# Finite resolution by semi-stable bundles

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.

• I think the standard proof that you have a resolution by direct sums of line bundles shows this: if $E$ is a vector bundle, we can harmlessly replace $E$ by its twist by a line bundle and thus assume $E$ is globally generated. There is then a surjection $\bigoplus \mathcal O_C\to E\to 0$. The term $\bigoplus \mathcal O_C$ is semistable, and by induction on projective dimension the kernel of this morphism is resolved by semistable bundles. Commented Jan 6, 2021 at 19:25
• I still don't get why the last element which appears to be a kernel has to be semi-stable? is it going to be in the form of $\oplus \mathcal{O}(n)$ too? Commented Jan 6, 2021 at 20:52
• The kernel will probably not be semistable (or even locally free), but you can surject onto it from a semistable/locally free sheaf (by the same reasoning as for $E$), and thus continuing inductively you're done. Commented Jan 7, 2021 at 0:09
• But the resolution needs to be finite. At some point you need to stop and the kernel needs to be semi stable. Commented Jan 7, 2021 at 0:14
• The resolution will be finite by Hilbert's syzygy theorem- if you repeat the above process, it will terminate after a finite number of steps (with last term semistable). Commented Jan 7, 2021 at 5:50

## 1 Answer

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.

• Regarding your first sentence about the approach suggested in the comments above being essentially correct. Does it mean if we repeat the generic resolution, which means tensors it with $\mathcal{O}(1)$ until it becomes globally generated and then surject $\mathcal{O}^n$ for n the dimension of the global sections, eventually lead to a kernel that is semi-stable? Commented Jan 11, 2021 at 1:40
• @user127776 I meant the general idea of twisting up, choosing a bunch of sections, and then looking at the kernel. I'm not very knowledgable about Hilbert's syzygy theorem. Commented Jan 11, 2021 at 2:06