Let $X$ be a smooth projective curve and $\mathcal{F}$ a locally free sheaf on $X$ of rank $2$ and negative degree. Assume further that $\mathcal{F}$ is irreducible in the sense, $\mathcal{F}$ cannot be written as a direct sum of two line bundles. Is it then true that $\mathcal{F}$ has no global sections?
Let $X$ be a smooth projective curve of genus 2 defined over an algebraically closed field $k$ of characteristic 0, and let $p \in X$ be given. Then $${\rm Ext}^{1}(\mathcal{O}_{X}(p),\mathcal{O}_{X}) \cong H^{1}(\mathcal{O}_{X}(p)) \cong H^{0}(\omega_{X}(p))^{\ast} \cong k$$ It follows that there is a nonsplit exact sequence $$0 \rightarrow \mathcal{O}_{X} \rightarrow E \rightarrow \mathcal{O}_{X}(p) \rightarrow 0$$ where $E$ is a rank2 vector bundle on $X.$ Clearly $h^{0}(E) =1$ and $c_{1}(E) = 1.$ I claim that $E$ cannot be written as a direct sum of line bundles. Indeed, if $E \cong L_{1} \oplus L_{2}$ where $c_{1}(L_1)=c_{1}(L_2)1,$ (wlog, $c_{1}(L_{1}) < c_{1}(L_{2})$) then we must have $c_{1}(L_{2}) \geq 0.$ If $c_{1}(L_{2}) > 0,$ then the cokernel of $\mathcal{O}_{X} \rightarrow E$ has torsion, which is impossible; thus $c_{1}(L_2)=0$, and $L_2 \cong \mathcal{O}_{X}.$ In addition, we have that $c_{1}(L_{1})=1,$ and $E \rightarrow \mathcal{O}_{X}(p)$ induces a surjection $L_{1} \rightarrow \mathcal{O}_{X}(p),$ which must be an isomorphism. This contradicts the fact that our extension is nonsplit.
On the other hand, if your vector bundle is semistable, the answer to your question is yes.

$\begingroup$ Thanks a lot for the very nice answer. I have a question. Why in the case of semistable the answer is yes? $\endgroup$ – user43198 Mar 12 '15 at 7:13
