>> **Theorem.** Let $\mathcal E$ be a vector bundle on a smooth projective curve $C$ over an algebraically closed field $k$, and let $x \in C$. If $H^0(\mathcal End(\mathcal E))$ surjects onto $\mathcal End(\mathcal E)_x$, then $\mathscr E \cong \mathcal L^n$ for some line bundle $\mathcal L$ and some natural number $n$.

**Remark.** The converse is trivially true as well: in this case $\mathcal End(\mathcal E) \cong \mathcal O^{n^2}$. I treat the general coherent case in the corollary below.

Here is an auxiliary lemma.

>> **Lemma 1.** Let $\mathcal E$ as in the theorem. If $\mathcal F \subseteq \mathcal E$ is a subbundle such that $\operatorname{Hom}(\mathcal F, \mathcal E/\mathcal F) = 0$, then $\mathcal F$ is either $0$ or $\mathcal E$.

*Proof.* Suppose not. We can choose a basis $x_1,\ldots,x_n$ for $\mathcal E_x$ such that $x_1,\ldots,x_j$ form a basis of $\mathcal F_x \subseteq \mathcal E_x$ for some $1 < j < n$. Then consider the endomorphism $\phi_x$ of $\mathcal E_x$ that sends $x_1$ to $x_n$ and everything else to $0$. Suppose there exists a global endomorphism $\phi$ inducing $\phi_x$. Then the induced map $\mathcal F \to \mathcal E \stackrel\phi\to \mathcal E \twoheadrightarrow \mathcal E/\mathcal F$ is zero, since $\operatorname{Hom}(\mathcal F,\mathcal E/\mathcal F) = 0$. This clearly contradicts the choice of $\phi_x$. $\square$

**Remark.** We will use the language of stability of sheaves. See for example Huybrechts–Lehn, *The geometry of moduli spaces of sheaves*, **Chapter 1**. Maybe there is an easier proof; I would be very interested to see one.

>> **Lemma 2.** Let $\mathcal E$ and $\mathcal G$ be nonzero semistable bundles whose Jordan–Hölder filtrations only include the stable factor $\mathcal F$. If the map $\operatorname{Hom}(\mathcal E,\mathcal G) \to \mathcal Hom(\mathcal E,\mathcal G)_x$ is surjective, then it is an isomorphism and $\mathcal E$ and $\mathcal G$ both split as a direct sum of copies of $\mathcal F$. Moreover, this can only happen if $\mathcal F$ is a line bundle.

*Proof.* We proceed by induction on the sum of the lengths $r, s$ of $\mathcal E$ and $\mathcal G$. The case $r = s = 1$ follows from the fact that $\operatorname{End}(\mathcal F) = k$, since $k$ is algebraically closed and $\mathcal F$ is stable hence simple ("Schur's lemma"). If $\mathcal F$ is not a line bundle, the map $\operatorname{End}(\mathcal F) \to \mathcal End(\mathcal F)_x$ is visibly not surjective.

Let $r+s \geq 3$, and assume we know the result for total length $< r+s$. Now either $r \geq 2$ or $s \geq 2$. If $r \geq 2$, consider a Jordan–Hölder filtration $0 = \mathcal E_0 \subsetneq \ldots \subsetneq \mathcal E_r = \mathcal E$ of $\mathcal E$. By assumption, all the stable subquotients $\mathcal F_i = \mathcal E_i/\mathcal E_{i-1}$ are isomorphic to $\mathcal F$. The sequence $0 \to \mathcal E_{r-1} \to \mathcal E \to \mathcal F_r \to 0$ induces a commutative diagram with exact rows
$$\begin{array}{ccccccccc}0 & \to & \operatorname{Hom}(\mathcal F_r, \mathcal G) & \to & \operatorname{Hom}(\mathcal E,\mathcal G) & \to & \operatorname{Hom}(\mathcal E_{r-1},\mathcal G) & & \\ & & \downarrow & & \downarrow & & \downarrow & & \\ 0 & \to & \mathcal Hom(\mathcal F_r ,\mathcal G)_x & \to & \mathcal Hom(\mathcal E,\mathcal G)_x & \to & \mathcal Hom(\mathcal E_{r-1},\mathcal G)_x & \to &\ 0. \end{array}$$
The third vertical arrow is surjective because the bottom right arrow and the middle vertical arrow are. Hence, by the induction hypothesis the third vertical arrow is an isomorphism, and $\mathcal E_{r-1}$ and $\mathcal G$ are isomorphic to a direct sum of copies of $\mathcal F$. Finally, $\mathcal F$ has to be a line bundle by induction.

The left vertical map is an isomorphism as well, since $\mathcal G \cong \bigoplus \mathcal F$. Then the top right map is surjective by a diagram chase. The middle vertical arrow is an isomorphism by the snake lemma or the five lemma.

To prove the statement about $\mathcal E$ splitting as a direct sum of  copies of $\mathcal F$, note that we have this result for $\mathcal E_{r-1}$ by the induction hypothesis (see above). The sequence
$$0 \to \operatorname{Hom}(\mathcal F_r,\mathcal F) \to \operatorname{Hom}(\mathcal E, \mathcal F) \to \operatorname{Hom}(\mathcal E_{r-1},\mathcal F) \to 0$$
is exact, by the same argument as above (replace $\mathcal G$ with $\mathcal F$). Taking direct sums and using the fact that $\mathcal E_{r-1} \cong \bigoplus \mathcal F$, we get the short exact sequence
$$0 \to \operatorname{Hom}(\mathcal F_r,\mathcal E_{r-1}) \to \operatorname{Hom}(\mathcal E, \mathcal E_{r-1}) \to \operatorname{Hom}(\mathcal E_{r-1},\mathcal E_{r-1}) \to 0.$$
Surjectivity says exactly that the extension $0 \to \mathcal E_{r-1} \to \mathcal E \to \mathcal F_r \to 0$ splits. This finishes the proof if $r \geq 2$; the case $s \geq 2$ is similar. $\square$

**Remark.** The lemma is false if one of $\mathcal E$ and $\mathcal G$ is zero but the other isn't.

*Proof of Theorem.* Let $0 = \mathcal E_0 \subsetneq \ldots \subsetneq \mathcal E_r = \mathcal E$ be the Harder–Narasimhan filtration, and let $\mathcal F_i = \mathcal E_i/\mathcal E_{i-1}$ be the $i$-th graded piece. This means that the $\mathcal F_i$ are semistable vector bundles of slopes $\mu_1 > \ldots > \mu_r$. There are no nonzero homomorphisms $\mathcal F_i \to \mathcal F_r$ for $i < r$, and a simple induction shows that $\operatorname{Hom}(\mathcal E_{r-1},\mathcal F_r) = 0$. By **Lemma 1**, this implies that $\mathcal E_{r-1} = 0$, i.e. $r = 1$; that is, we know that $\mathcal E$ is semistable.

Then we have a (non-canonical) Jordan–Hölder filtration $0 = \mathcal E_0 \subsetneq \ldots \subsetneq \mathcal E_r = \mathcal E$, with stable subquotients $\mathcal F_i = \mathcal E_i/\mathcal E_{i-1}$. The isomorphism classes of stable objects $\mathcal F_i$ occurring agree with the isomorphism classes of stable factors of the polystable sheaf $\operatorname{gr}(\mathcal E_\bullet)$, and these are independent of the filtration chosen.

Firstly, assume that different stable factors occur. Let $\mathcal F_1$ be the first factor, and let $\mathcal G \subseteq \mathcal E$ be a maximal subsheaf containing only factors isomorphic to $\mathcal F_1$. Since $\mathcal F_1 \subseteq \mathcal E$, we know that $\mathcal G \neq 0$. If $\operatorname{Hom}(\mathcal G,\mathcal E/\mathcal G) \neq 0$, then the inverse image in $\mathcal E$ of the image of a nonzero map $\mathcal G \to \mathcal E/\mathcal G$ is a strictly larger subsheaf, all of whose factors are isomorphic to $\mathcal F_1$. This is a contradiction, so we get $\operatorname{Hom}(\mathcal G,\mathcal E/\mathcal G) = 0$. Then **Lemma 1** implies that $\mathcal G = \mathcal E$.

Thus, all stable factors of $\mathcal E$ are the same; say they equal some $\mathcal F$. By **Lemma 2**, the assumption forces $\mathcal E$ to split as a sum of copies of $\mathcal F$, where $\mathcal F$ is a line bundle. $\square$

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>> **Corollary.** Let $C$ be a smooth projective curve over an algebraically closed field, and let $x \in C$. Let $\mathcal F$ be a coherent sheaf on $C$. Then the map $\operatorname{End}(\mathcal F) \to \mathcal End(\mathcal F)_x$ is surjective if and only if $\mathcal F \cong \mathcal L^n \oplus \mathcal G$ for some torsion sheaf $\mathcal G$, some line bundle $\mathcal L$, and some natural number $n$. It is an isomorphism if and only if $\mathcal G$ is supported at $x$.

*Proof.* We can always write $\mathcal F \cong \mathcal E \oplus \mathcal G$ for some vector bundle $\mathcal E$ and some torsion sheaf $\mathcal G$. Then $\mathcal End(\mathcal F) = \mathcal End(\mathcal E) \oplus \mathcal Hom(\mathcal E, \mathcal G) \oplus \mathcal End(\mathcal G)$. (Note that $\operatorname{Hom}(\mathcal G, \mathcal E) = 0$.) The map from global sections to the stalk is surjective if and only if it is so for each piece of the above decomposition, and by the theorem above we easily see this to be equivalent to the statement that $\mathcal E \cong \mathcal L^n$. $\square$