Dual of slope semistable vector bundle on higher dimensional variety

Question

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...

Answer 1

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.