Over a smooth algebraic curve, do all vector bundles admit a finite resolution by semi-stable bundles? Or is there a characterization of the vector bundles that do?

Edit: As an example on $\mathbb{P}^1$, it is possible. For a vector bundle tensor it with the right $\mathcal{O}(n)$ so that the smallest summand in the decomposition of the vector bundles to line bundles is $\mathcal{O}$. So we have $E=\mathcal{O}^{\oplus a_0}\oplus \ldots \mathcal{O}(m)^{\oplus a_m}$ for $m>0$. Now surject to this an $\mathcal{O}^{\oplus k}$ for the right $k$ that each $\mathcal{O}$ corresponds to the generators of the global sections. Looking at the kernel, it is the direct sum of kernels of $\mathcal{O}^{\oplus k_i}\rightarrow \mathcal{O}(i)$. If $i=0$ this becomes bascially the identity $\mathcal{O}\rightarrow \mathcal{O}$. If $k_i$ is given by the exact right value that is the dimension of $\Gamma(\mathcal{O}(i))$, then the kernel of $\mathcal{O}^{\oplus k_i}\rightarrow \mathcal{O}(i)$ will have summands of the form $\mathcal{O}(j)$ where $j<0$ (Because the kernel doesn't have a global section since we chose the right $k_i$). But since $j$'s sum up to $-i$ so all of them are between $-i$ and $-1$. Now direct summing all these kernels tells us that the range of the $j$'s appearing in the kernel is between $-\mu_{max}$ and $-\mu_{min}-1$. Note that we assumed initially that $\mu_{min}=0$. This implies that the value $\mu_{max}-\mu_{min}$ for the kernel is strictly less that the original vector bundle. Continuing this way implies that we have to stop at somewhere and when we do we have written a finite resolution of the vector bundle by the semi-stable ones.

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