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The following question arose while thinking about a step in the proof of Huybrechts-Lehn, Theorem 1.3.1 (the Harder-Narasimhan filtration for the projective line $\mathbb{P}^{1}$):

Setup: Let $k$ be a field, let $X$ be a projective $k$-scheme, let $\mathcal{O}_{X}(1)$ be a fixed very ample line bundle on $X$, let $E$ be a vector bundle on $X$. For an integer $b \in \mathbb{Z}$, let $\mathrm{ev}_{E(b)} : \mathrm{H}^{0}(X,E(b)) \otimes_{k} \mathcal{O}_{X} \to E(b)$ be the evaluation map and let $E_{b} \subseteq E$ be the image of the twist $\mathrm{ev}_{E(b)}(-b) : \mathrm{H}^{0}(X,E(b)) \otimes_{k} \mathcal{O}_{X}(-b) \to E$.

Question: Why is $E_{b} \subset E_{b+1}$?

(And does this inclusion depend on a choice of global section $s \in \Gamma(X,\mathcal{O}_{X}(1))$?)

Remark: Note that $\mathrm{Hom}_{\mathcal{O}_{X}}(\mathcal{O}_{X}(-b),\mathcal{O}_{X}(-b-1)) = 0$ so there is no map $\varphi : \mathrm{H}^{0}(X,E(b)) \otimes_{k} \mathcal{O}_{X}(-b) \to \mathrm{H}^{0}(X,E(b+1)) \otimes_{k} \mathcal{O}_{X}(-b-1)$ satisfying $\mathrm{ev}_{E(b+1)}(-b-1) \circ \varphi = \mathrm{ev}_{E(b)}(-b)$.

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There is a commutative diagram $$\require{AMScd} \begin{CD} H^0(E(b)) \otimes H^0(\mathcal{O}(1)) \otimes \mathcal{O}(-b-1) @>{\mathrm{ev}_{\mathcal{O}(1)}}>> H^0(E(b)) \otimes \mathcal{O}(-b) \\ @VVV @V{\mathrm{ev}_{E(b)}}VV \\ H^0(E(b+1)) \otimes \mathcal{O}(-b-1) @>{\mathrm{ev}_{E(b+1)}}>> E \end{CD} $$ where the left vertical arrow is the multiplication of global sections. The top arrow is surjective (because $\mathcal{O}(1)$ is very ample), hence the composition of the top and right arrows has image $E_b$. By commutativitity of the diagram it is contained (canonically) in the image $E_{b+1}$ of the bottom arrow.

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  • $\begingroup$ Thanks very much. I think your argument shows that we only really need $\mathcal{O}(1)$ to be globally generated: Let $X$ be a scheme, let $E$ be a quasicoherent $\mathcal{O}_{X}$-module, let $\mathcal{L}$ be a globally generated line bundle. Then $\operatorname{im} \mathrm{ev}_{E} \subseteq \operatorname{im} (\mathrm{ev}_{E \otimes \mathcal{L}} \otimes \mathcal{L}^{-1})$. $\endgroup$ – user2831784 Sep 23 at 18:29
  • $\begingroup$ Yes, that's correct. $\endgroup$ – Sasha Sep 23 at 18:32

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