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Let $X$ be a smooth projective curve, and $E$ an indecomposable vector bundle on $X$ with $\mathrm{deg} E<0$. Is it true that $H^0(X,E)=0$?

This is true if $E$ is a line bundle, which means it is also true whenever $X$ is $\mathbb{P}^1$, since all vector bundles split here.

It is also true by results of Atiyah if $X$ is an elliptic curve. What about for curves of higher genus?

The assumption that $E$ is indecomposable is of course necessary.

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    $\begingroup$ If $E$ you replace indecomposability by semistability, this is true. $\endgroup$
    – Sasha
    Commented Jun 23, 2019 at 17:41

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Let $X$ be a smooth projective curve of genus $2$. First, show that $\mathrm{Ext}^1(\mathcal{O}_X(-p),\mathcal{O}_X)=1$. Then, we get an exact sequence $$ 0\rightarrow \mathcal{O}_X\rightarrow E\rightarrow \mathcal{O}_X(-p)\rightarrow 0 $$ We have that $h^0(E)=1$, but $E$ has degree $-1$.

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