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Recall the definition of slope semistability, taken from section 1.2 of Huybrechts and Lehn's "Geometry of Moduli Spaces of Sheaves" book. Let $X$ be a projective $\mathbb{C}$-scheme and $E \in Coh(X)$ of dimension $d = \dim(X)$. The rank of $E$ is defined to be \begin{equation} rk(E) = \frac{\alpha_d(E)}{\alpha_d(\mathcal{O}_X)}, \end{equation} and the degree of $E$ is defined to be \begin{equation} \deg E = \alpha_{d-1}(E) - rk(E) \alpha_{d-1}(\mathcal{O}_X). \end{equation} The slope of $E$ is defined to be $\mu(E) := \frac{\deg(E)}{rk(E)}$. Then we say $E$ is slope semistable iff for all subsheaves $F \subset E$ with $0 < rk(F) < rk(E)$ we have $rk(E) \deg(F) \leq rk(F) \deg (E)$.

My question is the following. Let $E$ be a slope semistable vector bundle on, say, nonsingular projective variety $X$. Then is the dual $E^*$ also slope semistable? The natural thing to do here is to take a $F \subset E^*$, then dualize, and use additivity of Hilbert polynomials (note the above inequality for slope semistable is equivalent to the inequality $\alpha_d(E) \alpha_{d-1}(F) \leq \alpha_d(F) \alpha_{d-1}(E)$). But when $F$ is not locally free, you run into issues of the dual not being a surjection of sheaves...

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One can do the following. Take a destabilizing $F \subset E^*$. First replace $F$ with the saturation $F'$ of $F$, i.e. the sheaf whose sections consist of those sections of $E^*$ that generically lie in $F$. There is a natural map $F \to F'$ whose cokernel is supported in codimension $1$. It follows that $\alpha_d(F')=\alpha_d(F)$ and $\alpha_{d-1}(F') \geq \alpha_{d-1}(F)$. So if $F$ is destabilizing then $F'$ is as well.

Now since $F'$ is saturated, $F'$ and $E^* /F'$ are locally free in codimension $1$. Since $\alpha_d$ and $\alpha_{d-1}$ may be calculated by ignoring any codimension $2$ locus, you can ignore the locus where they are not locally free and then use the argument for sub-vector bundles.

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    $\begingroup$ @maxo "destabilizing" just means that $F$ contradicts the inequality which all subsheaves need to satisfy if $E$ is stable. This is the same sense it is used in "maximal destabilizing subsheaf". One could probably run the argument with a maximal destabilizing subsheaf in which case I guess one would check that this sheaf is already saturated. $\endgroup$
    – Will Sawin
    Commented Jul 11 at 11:18
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    $\begingroup$ @maxo For your second question, we work locally, and the local rings in codimension $1$ are discrete valuation rings (because $X$ is regular). Submodules of free modules over DVRs are free, and modules over DVRs are free unless they contain torsion, but torsion elements would contradict saturatedness. $\endgroup$
    – Will Sawin
    Commented Jul 11 at 11:19

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