# Simple vector bundle isomorphic to one of its twistings

Let $$V$$ be a vector bundle over an algebraic curve $$C$$, ad assume that that $$V \cong V \otimes L$$ for some line bundle $$L$$.

If $$V$$ is decomposable this is clearly possible, for example take $$V \cong \mathcal{O}_C \oplus L$$ with $$L^2 \cong \mathcal{O}_C$$.

Q. Does $$V \cong V \otimes L$$ imply $$L \cong \mathcal{O}_C$$ if $$V$$ is simple?

• If $L$ has a non-trivial section $s$, then don't we have a map $V \otimes \mathcal{O}_{C} \to V \otimes L$ induced by $Id \otimes s$, which is not multiplication by a constant? By your assumption, this would give an endomorphism of $V$ which is not simply multiplication by a constant. This contradicts simplicity. Since either $L$ or $L^{-1}$ has sections, I think this suffices. Is there a hole in this argument? – Benighted Aug 17 at 6:12
• @Benighted: if $V\cong V\otimes L$, $L$ is torsion, so it has no global sections unless it is trivial. – abx Aug 17 at 7:19
• Let $L$ be a 2-torsion line bundle on $C$, and $\pi :\tilde{C}\rightarrow C$ the associated étale 2-sheeted covering. Let $V=\pi _*M$ for any line bundle $M$ on $\tilde{C}$. Then $V\otimes L\cong V$; if $M$ is not the pull back of a line bundle $M$ on $C$, it is easy to see that $V$ is simple. – abx Aug 17 at 7:48
• Thanks for a nice counterexample. It saved me from wasting time for wrong line of argument. – Insong Choe Aug 18 at 3:15

Take an elliptic curve $$E$$, choose a point $$p \in E$$ and consider the unique non-split extension $$0 \longrightarrow \mathcal{O}_E \longrightarrow V_p \longrightarrow \mathcal{O}_E(p) \longrightarrow 0.$$ Then $$V_p \simeq V_p \otimes L$$ for every $$2$$-torsion line bundle $$L$$ on $$E$$, see Remark p. 35 of [F].
Note that $$V_p$$ is stable and thus simple, see [F], Theorem 9 p. 89.