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Let $A := \textrm{Jac}(C)$ be the Jacobian of a genus $2$ curve $C$, with principal polarization $\Theta$.

Studying some branched covers of $A$, I was led to consider rank $2$ holomorphic vector bundles $\mathscr{E}$ on $A$ defined as non-trivial extensions of the form $$1 \longrightarrow \mathscr{O}_A \longrightarrow \mathscr E \longrightarrow \mathscr{O}_A(2 \Theta) \otimes \mathscr{I}_Z \longrightarrow 1,$$ where $Z$ is a $0$-dimensional sub-scheme of $A$ having length $2$.

Since $\mathscr{E}$ is locally free, then $Z$ satisfies the Cayley-Bacharach condition with respect to the linear system $|2 \Theta|$, namely if $\textrm{supp}(Z) = \{p, \, q\}$ then any divisor of $|2 \Theta|$ containing $p$ must also contain $q$, and conversely. In other words, $p$ and $q$ are identified under the generically $2: 1$ Kummer map $$\varphi_{|2 \Theta|} \colon A \longrightarrow \textrm{Kum}(A) \subset \mathbb{P}^3.$$

Question. Is $\mathscr{E}$ a simple vector bundle, i.e. $\textrm{Hom}(\mathscr E, \, \mathscr E ) = \mathbb{C}?$

Let $A := \textrm{Jac}(C)$ be the Jacobian of a genus $2$ curve $C$, with principal polarization $\Theta$.

Studying some branched covers of $A$, I was led to consider some rank $2$ holomorphic vector bundles $\mathscr{E}$ on $A$, defined as non-trivial extensions of the form $$1 \longrightarrow \mathscr{O}_A \longrightarrow \mathscr E \longrightarrow \mathscr{O}_A(2 \Theta) \otimes \mathscr{I}_Z \longrightarrow 1,$$ where $Z$ is a $0$-dimensional sub-scheme of $A$ having length $2$.

Since $\mathscr{E}$ is locally free, then $Z$ satisfies the Cayley-Bacharach condition with respect to the linear system $|2 \Theta|$, namely if $\textrm{supp}(Z) = \{p, \, q\}$ then any divisor of $|2 \Theta|$ containing $p$ must also contain $q$, and conversely. In other words, $p$ and $q$ are identified under the generically $2: 1$ Kummer map $$\varphi_{|2 \Theta|} \colon A \longrightarrow \textrm{Kum}(A) \subset \mathbb{P}^3.$$

Question. Is $\mathscr{E}$ a simple vector bundle, i.e. $\textrm{Hom}(\mathscr E, \, \mathscr E ) = \mathbb{C}?$

Let $A := \textrm{Jac}(C)$ be the Jacobian of a genus $2$ curve $C$, with principal polarization $\Theta$.

Studying some branched covers of $A$, I was led to consider rank $2$ vector bundles $\mathscr{E}$ on $A$ defined as non-trivial extensions of the form $$1 \longrightarrow \mathscr{O}_A \longrightarrow \mathscr E \longrightarrow \mathscr{O}_A(2 \Theta) \otimes \mathscr{I}_Z \longrightarrow 1,$$ where $Z$ is a $0$-dimensional subscheme of $A$ having length $2$.

Since $\mathscr{E}$ is locally free, then $Z$ satisfies the Cayley-Bacharach condition with respect to the linear system $|2 \Theta|$, namely if $\textrm{supp}(Z) = \{p, \, q\}$ then any divisor of $|2 \Theta|$ containing $p$ must also contain $q$, and conversely. In other words, $p$ and $q$ are identified under the generically $2:1$ Kummer map $$\varphi_{|2 \Theta|} \colon A \longrightarrow \textrm{Kum}(A) \subset \mathbb{P}^3.$$

Question. Is $\mathscr{E}$ a simple vector bundle, i.e. $\textrm{Hom}(\mathscr E, \, \mathscr E ) = \mathbb{C}?$

Let $A := \textrm{Jac}(C)$ be the Jacobian of a genus $2$ curve $C$, with principal polarization $\Theta$.

Studying some branched covers of $A$, I was led to consider rank $2$ holomorphic vector bundles $\mathscr{E}$ on $A$ defined as non-trivial extensions of the form $$1 \longrightarrow \mathscr{O}_A \longrightarrow \mathscr E \longrightarrow \mathscr{O}_A(2 \Theta) \otimes \mathscr{I}_Z \longrightarrow 1,$$ where $Z$ is a $0$-dimensional sub-scheme of $A$ having length $2$.

Since $\mathscr{E}$ is locally free, then $Z$ satisfies the Cayley-Bacharach condition with respect to the linear system $|2 \Theta|$, namely if $\textrm{supp}(Z) = \{p, \, q\}$ then any divisor of $|2 \Theta|$ containing $p$ must also contain $q$, and conversely. In other words, $p$ and $q$ are identified under the generically $2: 1$ Kummer map $$\varphi_{|2 \Theta|} \colon A \longrightarrow \textrm{Kum}(A) \subset \mathbb{P}^3.$$

Question. Is $\mathscr{E}$ a simple vector bundle, i.e. $\textrm{Hom}(\mathscr E, \, \mathscr E ) = \mathbb{C}?$

Let $A := \textrm{Jac}(C)$ be the Jacobian of a genus $2$ curve $C$, with principal polarization $\Theta$.

Studying some branched covers of $A$, I was led to consider rank $2$ vector bundles $\mathscr{E}$ on $A$ defined as non-trivial extensions of the form $$1 \longrightarrow \mathscr{O}_A \longrightarrow \mathscr E \longrightarrow \mathscr{O}_A(2 \Theta) \otimes \mathscr{I}_Z \longrightarrow 1,$$ where $Z$ is a $0$-dimensional subscheme of $A$ having length $2$. Note that, since $\mathscr{E}$ is locally free, then $Z$ satisfies the Cayley-Bacharach condition with respect to the linear system $|2 \Theta|$, namely if $\textrm{supp}(Z) = \{p, \, q\}$ then any divisor of $|2 \Theta|$ containing $p$ must also contain $q$, and conversely. In other words, $p$ and $q$ are identified under the generically $2:1$ Kummer map $$\varphi_{|2 \Theta|} \colon A \longrightarrow \textrm{Kum}(A) \subset \mathbb{P}^3.$$

Question. Is $\mathscr{E}$ a simple vector bundle, i.e. $\textrm{Hom}(\mathscr E, \, \mathscr E ) = \mathbb{C}?$

Let $A := \textrm{Jac}(C)$ be the Jacobian of a genus $2$ curve $C$, with principal polarization $\Theta$.

Studying some branched covers of $A$, I was led to consider rank $2$ vector bundles $\mathscr{E}$ on $A$ defined as non-trivial extensions of the form $$1 \longrightarrow \mathscr{O}_A \longrightarrow \mathscr E \longrightarrow \mathscr{O}_A(2 \Theta) \otimes \mathscr{I}_Z \longrightarrow 1,$$ where $Z$ is a $0$-dimensional subscheme of $A$ having length $2$.

Since $\mathscr{E}$ is locally free, then $Z$ satisfies the Cayley-Bacharach condition with respect to the linear system $|2 \Theta|$, namely if $\textrm{supp}(Z) = \{p, \, q\}$ then any divisor of $|2 \Theta|$ containing $p$ must also contain $q$, and conversely. In other words, $p$ and $q$ are identified under the generically $2:1$ Kummer map $$\varphi_{|2 \Theta|} \colon A \longrightarrow \textrm{Kum}(A) \subset \mathbb{P}^3.$$

Question. Is $\mathscr{E}$ a simple vector bundle, i.e. $\textrm{Hom}(\mathscr E, \, \mathscr E ) = \mathbb{C}?$