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We know that a coherent sheaf on a scheme is determined by its restriction on certain open coverings (satisfying compatibility condition). Now I wonder how about a closed covering. To do so I started with simple cases on a smooth complex projective varieity $X$ of dimension $n$.

Taking $X$ itself to be the covering is trivial, so I want to start with coverings of smalll dimensions. Firstly I think about the closed points (covering of dimension zero), but they are not a covering; also coherent sheaf with same stalks can be different.

Then it comes to dimension one coverings: let $X$ a smooth projective variety which can be covered by lines. Given two coherent sheaves $\mathcal{F}$ and $\mathcal{G}$ on $X$ such that $$\iota^*_L\mathcal{F}\cong\iota^*_L\mathcal{G}$$ for any line $\iota_L:L\subset X$. What could one say about the relation between $\mathcal{F}$ and $\mathcal{G}$?

I think we should have $\mathcal{F}\cong\mathcal{G}$ for $X=\mathbb{P}^n$ by the structure of $\textbf{Coh}(\mathbb{P}^n)$ (edited: no we do not, here I should write $\mathbb{P}^1$ because the vector bundles on $\mathbb{P}^n$ are not necessarily decomposable).

What about the general case, for example a ruled surface?

I think such questions should be considered before. If so, any reference is welcome!

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  • $\begingroup$ Welcome new contributor. As Sasha wrote, this is false if $\mathcal{G}$ is not a direct sum of copies of the structure sheaf. However, by Biswas and dos Santos, it is true if $\mathcal{G}$ is a direct sum of copies of the structure sheaf and we use the family of rational curves in a rationally connected variety as the covering family. $\endgroup$
    – Jason Starr
    Commented Nov 7, 2022 at 22:52
  • $\begingroup$ @JasonStarr Thank you for your answer. Could you please offer me name of reference by Biswas and dos Santos? $\endgroup$
    – user494339
    Commented Nov 8, 2022 at 11:21
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    $\begingroup$ Here is the link: webusers.imj-prg.fr/~joao-pedro.dos-santos/… $\endgroup$
    – Jason Starr
    Commented Nov 8, 2022 at 12:50

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It is not true even for $\mathbb{P}^n$. For instance, the tangent bundle $T_{\mathbb{P}^n}$ restricts to each line as $$ T_{\mathbb{P}^n}\vert_L \cong \mathcal{O}_L(2) \oplus \mathcal{O}_L(1)^{\oplus (n-1)}, $$ and on the other hand $$ (\mathcal{O}_{\mathbb{P}^n}(2) \oplus \mathcal{O}_{\mathbb{P}^n}(1)^{\oplus (n-1)})\vert_L \cong \mathcal{O}_L(2) \oplus \mathcal{O}_L(1)^{\oplus (n-1)}, $$ however $T_{\mathbb{P}^n} \not\cong (\mathcal{O}_{\mathbb{P}^n}(2) \oplus \mathcal{O}_{\mathbb{P}^n}(1)^{\oplus (n-1)})$.

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  • $\begingroup$ how do you know about this property of the tangent bundle? It is unclear for me (sorry if the answer is trivial). $\endgroup$
    – Marsault Chabat
    Commented Nov 7, 2022 at 21:22
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    $\begingroup$ @Marsault Chabat It follows from tensoring the Euler-Sequence for $\mathbb{P}^n$ with $\mathcal{O}_L$ assuming that $L=V(x_2,\ldots,x_n)$, then dualizing and noticing that $0 \to S \to S(1) \oplus S(1) \to S(2) \to 0$ is the exact Koszul-complex for the regular sequence $x_0,x_1$ in $S = k[x_0,x_1]$. $\endgroup$
    – Jürgen Böhm
    Commented Nov 7, 2022 at 22:55

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