# The evaluation map on twists of a vector bundle and an induced filtration

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 $$\mathrm{ev}_{E(b+1)}(-b-1) \circ \varphi = \mathrm{ev}_{E(b)}(-b)$$.

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.
• 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 $\operatorname{im} \mathrm{ev}_{E} \subseteq \operatorname{im} (\mathrm{ev}_{E \otimes \mathcal{L}} \otimes \mathcal{L}^{-1})$. – user2831784 Sep 23 at 18:29