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Let $S$ be a smooth projective surface with an ample divisor $H$ so that $K_S \cdot H < 0$.

Let us consider the Mumford slope $\mu(E) = \frac{H \cdot c_1(E)}{\operatorname{rk}(E)}$ on $\operatorname{Coh}(S)$. Let $\mathcal A^{\tau}$, $\tau >> 0$ be the subcategory of semistable sheaves with the large slope.

Mozgovoy and Reineke claim (cf. Example 3.8) that the global sections functor $\Gamma_S$ is exact when restricted to $\mathcal A^{\tau}$ (without the proof). How to see this?

Is anything of this form known for Fano threefolds?

Evidently, the standard trick with Serre duality (which can be used for curves) no more works.

Thank you!

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    $\begingroup$ I don't think this is correct. If you fix your $r,c_1,ch_2$, then there are moduli spaces of semistable sheaves and Serre vanishing + Noetherian induction shows that a large twist $E(nH)$ has no higher cohomology for every sheaf in the moduli space. (This argument is a bit circular though as this fact is used to construct the moduli space in the first place; see e.g. Huybrechts and Lehn "The geometry of moduli spaces of sheaves"). If $r,c_1$ are fixed and $ch_2$ becomes sufficiently negative though, the Euler characteristic of $E$ becomes negative and there must be nonzero $H^1$. $\endgroup$
    – Jack Huizenga
    Commented Oct 2 at 21:12
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    $\begingroup$ @JackHuizenga I agree. For the fixed topological type this is, of course, the standard boundedness statement. When the topological type is not fixed, a friend of mine suggested a simple counterexample: twist by the given ample line bundle (of a sufficiently high slope) the sequence of ideal sheaves of $1$, $2$, ..., $n$, ... points: $H^1$ should finally be nonzero. Needless to say, you may turn your comment to an answer that will be immediately accepted. $\endgroup$
    – Secondflooroffice
    Commented Oct 2 at 21:16
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    $\begingroup$ Yes that's a nice example too that is easy to understand. $\endgroup$
    – Jack Huizenga
    Commented Oct 2 at 21:22

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I don't think this is correct. If you fix your $r,c_1,ch_2$, then there are moduli spaces of semistable sheaves and Serre vanishing + Noetherian induction shows that a large twist $E(nH)$ has no higher cohomology for every sheaf in the moduli space. (This argument is a bit circular though as this fact is used to construct the moduli space in the first place; see e.g. Huybrechts and Lehn "The geometry of moduli spaces of sheaves"). If $r,c_1$ are fixed and $ch_2$ becomes sufficiently negative though, the Euler characteristic of $E$ becomes negative and there must be nonzero $H^1$.

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