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Let $X$ be a smooth projective variety over $\mathbb{C}$, and fix $A$ an ample divisor as the polarization. We say a vector bundle $E$ to be (semi-)stable, if for any proper subsheaf $F$ of $E$, $\mu(F)<\mu(E)$ (resp. $\leq$).

I guess it is not sufficient to check just subbundles $F$ of $E$. But is there a counterexample? Or more precisely, is there an example of $(X, A, E, F)$ satisfying the following conditions?

(1)$X,A,E$ are as above, and $F$ is a proper subsheaf of $E$ breaking the stability condition, i.e., $\mu(F)\geq \mu(E)$.

(2)There exists no vector bundle which break the stability condition, i.e., $\mu(F')<\mu(E)$ for any subbundle $F'$ of $E$.

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up vote 6 down vote accepted

There are many examples of unstable bundles on a projective surface that have no non-trivial subbundles. For example, if $k$ is an integer with $k < 3$ and $I$ is the sheaf of ideal of $m$ distinct points in $\mathbb P^2$, with $m > 0$, there exists an extension $$ 0 \longrightarrow \mathcal O \longrightarrow E \longrightarrow I(k) \longrightarrow 0 $$ on $\mathbb P^2$ in which $E$ is locally free. Furthermore, the Chern classes of $E$ are $c_1(E) = k$ and $c_2(E) = m$ (for this, see page 103 of "Vector bundles on complex projective spaces", by Okonek, Schneider and Spindler). If $k < 0$, this vector bundle is clearly unstable; but for most values of $k$ and $m$ it can not split as a direct sum of line bundles, hence it cannot contain a line subbundle (since every extension of line bundles on $\mathbb P^2$ splits).

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The reason for the condition to be on subsheaves is that the slope of a sheaf E on X is unchanged by removing subschemes of X of cod >1. Consequently the Harder-Narasimhan filtration, etc. must allow torsion free subsheaves in its construction. It is enough to consider 'saturated' subsheaves meaning subsheaves F such that E/F is torsion free since passing from a subsheaf F' to its saturate raises the slope.

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