# FIltrations on a vector bundle on a curve

Let $X$ be a smooth connected projective curve of genus $g$ over an algebraically closed field. Let $E$ be a vector bundle on $X$ of rank $n$. Is it true that there exists a constand $N(g,n)$ such that each such $E$ has a filtration $$0\subset E_1\subset ...\subset E_n=E$$ by subbundles with $rank(E_i)=i$, such that $deg(E_i/E_{i-1})-deg(E_{i+1}/E_i)\geq N(g,n)$ ? What is the reference for this?

• This reminds me the Harder-Narashiman filtration, where $$\mu(E_i/E_{i-1}) > \mu(E_{i+1}/E_i)$$ and $\mu(F) :=\deg(F) / \textrm{rank}(F)$ is the slope. – Francesco Polizzi Aug 16 '14 at 11:32
• Well, it has something to do with Harder-Narasimhan, but it is not quite that. For example, if $n=2$ then the statement is that any rank 2 bundle has a line sub-bundle whose degree is not too small. – Alexander Braverman Aug 16 '14 at 11:38
• Do you mean to assume $n > 1$? – user54268 Aug 16 '14 at 14:26
• Yes, if you wish (otherwise the question is empty) – Alexander Braverman Aug 16 '14 at 15:10
• I must be missing something, we have rank two (semi-)stable bundles of arbitrary negative degree. So you can not bound the degree of the line sub-bundles from below universally. Do you want a bound that depends on $g$, $n$, and the degree? – Tony Pantev Aug 17 '14 at 1:33

## 1 Answer

According to lemma 4 of

M.F. Atiyah, Vector bundles over an elliptic curve, Proc. London Math. Soc. (3) 7 (1957), 414-452.

you could take $N(g,n) = - 2 g$.

• You can think of this as an example of "bend-and-break". If you denote by $(\pi:P\to X,\pi^*E\to \mathcal{O}_P(1))$ the universal invertible quotient of the pullback of $E$, then the problem is to give an upper bound on the $\mathcal{O}_P(1)$-degree of a section. Since $\omega_\pi$ is $\pi^*\text{det}(E)\otimes\mathcal{O}_P(-\text{rank}E)$, and since bend-and-break breaks sections until their anticanonical degree is at most $2$, this precisely gives that the minimal $\mathcal{O}_P(1)$-degree of a section is no greater than $(\text{deg}(E)+2g)/\text{rank}(E)$. – Jason Starr Aug 17 '14 at 11:05
• Thanks! This was exactly the reference I was looking for. – Alexander Braverman Aug 18 '14 at 9:56