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When a closed subvariety $Y$ of $X$ satisfy effective Lefschetz condition $Leff(X,Y)$, it implies there is an equivalence if categories between the vector bundles on the formal neighborhood of $Y$ in $X$ and vector bundles on a neighborhood of $Y$. One of functors inducing this equivalence is clear, it is the restriction functor which is exact. Is there any explicit description of the other functor? Is it exact?

A theorem Hartshorne implies that when $X$ and $Y$ are smooth projective varieties and $Y$ is a complete intersection with $\text{dim}(Y)\geq 2$, this property holds. I wonder whether there are any interesting examples of pairs that $X$ is a surface and $Y$ is a curve, I'm more interested in having an interesting $X$. (in char $p$ preferably)

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  • $\begingroup$ If $X$ is smooth, $\operatorname{Leff}(X,Y) $ implies that the embedding induces an isomorphism $\pi _1(Y)\rightarrow \pi _1(X)$. If $X$ is a surface and $Y$ a curve, this is rather unlikely. $\endgroup$
    – abx
    Commented Jan 10, 2021 at 5:02
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    $\begingroup$ @abx There are some examples where this isomorphism holds, like ruled surfaces, or $\mathbb P^1$ on rational surfaces. Could the effective Lefschetz be true for those? $\endgroup$
    – Will Sawin
    Commented Jan 10, 2021 at 15:10
  • $\begingroup$ I think there will rarely be an explicit description of the inverse, except when you have an explicit description of the category of vector bundles on a formal neighborhood. However, since it is an equivalence of categories, it is necessarily exact. $\endgroup$
    – Will Sawin
    Commented Jan 10, 2021 at 15:11
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    $\begingroup$ @Will Sawin: you are right, sorry. I still suspect that these do not satisfy Lefschetz effective, but I'd have to check. $\endgroup$
    – abx
    Commented Jan 10, 2021 at 20:55
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    $\begingroup$ Good point. However, a sequence of vector bundles is exact on some open neighborhood of $Y$ as long as it is exact on a formal neighborhood of $Y$, or even on $Y$, because exactness can be expressed as the vanishing of some coherent sheaf and thus is an open condition. So as long as you define exact sequences of vector bundles on a neighborhood of $Y$ to be those that are exact on at least one neighborhood, exactness seems clear. $\endgroup$
    – Will Sawin
    Commented Jan 10, 2021 at 22:09

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