I have the following problem: consider a complex algebraic curve $C$ and $L,E,M$ slope semi stable vector bundles on $C$ that fits in a short exact sequence as following:
$$ 0 \rightarrow L \rightarrow E \rightarrow M \rightarrow 0 $$
with $ \mu(L) < \mu (E) < \mu (M) $ and $L$ linear,
If $L,E$ are slope-stables, can I make sure that $M$ is slope-stable with this hypothesis?
Let $C$ be an elliptic curve and $P \ne Q \in C$ two points. Consider an exact sequence $$ 0 \to \mathcal{O} \to E \to \mathcal{O}(P) \oplus \mathcal{O}(Q) \to 0 $$ with extension class $$ \epsilon \in \mathrm{Ext}^1(\mathcal{O}(P) \oplus \mathcal{O}(Q), \mathcal{O}) = \mathrm{H}^1(C,\mathcal{O}(-P)) \oplus \mathrm{H}^1(C,\mathcal{O}(-Q)), $$ such that both components of $\epsilon$ are nonzero. Then $E$ is stable, so the sequence is a counterexample.
Indeed, the slope of $E$ is $2/3$, so to check stability it is enough to show that $E(-R)$ has no global sections for any point $R$ on $C$. For this consider the following exact sequence $$ 0 \to \mathcal{O}(-R) \to E(-R) \to \mathcal{O}(P-R) \oplus \mathcal{O}(Q-R) \to 0. $$ If $R \ne P$ and $R \ne Q$ then $H^0(E(-R)) = 0$ is obvious. If $R = P$ we obtain an exact sequence $$ 0 \to H^0(E(-R)) \to H^0(\mathcal{O} \oplus \mathcal{O}(Q-P)) \to H^1(\mathcal{O}(-P)) $$ and $H^0(E(-P)) = 0$ follows from the nontriviality of the first component of $\epsilon$. Similarly, $H^0(E(-Q)) = 0$, hence $E$ is stable.