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.

$\textbf{Edit:}$

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?

everysheaf is semistable of slope zero. The moduli space of all sheaves on $X$ is not a reasonable object, for several reasons. $\endgroup$ – potentially dense Nov 10 '15 at 16:45