This is a general question that I have. Let $X$ be a projective variety over an algebraically closed field $k$. Let $L$ be an ample line bundle over $X$. Let $F$ be a vector bundle on $X$. We say that $F$ is $L$-semistable if for any coherent subsheaf $0\neq E\subset F$ of strictly lesser rank, $\mu_L(E)\leq\mu_L(F)$. Here $\mu_L(F)=\frac{c_1(F).L^{n-1}}{r}$, where $r=$ rank$(F)$ and $n=dim\ X$.

My question is, why do we need $L$ to be ample? The definitions and basic properties seem to go through without ampleness. Even the fact that semistability is an open condition does not seem to depend on the fact that $L$ is ample, and existence of Harder-Narasimhan filtration/Jordan Holder filtration too seems to hold without ampleness. Why can't we just take $L$ to be a globally generated or nef and big line bundle for example.


If $L$ is a big and nef bundle, it seems like the following will go through still,

a) Definition and basic properties of like there is no nontrivial homomorphism of semi stable sheaves from one with higher slope to one with lesser slope etc.

b) Existence and uniqueness of Harder-Narasimhan and Jordan-Holder filtrations

c) stability/semistability is an open condition.

Am I mistaken. Is there some subtle/obvious point where we need ampleness for this?

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    $\begingroup$ I am far from an expert, and Jason Starr will be along at some point to give a full answer, but here's a basic point that occurs to me: the point of (semi-)stability is to construct reasonable moduli spaces. If you weaken ampleness to basepoint-freeness then you could take $L=O_X$; then every sheaf is semistable of slope zero. The moduli space of all sheaves on $X$ is not a reasonable object, for several reasons. $\endgroup$ Nov 10, 2015 at 16:45
  • $\begingroup$ My previous comment was written before the words "and big" appeared in the last line... $\endgroup$ Nov 10, 2015 at 19:31
  • $\begingroup$ @potentially dense, sorry. I wanted to write nef and big as a generalization of ample, earlier. But I missed big. $\endgroup$ Nov 10, 2015 at 20:48
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    $\begingroup$ Fair enough. I don't think the extra condition will change things, though: even if $L$ is nef and big, it may well be trivial on some subvariety (e.g. the exceptional divisor of a birational morphism). Then $L$ still cannot "see" what is happening on that subvariety. I'll let someone else give you a precise answer, though. $\endgroup$ Nov 10, 2015 at 22:48

1 Answer 1


You are right, the condition that $L$ be ample can be weakened. In fact, on an $n$-dimensional normal projective variety $X$ one can measure (semi-)stability with respect to an arbitrary movable curve class $\alpha \in N_1(X)_{\mathbb R}$. A numerical curve class $\alpha$ is said to be movable if $D \cdot \alpha \ge 0$ for all effective divisors $D$ on $X$. Then most of the basic properties you mentioned continue to hold.

The classical case ($L$ ample) is recovered by taking $\alpha$ to be the class of a complete intersection curve $C = H_1 \cap \cdots \cap H_{n-1}$, where the $H_i$ are general members of some very ample linear system $|mL|$, $m$ sufficiently large.

This (along with several applications) is worked out in the paper "Movable curves and semistable sheaves" by Greb, Kebekus, and Peternell.

You might also be interested in the survey "Moduli of vector bundles on higher-dimensional base manifolds - Construction and Variation" by Greb, Ross, and Toma, where this machinery is applied.


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