# Why does this vector bundle on the surface sit in this exact sequence?

Let $X$ be a K3 surface. Let $E$ be a semistable rank 3 vector bundle. Now suppose $0 = E_0\subset E_1\cdots\subset E_s=E$ be the Harder-Narasimhan filtration. Suppose $E_1$ is $\mu$-stable and rank $E_1$ is 2. Then the paper says that $E$ sits in the following short exact sequence :

$0\longrightarrow E_1\longrightarrow E\longrightarrow N\otimes I_\xi\longrightarrow 0$ where $N$ is a line bundle and $I_\xi$ is the ideal sheaf of a 0-dimensional subscheme $\xi\subset X$.

How do we get this exact sequence? What are $N$ and $\xi$ exactly? Any help will be appreciated! Thanks in advance!

Any torsion free sheaf on a smooth surface can be written in the form $N \otimes I_\xi$, in particular this holds for $E/E_1$. Explicitly, one can write $N = \det(E)\otimes \det(E_1)^{-1}$, and also one can figure out the length of $\xi$ from Chern classes of $E_1$ and $E$. The actual subscheme $\xi$ is more difficult to understand.
• We have an injective map $E/E_1\longrightarrow {E/E_1}^{\vee\vee}=F$. Now $F$ is a line bundle, therefore, tensoring by $F^{\vee}$, we get an injective map $E/E_1\otimes F^{\vee}\longrightarrow F\otimes F^{\vee}=\mathcal{O}_X$. Therefore, $E/E_1\otimes F^{\vee}$ is an ideal sheaf of a closed subscheme say $Z$. So $E/E_1 = F\otimes I_Z$. So I get what you said. Why is the codimension of $Z$ two? – gradstudent Apr 20 '15 at 12:28
• Correct. If $codim Z = 1$ then $I_Z$ is invertible and $N \otimes I_Z = N'$ with $N'$ a line bundle. – Sasha Apr 20 '15 at 14:48