# Jordan–Hölder sequence for $\mu$-semi stable sheaves

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.

• This is done in detail in the book by Huybrechts and Lehn.
– abx
Dec 13 '19 at 14:55
• @abx I don't see this in the book. They prove in detail the existence of a Harder-Narasimhan filtration, but the Jordan-Hölder filtration is really short. I added a paragraph about my confusion. Or am I missing something here? Dec 13 '19 at 16:12
• The final blockquote seemed to be directly cut-and-pasted, as it had Unicode symbols in place of TeX. I edited, except that I wasn't sure what to make of "$Z(E) ∈ R >0$". I guessed it meant "$Z(E) \in \mathbb R_{> 0}$". If this was wrong, then I apologise; please feel free to (re-)correct it. Dec 13 '19 at 17:34
• @LSpice You are correct, thanks! I figured there would be no harm leaving the Unicode symbols, but $R > 0$ slipped my sight. Dec 13 '19 at 17:36

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.