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Recall Schur's Lemma for Gieseker-semistable sheaves, in particular the injectivity statement:

Let $\psi : F \to G$ be a morphism of Gieseker-semistable sheaves. If $p(F)=p(G)$ and $F$ is stable, then $\psi$ is either injective or zero.

where $p(-)$ is the reduced Hilbert polynomial. My question is: is there a variant/similar statement for sheaves with different Hilbert polynomials?

I have heard that one can use $p_{-}(-)$ and $p_{+}(-)$ coming from the Harder-Narasimhan filtration. Does anyone have a reference for this/know the statement? In the book of Huybrechts-Lehn, there is the statement $p_-(F)>p_+(G)$ implies $\psi=0$, but there doesn't seem to be a statement regarding injectivity.

Thanks!

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    $\begingroup$ When $\mathscr F$ is stable (or even semistable), we have $p_-(\mathscr F) = p(\mathscr F) = p_+(\mathscr F)$, so $p(\mathscr F) > p_+(\mathscr G)$ implies $\psi = 0$. I don't know what you mean by "there doesn't seem to be a statement regarding injectivity" ― the conclusion is actually stronger in the case of different slopes. (Compare this with Schur's lemma for representations of finite groups, where there are no nonzero maps between different irreducible characters.) $\endgroup$
    – R. van Dobben de Bruyn
    Commented Mar 10, 2021 at 20:35
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    $\begingroup$ By contrast, if $\mathscr F$ and $\mathscr G$ are stable but $p(\mathscr F) < p(\mathscr G)$, then there are many nonzero maps $\mathscr F \to \mathscr G$, which may even have kernels. For example, if $\mathscr F$ is stable of rank $2$ and $s \in H^0(\mathscr F^\vee(d))$ for $d \gg 0$, then $s$ gives a map $\mathscr F \to \mathcal O(d)$ whose kernel must be a line bundle. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Mar 10, 2021 at 20:41
  • $\begingroup$ @R.vanDobbendeBruyn Thanks for your comments. I suppose my question is whether there's a statement like: $p_-(F)=p_+(G)$ and $F$ stable implies $\psi$ is injective or zero? Because if this were true, then instead of requiring $p(F)=p(G)$ for injectivity, I suppose we could require $p(F)=p_+(G)$ $\endgroup$
    – alg_et_geom
    Commented Mar 10, 2021 at 21:24
  • $\begingroup$ Ah, that's certainly true, and follows from the stated version together with the version in my first comment. Indeed, if $0 \to \mathscr G_+ \to \mathscr G \to \mathscr G' \to 0$ is the semistable subobject of highest slope (for whatever slope function you're using), then $\operatorname{Hom}(\mathscr F, \mathscr G') = 0$ so $\psi$ lands in $\mathscr G_+$, which is semistable with $p(\mathscr G_+) = p(\mathscr F)$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Mar 10, 2021 at 23:29
  • $\begingroup$ I see, thanks for your help! $\endgroup$
    – alg_et_geom
    Commented Mar 11, 2021 at 1:12

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