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Let's say that $X$ is an integral scheme of finite type over a field and $Y \subset X$ is a closed subscheme. Given a vector bundle $E$ on $Y$, is $E$ the restriction to $Y$ of a vector bundle on a neighborhood $U$ of $Y$ in $X$?

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Isn't this exactly the Leff(X,Y) condition? This is true if $Y$ is a complete intersection of dimension at least 2 in a smooth projective variety $X$. Given $E$ on $Y$, you take $\hat{E}$ on the completion $\hat{X}$ of $X$ along $Y$, and Leff(X,Y) allows you to lift it to a bundle $E'$ in a neighborhood of $Y$ in $X$. –  Parsa Apr 8 '12 at 17:14
    
What is Leff(X,Y)? –  Ian Shipman Apr 8 '12 at 17:24
    
Ok, I found it. Even if the pair (X,Y) does not satisfy Leff, is there a way to understand when E on Y extends to a bundle on the formal completion? Some kind of obstruction perhaps to extending to various infinitesimal thickenings of Y... –  Ian Shipman Apr 8 '12 at 17:29

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No, this is not even true for line bundles.

For an example, let $X = \mathbb{P}^2$ and $Y$ a smooth curve in $X$ of genus $>0$ (over an algebraically closed field). Since $X$ is smooth, any line bundle on an open set $U$ extends to a line bundle on $X$ so the map $\operatorname{Pic}(X) \to \operatorname{Pic}(U)$ is surjective. Since $\operatorname{Pic}(X) \cong \mathbb{Z}$, it follows that the image of $\operatorname{Pic}(U)$ in $\operatorname{Pic}(Y)$ is of rank $1$ and is independent of $U \supset Y$. Since $\operatorname{Pic}(Y)$ is not even finitely generated we see that there exist (many) line bundles on $Y$ which do not extend to any open $U \supset Y$.

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