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Let $\mathcal{E}$ be a locally free sheaf on a smooth projective $X$ which is generated by global sections. I.e. there is a surjection $H^0(\mathcal{E})\otimes \mathcal{O}_X\twoheadrightarrow \mathcal{E}$. If $f$ is an automorphism acting on $X$, then we have a natural isomorphism of vector spaces $H^0(\mathcal{E})\xrightarrow{\sim} H^0(f^*\mathcal{E})$. So I think its not true that for a general coherent sheaf $\mathcal{E}$ such an isomorphism induces a map $\mathcal{E} \rightarrow f^*\mathcal{E}$, not even when its generated by global sections. But my question is: Does the isomorphism $H^0(\mathcal{E}) \xrightarrow{\sim} H^0(f^*\mathcal{E})$ for $\mathcal{E}$ globally generated and locally free induce an iso $\mathcal{E} \rightarrow f^*\mathcal{E}$?
Idea: We have an exact sequence $0 \rightarrow K\rightarrow H^0(\mathcal{E})\otimes \mathcal{O}\rightarrow \mathcal{E} \rightarrow 0$. Now $K$ is also locally free and I think it's also globally generated, i.e. can be written as $K=V\otimes \mathcal{O}$ for $V \subset H^0(\mathcal{E})$. Then I can I apply $f^*$ to the middle and left term to get another exact sequence and a commutative diagram of short exact sequences with a map of co-kernels $\mathcal{E} \rightarrow f^*\mathcal{E}$...?

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    $\begingroup$ In general global generation is not helpful. Try looking up non-homogeneous vector bundle on projective space. $\endgroup$
    – Mohan
    Commented Jul 19, 2023 at 14:23
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    $\begingroup$ Even in the simplest case of a line bundle, If $ X $ has ample canonical bundle then it is preserved under any automorphism and globally generated therefore the statement you want to prove is utterly wrong. $\endgroup$ Commented Jul 19, 2023 at 16:30
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    $\begingroup$ It is not true that $K$ is globally generated. Actually, $H^0(K) = 0$. $\endgroup$
    – Sasha
    Commented Jul 20, 2023 at 6:29
  • $\begingroup$ I agree you're right. If you apply H^0 then the kernel is zero, thanks! $\endgroup$
    – Simonsays
    Commented Jul 20, 2023 at 11:59
  • $\begingroup$ @CraniumClamp why is the canonical sheaf preserved under any automorphism? $\endgroup$
    – Simonsays
    Commented Jul 20, 2023 at 12:01

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A simple counterexample is when $X = \mathbb{P}^1 \times \mathbb{P}^1$, $f$ swaps the factors, and $\mathcal{E} \cong \mathcal{O}(1,0)$.

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