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Sep 17, 2018 at 0:24 comment added Dan Petersen If the codimension is $c$, then $H^i(\overline X) \to H^i(X)$ is an isomorphism for $i<2c-1$ and injective for $i=2c-1$ (use the long exact sequence of a pair). This gives cohomological obstructions to the existence of such a compactification; for example, the Hodge structure on $H^i(X)$ must be pure of weight $i$ in low degrees.
Oct 11, 2010 at 6:49 comment added Lars Ah, yes, of course!
Oct 10, 2010 at 18:08 answer added Torsten Ekedahl timeline score: 6
Oct 10, 2010 at 12:52 comment added Torsten Ekedahl Vector bundles extend to vector bundles in codimension $2$ but not necessarily beyond that so your equivalence of categories is OK for surfaces but not for higher dimension.
Oct 10, 2010 at 12:38 comment added Lars Yes, I should have mentioned this, there are stronger restrictions. For example, at least if $\overline{X}$ is smooth, then we must have $\Gamma(X,\mathcal{O}_X)=k$, and the category of vector bundles on is equivalent to the category of vector bundles on $X$.
Oct 10, 2010 at 11:57 comment added Daniel Loughran Just a small remark to get things going, perhaps you knew this already, but the complement of an affine variety is always a divisor. So if you want larger codimension $X$ must not be affine.
Oct 10, 2010 at 10:06 history asked Lars CC BY-SA 2.5