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.

  • 2
    $\begingroup$ This is done in detail in the book by Huybrechts and Lehn. $\endgroup$
    – abx
    Commented Dec 13, 2019 at 14:55
  • 3
    $\begingroup$ @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? $\endgroup$ Commented Dec 13, 2019 at 16:12
  • 1
    $\begingroup$ 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. $\endgroup$
    – LSpice
    Commented Dec 13, 2019 at 17:34
  • $\begingroup$ @LSpice You are correct, thanks! I figured there would be no harm leaving the Unicode symbols, but $R > 0$ slipped my sight. $\endgroup$ Commented Dec 13, 2019 at 17:36

1 Answer 1


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 :)


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