Let $X$ be a compact Riemann surface, and let $P \rightarrow X$ be a holomorphic $\mathbb P^1$-bundle over $X$. Then we know that $P$ is of form $\mathbb P(E)$ for some vector bundle $E \rightarrow X$ of rank $2$. At [the end of section 5](http://www.numdam.org/article/SB_1958-1960__5__193_0.pdf), Grothendieck proved that we have the following exact sequence:
$$ 1 \rightarrow \operatorname{Aut}(E)/\mathbb{C}^\* \rightarrow \operatorname{Aut}(P) \rightarrow \Gamma \rightarrow 1, $$
where $\Gamma = \{ [T] \mid [T]\text{ is isomorphism class of a line bundle }T \text{ such that } E \cong E \otimes T\}$. Hence $\Gamma$ is a finite group.


My question is *what will happen if $E$ is a stable vector bundle? Can we expect $\operatorname{Aut} (P) = e$, i.e. $\Gamma = e$?*



Automorphism of ruled surfaces associated to stable vector bundles
Not necessarily, but the counter-examples are quite particular. If $E\cong E\otimes T$, taking determinants give $T^{\otimes r}\cong \mathcal{O}\_X$, with $r=\operatorname{rk}(E) $. Let $\pi :\tilde{X}\rightarrow X $ be the degree $r$ cyclic étale covering associated to $T$. The isomorphism $E\rightarrow E\otimes T$ defines a structure of $\pi \_\*\mathcal{O}\_{\tilde{X} }$-module on $E$; this means that $E$ is of the form $\pi \_\*L$ for some line bundle $L$ on $\tilde{X} $. Conversely, for any $L\in\operatorname{Pic}(\tilde{X} ) $ the vector bundle $E:=\pi \_\*L$ satisfies $E\cong E\otimes T$. It is not difficult to show that $E$ is semi-stable, and stable if $L$ is general enough.


