TITLE: Descent of sheaves under galois covering
QUESTION [10 upvotes]: 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!

REPLY [12 votes]: 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.