Let $X$ be a projective reduced (not necessarily irreducible) curve over an algebraically closed field and $\mathcal{F}$ be a pure coherent sheaf on $X$. Is it true that $\mathcal{F}$ is Gieseker semistable if and only if it is slope semistable? If so, does the same conclusion holds if $X$ is of higher dimension?



Difference between Gieseker semistable and slope semistable
**Proposition:** Let $X$ be a smooth projective surface for which $H^{0}(\omega\_{X})=0.$ Then there exists a vector bundle $E$ on $X$ satisfying the property that for any ample divisor $H$ on $X,$ $E$ is slope-semistable with respect to $H$ but not Gieseker-semistable with respect to $H.$ 


**Proof:** Let $x \in X$ be a closed point, and consider the exact sequence


$$0 \rightarrow \omega\_{X} \otimes \mathcal{I}\_{x|X} \rightarrow \omega\_{X} \rightarrow \omega\_{X}|\_{x} \rightarrow 0$$


Taking cohomology and applying Serre duality, we have from the vanishing $H^{0}(\omega\_{X})=0$ that


$$0 \neq H^{0}(\omega\_{X}|\_{x}) \subset H^{1}(\omega\_{X} \otimes \mathcal{I}\_{x|X}) \cong {\rm Ext}^{1}(\mathcal{I}\_{x|X},\mathcal{O}\_{X})^{\ast}$$


It follows that there is a non-split extension 


$$0 \to \mathcal{O}\_{X} \to E \to \mathcal{I}\_{x|X} \rightarrow 0$$


where $E$ is a torsion-free sheaf of rank 2. Combining Theorem 5.1.1 in Huybrechts-Lehn with the vanishing $H^{0}(\omega\_{X})=0,$ we may take $E$ to be a vector bundle (i.e. locally free). For the rest of the proof, we fix an ample divisor $H$ on $X.$


We now verify that $E$ is slope-semistable with respect to $H.$ First, observe that $c\_{1}(E)=c\_{1}(\mathcal{O}\_{X})+c\_{1}(\mathcal{I}\_{x|X})=0$; in particular, $\mu\_{H}(E)=0.$ If $E$ is not slope-semistable, then since $E$ has rank 2, there exists a line bundle $L$ on $X$ satisfying $\mu\_{H}(L)=c\_{1}(L) \cdot H > 0$ and $L \subset E$; the latter implies that $0 \neq {\rm Hom}(L,E)=H^{0}(L^{-1} \otimes E).$ Moreover, $c\_{1}(L^{-1}) \cdot H = -c\_{1}(L) \cdot H < 0,$ so we have that $H^{0}(L^{-1})$ and its subspace $H^{0}(L^{-1} \otimes \mathcal{I}\_{x|X})$ are both zero. Twisting by $L^{-1}$ and taking cohomology, we have that $H^{0}(L^{-1} \otimes E)=0,$ a contradiction. 


To show that $E$ is not Gieseker-semistable with respect to $H,$ it suffices to show that $\mathcal{I}\_{x|X}$ is a destabilizing quotient of $E,$ i.e. that the reduced Hilbert polynomial of $\mathcal{I}\_{x|X}$ with respect to $H$ is greater than that of $E.$ To compare these, it suffices in turn to look at $h^{0}(E(mH))/2$ and $h^{0}(\mathcal{I}\_{x|X}(mH))$ for sufficiently large $m.$ Fix $m$ large enough such that this is valid, $\mathcal{O}\_{X}(mH)$ is globally generated, and $H^{1}(\mathcal{O}\_{X}(mH))=0.$ The fact that $\mathcal{O}\_{X}(mH)$ is globally generated implies that $h^{0}(\mathcal{I}\_{x|X}(mH))=h^{0}(\mathcal{O}\_{X}(mH))-1,$ so we have
$$h^{0}(E(mH))/2 = (h^{0}(\mathcal{O}\_{X}(mH))+h^{0}(\mathcal{I}\_{x|X}(mH)))/2 = (2h^{0}(\mathcal{O}\_{X}(mH))-1)/2 > h^{0}(\mathcal{I}\_{x|X}(mH))$$
This concludes the proof.


