Difference between Gieseker semistable and slope semistable 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?
 A: 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.
A: Each of the implications
$$ \text{Slope stability} \Rightarrow \text{Gieseker stability} \Rightarrow \text{Gieseker semistability} \Rightarrow \text{slope semistability} $$
is strict, i.e. the converses do not hold.
Take $X$ to be a smooth projective surface of Picard number $\rho(X) \ge 2$ and with $K_X$ numerically trivial.
Pick an ample divisor $H$ on $X$ and a divisor $D$ such that $D \cdot H = 0$, but $D$ is not numerically trivial.
By Riemann-Roch and the Hodge Index Theorem, the self-intersection $D^2$ is even and negative.
Let $Z \subset X$ be a zero-dimensional locally complete intersection subscheme.
We have $H^0(X, \mathcal O_X(K_X - D)) = 0$, hence by the Serre correspondence (Thm.~5.1.1 in Huybrechts-Lehn) there exists an extension
$$ 0 \to \mathcal O_X(D) \to \mathscr E \to \mathscr I_Z \to 0 $$
where $\mathscr E$ is a rank two vector bundle and $\mathscr I_Z$ is the ideal sheaf of $Z \subset X$.
If the length $\ell(Z) > -\frac 12 D^2$, then $\mathscr E$ is slope semistable but not Gieseker semistable.
If $\ell(Z) = -\frac 12 D^2$, then $\mathscr E$ is Gieseker semistable but not Gieseker stable.
If $0 < \ell(Z) < -\frac 12 D^2$, then $\mathscr E$ is Gieseker stable but not slope stable.
A: I cannot comment due to low reputation, but there is a whole chapter on this in "Geometry of Moduli Spaces of Sheaves" by Huybrechts and Lehn.
There are also some explicit computations to compare these notions for curves and surfaces in Friedmann's book called "Algebraic surfaces and holomorphic vector bundles".
Okay: here is a example of a Gieseker stable but not stable sheaf:
Pick a stable bundle $E$ of rank two on $\mathbb{P}^2$ with $c_1(E)=0$ and $H^1(E)\neq 0$. A nontrivial element $t\in H^1(E)$ gives an extension $0\rightarrow E\rightarrow F\rightarrow O_{\mathbb{P}^2}\rightarrow 0$. Then one can compute $H^0(F)=0$ and $H^0(F^{\vee})\neq 0$. Since $c_1(F)=0$ one sees that $F$ cannot be stable. But $F$ is Gieseker stable.
This example is due to Maruyama. A complete proof and can be found in Okonek/Spindler/Schneider - Vector bundles on complex projective paces.
A: Let's fix an ample line bundle and ask about (semi)stability with respect to this line bundle. For pure sheaves on curves, slope stability and Gieseker stability will be the same. If you think about the Hilbert polynomial of $\mathcal{F}$, it is $\chi(\mathcal{F}(m))=\mathrm{rk}(\mathcal{F}) m + \mathrm{deg}(\mathcal{F})$. The reduced Hilbert polynomial is then essentially the same as the slope.
For higher-dimensional varieties, this is no longer necessarily true. You always have the following implications:
Slope stability $\Rightarrow$ Gieseker stability $\Rightarrow$ Gieseker semistability $\Rightarrow$ Slope semistability.
There definitely are sheaves which are slope semistable but not Gieseker semistable, but I can't think of an explicit example off-hand.
A: Here is an naive counterexample on higher dimension: Take two slope semi-stable sheaves $E,F$ with the same slope but coefficients of higher degree terms in their reduced Hilbert polynomial being different. Then $E\oplus F$ is slope semi-stable but not Gieseker semi-stable.
Here is a another natural counterexample for higher dimension: On $\mathbb{P}^{2}$ say, tensor product of two slope semi-stable bundles is slope semi-stable. Now take a slope semi-stable vector bundle $E$, and consider $E\otimes E^{*}$, it admits a map both from and to $\mathcal{O}$, hence $\Delta(E\otimes E^{*})=\Delta(E)^{2}=\Delta(\mathcal{O})=0$, but there are (many) $E$ with $\Delta\neq 0$. (If you are not familiar with discriminant, you can just directly compare the reduced Hilbert polynomial.)
