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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, 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. $\Gamma = e$?

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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.

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