5
$\begingroup$

Let $X$ be a projective variety with at worst (analytic) normal crossings singularities and $\pi:\tilde{X} \to X$ be the normalisation. Is there a "nice" description relating the picard group of $X$ and $\tilde{X}$, for example through a short exact sequence. We know that, in the case $X$ is a curve we have such a short exact sequence.

Any idea/reference will be most helpful.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Since $X$ is $S2$, the sheaf $\mathcal{O}_X^\times$ equals the pushforward of $\mathcal{O}_U^\times$ for every open subset $U$ of $X$ whose complement has codimension $2$. Thus, you can use the same short exact sequence as in the case of a nodal curve to compute the relation between the Picard group of $X$ and the Picard group of $\widetilde{X}$ via the restrictions to the "double locus" in each of $X$ and $\widetilde{X}$. $\endgroup$ Commented Dec 1, 2017 at 17:16
  • $\begingroup$ I believe my previous comment is wrong. It is true that rational sections of an invertible sheaf on $X$ that are regular away from codimension $\geq 2$ are everywhere regular. However, that is not enough to extend invertible sheaves. By Grothendieck's proof of Samuel's conjecture, if the complement of $U$ has codimension $\geq 4$, then every invertible sheaf on $U$ extends uniquely to $X$. However, I do not see at the moment whether this holds if the codimension of $U$ equals $2$, resp. $3$. $\endgroup$ Commented Dec 1, 2017 at 19:16
  • 1
    $\begingroup$ If $X$ is the push out of the maps $D\leftarrow \overline{D}\to \overline{X}$ where $D\subset X$ and $\overline{D}\subset\overline{X}$ are the conductor loci, e.g., if $X$ is demi-normal, then you can use Milnor pasting to show that $\mathrm{Pic}(X)$ is the fibre product of $\mathrm{Pic}(\overline{X})$ and $\mathrm{Pic}(D)$ over $\mathrm{Pic}(\overline{D})$. This is usually very useful. For certain surfaces, there is an exact sequence in an article by Hartshorne and Polini, if I recall correctly. $\endgroup$ Commented Dec 1, 2017 at 22:25

0

Reset to default

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.