10
$\begingroup$

Let $\pi: Y\rightarrow X$ be a finite Galois covering between normal projective varieties with Galois group $G$. Let $E$ be a coherent sheaf on $Y$ with a $G$-linearisation, i.e., there are isomorphisms $\lambda_g:E\cong g^*E$ for all $g\in G$ satisfying $\lambda_1=id$ and $\lambda_{gh}=h^*\lambda_g\circ\lambda_h$. Does there always exist a coherent sheaf $F$ on $X$ such that $\pi^*F\cong E$?

I know that when $\pi$ is etale, the answer is yes by descent theory along torsors (see Vistoli: "Notes on Grothendieck topologies, fibered categories and descent theory," arXiv preprint math/0412512). But how about the general case? In the proof of Lemma 3.2.2 in the book "D Huybrechts, M Lehn: The Geometry of Moduli Spaces of Sheaves", the authors claim that the answer is yes by descent theory, but don't give us a reference. Which descent theory they used? Could someone show me a proof? Thank you very much!

Improve this question
$\endgroup$

1 Answer 1

Reset to default
12
$\begingroup$

The answer is no. First note that you misquote the Lemma of Huybrechts-Lehn: they talk about a subsheaf of a sheaf on $Y$ which is already of the form $\pi ^*\mathcal{F}$.

For a counter-example, take for $\pi :Y\rightarrow X$ a double covering of smooth projective curves, ramified at a point $p\in Y$, and take $E=\mathcal{O}_Y(p)$. Let $\sigma $ be the nontrivial involution of $Y$ such that $\pi \circ \sigma =\pi $. Clearly $\sigma ^*E\cong E$; choose any $\lambda :\sigma ^*E\stackrel{\sim}{\longrightarrow }E$. Then $\sigma ^*\lambda \circ \lambda $ is an automorphism of $E$, hence multiplication by some $t\in\mathbb{C}^*$; dividing $\lambda $ by $\sqrt{t}$ we may assume $t=1$, so that $\lambda $ gives a $\sigma $-linearization of $E$. But obviously $E$ does not descend since its degree is odd.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Could you show me how to prove an invariant subsheaf of $\pi^*F$ can be descent? $\endgroup$
    – Diego Maradona
    Commented Jan 10, 2016 at 3:56
  • $\begingroup$ I think the statement in Huybrechts-Lehn is slightly incorrect: one should assume that the subsheaf $\mathcal{G}$ of $\pi ^*\mathcal{F}$ is saturated, that is, the quotient $\pi ^*\mathcal{F}/\mathcal{G}$ is torsion-free. A maximum destabilizing subsheaf has this property, so their proof still works. You want to prove that the natural homomorphism $\pi ^*((\pi _*\mathcal{G})^G)\rightarrow \mathcal{G}$ is an isomorphism in codimension 1. I think this follows by considering the exact sequence $0\rightarrow \mathcal{G}\rightarrow \pi ^*\mathcal{F}\rightarrow \mathcal{Q}\rightarrow 0$. $\endgroup$
    – abx
    Commented Jan 10, 2016 at 18:21

You must log in to answer this question.

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