# Harder Narasimhan filtration for the endomorphism bundle

Let $E$ be a vector bundle over a compact Riemann surface $X$, and let $$0=E_0\subsetneq E_1\subsetneq \ldots \subsetneq E_n=E$$ be its Harder-Narasimhan filtration: we have $V_i:=E_i/E_{i-1}$ semistable and $\mu_i:=\mu(V_i) > \mu_{i+1}$ for all $i>0$. My question is how to construct the Harder-Narasimhan filtration for $End(E)$ from there. Here is what I have from Atiyah-Bott "Yang-Mills equations over Riemann surfaces" (p.590). Setting $W_i:=\{\phi \in End(E)~|~\phi(E_k)\subset E_{k+i}~ \forall k\}$, one obtains a filtration $$(*)\quad \quad 0=W_{-n}\subsetneq \ldots \subsetneq W_0\subsetneq \ldots \subsetneq W_{n-1}=End(E).$$ We have that $W_{i}/W_{i-1}=\bigoplus_k Hom(V_k,V_{k+i})$ is a direct sum of semistable vector bundles of respective weights $\mu_{k+i}-\mu_k$.

The paper concludes that

1. Any subbundle $0\neq F\subset End(E)/W_0$ satisfies $\mu(F)<0$.
2. The Harder-Narasimhan filtration for $End(E)$ is a refinement of $(*)$.

Could anybody help me understand this conclusion ? I see that $W_0/W_{-1}$ is semistable of slope $0$ and that we may refine the filtration $(*)$ using Harder-Narasimhan for $W_i/W_{i-1}$ to obtain semistable successive quotients, but then the weights will not necessarily decrease.

For 2. let's consider the quotient $W_i/W_{i-1} = \bigoplus_{\mu} E_\mu$, where $E_\mu$ is the direct sum of those semistable bundles $Hom(V_k, V_{k+i})$ which have slope $\mu$. Then you get a Harder-Narasimhan filtration (=HNF) by taking $W_i/W_{i-1} \supset \bigoplus_{\mu \neq \mu_{min}} E_\mu$, where $\mu_{min}$ is the smallest slope (this yields a decreasing filtration of vector bundles where the quotients are $E_\mu$ so semistable and have strictly decreasing slopes thus it's the HNF).

Refining the filtration $W_i$ by $W_i \supset \bigoplus_{\mu \neq \mu_{min}} E_\mu + W_{i-1}$ etc. we should get the HNF. As Beth pointed out in the comments there is a gap here: We still need to check that our refinement fits together: That is we need that the slope of the last graded piece of the HNF of $W_i/W_{i-1}$ is smaller than the slope of the first graded piece of the HNF of $W_{i-1}/W_{i-2}$. In other words, we need that

$$\max \{\mu_{k+i} - \mu_k \vert k \} < \min \{\mu_{k+i-1} - \mu_k \vert k \}$$

As for 1. We need to check that the first non-trivial piece of the HNF of $End(E)/W_0$ has slope $< 0$. Again the HNF is obtained by refining as above. The slope $\mu$ of any $E_\mu$ as occuring above is of the form $\mu_{k+1} - \mu_k < 0$ as desired.

• Thank you very much for the answer. There still is a problem though: the weights occuring in the refined filtration you decribe are all possible differences of weights, but ordered in the following way. For some fixed $i$, the set $\{\mu_{k+i}-\mu_k~|~k\}$ in a decreasing order. Then for $i+1$, the set $\{\mu_{k+i+1}-\mu_k~|~k\}$ in a decreasing order. But it is very well possible that $\min_k (\mu_{k+i}-\mu_k)<\max_k(\mu_{k+i+1}-\mu_k)$. – Beth Jul 16 '18 at 18:22
• @Beth, okay I agree there's a gap here in the argument. I'll edit again if I come up with something. I also had the wrong order in the filtration, one should start by excluding the $E_\mu$ with minimal slope. – Axel Stäbler Jul 17 '18 at 8:11
• Maybe it's also worth mentioning that the filtration I described has to be the correct one provided one can obtain the HNF by a refinement of the $W_i$. It's clearly the HNF of the last non-trivial piece $W_{-n+1}$. By way of construction of the HNF we then may proceed by construction a HNF for $W_{-n}/W_{-n+1}$ since, by assumption, both $W_{-n}$ and $W_{-n+1}$ occur in the HNF. – Axel Stäbler Jul 17 '18 at 11:41
• I think you're right: 2. is correct if and only if $\min_k (\mu_{k+i}-\mu_k)>\max_k(\mu_{k+i+1}-\mu_k)$ for all $i$. And one can easily construct examples where this is not the case. – Beth Jul 17 '18 at 15:09

I found a direct proof for 1. For all $i\geq 0$, denote $$\overline{W}_i:=W_i/W_0\, .$$ Let $0\neq F\subset \overline{W}_n$. Consider the exact sequence $$0\longrightarrow F\cap \overline{W}_i\longrightarrow F\oplus \overline{W}_i \longrightarrow F+\overline{W}_i \longrightarrow 0$$ which injects into the corresponding exact sequence for $i+1$. Considering the quotient we get $0\longrightarrow (F\cap \overline{W}_{i+1} ) / ( F\cap \overline{W}_{i})\longrightarrow \overline{W}_{i+1}/ \overline{W}_i=W_{i+1}/W_i$. In particular, either $(F\cap \overline{W}_{i+1} ) / ( F\cap \overline{W}_{i})$ is zero or $$\mu\left( (F\cap \overline{W}_{i+1} ) / ( F\cap \overline{W}_{i}) \right)\, \leq \,\mu_\max(W_{i+1}/W_i)= \max_k (\mu_{k+i+1}-\mu_k) \, <\, 0\, .$$ Now consider only $i\geq i_0$ where $i_0$ is the first index such that $F\cap \overline{W}_{i_0}\neq 0$, and the short exact sequences $$0\longrightarrow F\cap \overline{W}_i\longrightarrow F\cap \overline{W}_{i+1} \longrightarrow (F\cap \overline{W}_{i+1} ) / ( F\cap \overline{W}_{i}) \longrightarrow 0.$$ By induction, we have $\mu (F\cap \overline{W}_{i} )<0$ for all $i\geq i_0$.