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.



Finite resolution by semi-stable bundles
Devlin Mallory's approach is essentially correct, and in fact the situation is even a little better than he suggested.


Let $V$ be an arbitrary vector bundle of rank $n$. Let $W$ be a stable bundle of rank $n+1$ (if one exists) or a semistable bundle. Fix an ample bundle $\mathcal O(1)$.


I claim that for $m$ sufficiently large, there is a map $W(-m) \to V$ that is surjective and whose kernel is a line bundle (and thus is automatically stable).


It suffices to take $m$ large enough that $V\otimes W^\vee \otimes \mathcal O(m)$ is globally generated. We can view sections of this line bundle as maps $W(-m) \to V$, and the fiber of the section at a point corresponds to the fiber of the map at a point.


Maps from an $m+1$-dimensional vector space to an $m$-dimensional vector space have full rank outside of a codimension $2$ locus. Thus among global sections of $V \otimes W^\vee \otimes \mathcal O(m)^\vee$, those without full rank at a given point have codimension $2$. So those which fail to have full rank at one or more points of $C$ have codimension $1$. Thus we can find a section which has full rank at every point, which implies it is surjective and its kernel is a line bundle.


Over finite fields, this codimension argument doesn't work, but an analogous counting argument does.


