# Could certain closed covering determine a coherent sheaf?

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!

• 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. Commented Nov 7, 2022 at 22:52
• @JasonStarr Thank you for your answer. Could you please offer me name of reference by Biswas and dos Santos?
– user494339
Commented Nov 8, 2022 at 11:21
• Here is the link: webusers.imj-prg.fr/~joao-pedro.dos-santos/… Commented Nov 8, 2022 at 12:50

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 $$T_{\mathbb{P}^n} \not\cong (\mathcal{O}_{\mathbb{P}^n}(2) \oplus \mathcal{O}_{\mathbb{P}^n}(1)^{\oplus (n-1)})$$.

• how do you know about this property of the tangent bundle? It is unclear for me (sorry if the answer is trivial). Commented Nov 7, 2022 at 21:22
• @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 $S = k[x_0,x_1]$. Commented Nov 7, 2022 at 22:55