Let $X$ be a smooth, projective curve (over $\mathbb{C}$) of genus at least $2$ and $E$ be a globally generated sheaf on $X$. I am looking for conditions/examples such that there exists a closed point $x \in X$ for which the natural morphism from $H^0(\mathcal{E}nd(E))$ to $\mathcal{E}nd(E)\_x$ is surjective, where $\mathcal{E}nd(E)\_x$ denotes the fiber of the vector bundle. Any idea or reference regarding this is most welcome.



Endomorphism of globally generated sheaves on curves

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> > **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 $\mathcal E \cong \mathcal L^n$ for some line bundle $\mathcal L$ and some natural number $n$.
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**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.



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> > **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 $\mathcal E$.
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*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.



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> > **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.
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*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 $x$.
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*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 $\mathcal Hom(\mathcal G, \mathcal E) = 0$.) 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 $\mathcal E \cong \mathcal L^n$. $\square$


**Remark.** It seems that I have used the fact that we're on a curve only in the structure theorem of coherent sheaves on them, as in the corollary. A lot of the above carries through verbatim on arbitrary smooth projective varieties over algebraically closed fields, except one has to replace most occurrences of "bundle" by "sheaf" or "pure sheaf" (depending on the context).


