2
$\begingroup$

Let $X$ be a normal (possibly singular) projective surface over $\mathbb{C}$. Consider the set $M_X$ of all coherent sheaves $F$ on $X$ such that there exists a finite subset $Y\subset X$ such that $F$ restricted to $X\setminus Y$ is a line bundle. $M_X$ becomes a monoid via the tensor product. Now let $G_X$ be the set of equivalence classes of $M_X$ where two sheaves $F_1,F_2\in M_X$ are equivalent if there is a finite subset $Y\subset X$ such that $F_1$ and $F_2$ are isomorphic on $X\setminus Y$. This equivalence relation is compatible with the tensor product and so $G_X$ is a group.

In general, one has the group homomorphism $\textrm{Pic}(X)\to G_X$ that sends a line bundle to its equivalence class. If $X$ is smooth, then this is an isomorphism and thus $G_X$ is just the usual Picard group.

My hope is that in general we can understand $G$ in terms of a desingularisation $f: X'\to X$. Because $X$ is normal, it has only finitely many singularities. Away from these singularities $f$ is an isomorphism and the group $G_{X'}$ is just the Picard group of $X'$. So my hope is that we can identify $G_X$ with $\textrm{Pic}(X')$. Is something like that true?

$\endgroup$
  • 1
    $\begingroup$ I am confused. Why is $Pic(X) \to G_X$ an isomorphism? Doesn't $G_X$ parametrize coherent sheaves? What is the inverse of $[F]$ in $G_X$? $\endgroup$ – Ariyan Javanpeykar Apr 2 at 14:51
  • $\begingroup$ Not any coherent sheaves, but such $F$ that are a line bundle on $U=X\setminus Y$ for a finite set $Y$. Then the inverse of $[F]$ is any extension of the inverse of $F|_U$ to $X$. $\endgroup$ – Hans Apr 2 at 16:20
  • $\begingroup$ Ok, thank you. I misread the definition, sorry. $\endgroup$ – Ariyan Javanpeykar Apr 2 at 19:04
6
$\begingroup$

The group $G_X$ can be identified with the group of rank 1 reflexive sheaves on $X$ ($F$ is reflexive if the canonical morphism $F \to F^{\vee\vee}$ is an isomorphism) by taking a sheaf $F$ to the reflexive sheaf $F^{\vee\vee}$. The monoidal structure on the set of all reflexive sheaves is given by $$ (F,G) \mapsto (F \otimes G)^{\vee\vee}. $$ Furthermore, the group $G_X$ can be identified with the class group $\operatorname{Cl}(X)$ of Weil divisors on $X$.

The relation of $\operatorname{Pic}(X')$ to $\operatorname{Cl}(X)$ is given by the following exact sequence $$ 0 \to \bigoplus \mathbb{Z}[E_i] \to \operatorname{Pic}(X') \to \operatorname{Cl}(X) \to 0, $$ where $E_i$ are the components of the exceptional divisor of $X' \to X$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks! So the situation is even better than I have expected. Am I right in assuming that the reflexive sheaf corresponding to a Weil divisor is just the subsheaf of the quotient field of all functions with the prescribed pole and zero loci? $\endgroup$ – Hans Apr 2 at 16:33
  • 1
    $\begingroup$ There are two nice constructions. 1) Restrict the divisor to the smooth locus of the surface, take the corresponding line bundle, and then push it forward to the surface. 2) If the divisor is effective, take the dual of its ideal sheaf; and in general extend this association by linearity. $\endgroup$ – Sasha Apr 2 at 16:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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