Is there a non-semistable simple sheaf?

Question

Let $C$ be a smooth projective (irreducible) curve over an algebraically closed field $k$. A sheaf is said to be simple if its endomorphism algebra is isomorphic to $k$. It is known that a stable sheaf is simple, but the converse does not holds. There is a simple sheaf which is not stable (but it is still semistable). See, for example, Exercise 26.23 of Gortz and Wedhorne's Algebraic Geometry II.

Question: Is there a non-semistable simple sheaf on a smooth projective curve?

I also would like to know references for simple sheaves on a smooth projective curve (standard properties, classifications, moduli spaces).

Answer 1

Let $C$ be a smooth curve of genus at least $3$. Then $C$ has a degree $1$ line bundle $L$ which is not effective, and by Riemann-Roch $$h^1(L) = -\chi(L) = g-2 > 0.$$ We have $$Ext^1(O_C,L) \cong H^1(L)\neq 0,$$ so we can let $E$ be a rank $2$ bundle given by a non-split extension $$0\to L \to E \to O_C\to 0.$$ This sequence destabilizes $E$, so $E$ is not semistable. But it is also simple. We want to compute $Hom(E,E)$, so we apply $Hom(E,-)$ to get $$0\to Hom(E,L)\to Hom(E,E)\to Hom(E,O_C).$$

To compute $Hom(E,L)$, we apply $Hom(-,L)$ to the original sequence and get $$0\to Hom(O_C,L) \to Hom(E,L) \to Hom(L,L)\to Ext^1(O_C,L).$$ Here $Hom(O_C,L) = H^0(L) = 0$ since $L$ is not effective, and the map $k\cdot id = Hom(L,L)\to Ext^1(O_C,L)$ is injective as it sends $id$ to the extension class defining $E$. Therefore $Hom(E,L) = 0$.

To compute $Hom(E,O_C)$, we apply $Hom(-,O_C)$ to the original sequence and get $$0\to Hom(O_C,O_C) \to Hom(E,O_C) \to Hom(L,O_C).$$ Here $Hom(L,O_C) = 0$ by degree considerations, and $Hom(O_C,O_C) = k$, so $Hom(E,O_C) = k$.

We conclude that $Hom(E,E) = k$.

I am not too aware of many sources that study much about simplicity. Stability or semistability is usually the correct notion to work with. There may be some happy accidents say on an elliptic curve where the notion of simplicity is better behaved.