# Dual of slope semistable vector bundle on higher dimensional variety

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 $$$$rk(E) = \frac{\alpha_d(E)}{\alpha_d(\mathcal{O}_X)},$$$$ and the degree of $$E$$ is defined to be $$$$\deg E = \alpha_{d-1}(E) - rk(E) \alpha_{d-1}(\mathcal{O}_X).$$$$ The slope of $$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...

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.
• @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. Commented Jul 11 at 11:18
• @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. Commented Jul 11 at 11:19