Let $X$ be a smooth variety over an algebraically closed field $k$. If it makes things easier, $X$ may be assumed to be quasi-projective. By Nagata (or quasi-projectivity) there exists a proper variety $\bar{X}$ which contains $X$ as a dense open subvariety, and by the smoothness of $X$ we may assume $\bar{X}$ to be normal. Are there any criteria/theorems which give information about the codimensio of the $\overline{X}\setminus X$?

The same question can be asked if we assume resolution of singularities, such that we may assume $\overline{X}$ to be smooth. Under which conditions can a smooth compactification $\overline{X}$ be found such that $X$ has complement of codimension $>1$ in $\overline{X}$?

Finally, to rephrase, how can one detect whether a given smooth variety $X$ arises by removing a codimemsion $>1$ closed subvariety from some proper variety?

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    $\begingroup$ 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. $\endgroup$ Oct 10, 2010 at 11:57
  • $\begingroup$ 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 $X$. $\endgroup$
    – Lars
    Oct 10, 2010 at 12:38
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    $\begingroup$ 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. $\endgroup$ Oct 10, 2010 at 12:52
  • $\begingroup$ Ah, yes, of course! $\endgroup$
    – Lars
    Oct 11, 2010 at 6:49
  • $\begingroup$ 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. $\endgroup$ Sep 17, 2018 at 0:24

1 Answer 1


I don't think you can hope for any kind of useful conditions, here are just some comments.

Even if $X$ is smooth, whether or not we demand the compactification matters. Let for instance $X'$ be a smooth projective surface containing a $(-2)$-curve $C$, i.e., $C$ is isomorphic to $\mathbb P^1$ with self-intersection $-2$. Put $X:=X'u C$. Then there is no smooth compactification of $X$ with complement of codimension $³2$ but there is a normal one. In fact $C$ can be contracted to a point on a normal surface but that point is singular with a non-trivial local fundamental group. This fundamental group is an invariant of $X$ at infinity and would be trivial if we could add a smooth point instead.

I think (haven't checked) that there is also for surfaces a distinction between proper and projective (non-smooth) compactifications. If one does the usual thing of blowing up $10$ general points on a smooth cubic, then the complement of the strict transform of the cubic has a proper compactification by adding just one point but I don't think it has a projective such compactification.

If $X$ has a projective normal compactification $\overline X$ then there are some necessary conditions such as the space of global functions being finite dimensional. In fact for a line bundle on $\overline X$, its space of section are the same on $X$ as on $\overline X$ and hence using an ample line bundle on $\overline X$ we see that $\overline X$ is the $\mathrm Proj$ of the algebra of sections of powers of some line bundle on $X$. If we furthermore assume that $\overline X$ is smooth, then any line bundle on $X$ extends to $\overline X$ and will in particular have a finite dimensional space of sections.

Hence a reasonable question is if the condition that all line bundle have finite dimensional spaces of sections is enough to get a (smooth?) compactification with complement of codimension $>1$. I think that for surfaces the answer may be positive but feel dubious about higher dimensions.

  • $\begingroup$ Thanks for the answer! I like your example. Do you know if the argument with the local fundamental group also works in positive characteristic? $\endgroup$
    – Lars
    Oct 11, 2010 at 13:45
  • $\begingroup$ In odd characteristics it does. $\endgroup$ Oct 12, 2010 at 4:25

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