r_s=0$. In particular there always exists such a form with all residues $\neq 0$, except when $\# S=1$.

If $C=\mathbb{P}^1$ and $S=\{s_1,\ldots ,s_n\} \subset \mathbb{C}$, just take $\omega =\sum_s \dfrac{r_s\,dz}{z-s} $; it has the required residues, and the condition $\sum r_s=0$ ensures that $\omega $ is regular at $\infty$.

  • $\begingroup$ I just thought about this also. It is correct as stated, since you can allow an additional singularity outside $S$ if necessary. $\endgroup$ Aug 13, 2014 at 11:49
  • $\begingroup$ Yes, right. It is necessary only in the case $S$ has only one element. $\endgroup$
    – abx
    Aug 13, 2014 at 12:41

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