Let $X$ be a smooth variety over $\mathbb{C}$, and let $\omega \in \operatorname{Pic}(X)\_\mathbb{R}$ be an ample class.
I would like to know if any $\mu\_\omega$-semistable sheaf $E \in \operatorname{Coh}(X)$ admits a Jordan-Hölder sequence, i.e. a finite sequence
$$0 = E\_0 \subset E\_1 \subset \dots \subset E\_{n-1} \subset E\_n = E,$$
such that the factors $E\_i / E\_{i+1}$ are $\mu\_\omega$-stable.


Lehn and Huybrechts state something similar, the claim the existence of a Jordan-Hölder sequence with stable factors, which, according to the same book, is a weaker notion than $\mu$-stability. Also, I don't understand the proof they give:



> 
> Proof. Any filtration of $E$ by semistable sheaves with reduced Hilbert polynomial $p(E)$ has a maximal refinement, whose factors are necessarily stable.
> 
> 
> 


Why is that true, and what exactly is a maximal refinement? Or am I missing something different here?


This is what I tried:


If $E$ is not already stable, the set of proper subsheaves $E' \subsetneq E$ with $\mu(E') = \mu(E)$ is non-empty. Then we can mod out a *maximal* subsheaf of that kind (which exists by Noetherianity), and obtain $F = E / E'$.


**Question 1:** Why is $F$ torsion free?


Equivalently, one could ask why $E'$ is saturated. It would also be sufficient if the saturation $\tilde{E''}$ of any submodule $E'' \subset E$ had the same slope, but I don't think this is true: If $\tilde{E''} / E''$ is supported in dimension $1$, then $c\_1(\tilde{E''} / E'') \cdot \omega > 0$, so $\mu(\tilde{E''}) < \mu(E'')$.


Now suppose $F$ is torsion free. Then $\mu(F) = \mu(E)$, and
$F$ is in fact $\mu$-stable, because any proper nonzero quotient $F \to Q \to 0$ with $\mu(Q) = \mu(F) = \mu(E)$ violates the maximality of $E'$. So inductively we can construct a sequence
$$ E = E^0 \supset E^1 \supset E^2 \supset \dots,$$
whose quotients $F^i = E^i / E^{i+1}$ are torsion-free, stable sheaves of slope $\mu(E)$. Because the quotients are torsion-free, the ranks of the $F^i$ are strictly decreasing, so the sequence has to stop.


**Question 2:** Is the rest of my reasoning correct?


A bit of context: I'm reading Bridgeland's *Stability conditions on K3 surfaces*, and in the proof of **Lemma 6.2**, he writes:



> 
> The only nontrivial part is to check that if a torsion-free $\mu\_\omega$-semistable sheaf $E$ satisfies $(\Delta − r\beta)\cdot\omega = 0$, then $Z(E) \in \mathbb R\_{ >0}$. It is enough to check this when $E$ is $\mu\_\omega$-stable.
> 
> 
> 


I figured that a Jordan–Hölder sequence might prove this reduction step, and I don't see any other way to do it.



Jordan–Hölder sequence for $\mu$-semi stable sheaves
So I talked to my advisor, and one of the misconceptions I had is about $\mu\_\omega$-semistability: A torsion-free sheaf $E$ is called $\mu\_\omega$-semistable, if for all subsheaves $0 \neq E' \subset E$ with $\operatorname{rk}(E') < \operatorname{rk}(E)$, one has $\mu\_{\omega}(E') \leq \mu\_\omega(E)$. Similar for stability.


With this definition, one can take a maximal subsheaf $E'$ with $\mu\_\omega(E') = \mu\_\omega(E)$ and $\operatorname{rk}(E') < \operatorname{rk}(E)$. Then the saturation $\tilde{E'}$ has the same rank, and $c\_1(E') + c\_1(\tilde{E'} / E') = c\_1(\tilde{E'})$. Because $\omega$ is ample, $c\_1(\tilde{E'} / E') \cdot \omega \geq 0$, so $$c\_1(\tilde{E'}) \cdot \omega \geq c\_1(E') \cdot \omega,$$ and the same is true for the slopes. But $E'$ already has maximal slope, so $\mu(\tilde{E'}) = \mu(E')$, and hence $E' = \tilde{E'}$ by the maximality.


Would be nice if someone could check this :)


