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Let $U$ be a smooth quasi-projective variety. Does there always exist a smooth compactification of $U$? If not always when can we have smooth compactification? In particular, suppose $X$ is a singular projective variety and $U$ is the smooth locus. The question is does there always exist a smooth projective variety $Y$ containing $U$? If we add the condition that $U$ is open in $Y$ is $Y$ uniquely determined upto isomorphism?

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You should search for "resolution of singularities" to answer the question about existence. About uniqueness: in dimension at least 2, no, because of blow-ups. –  Artie Prendergast-Smith Mar 11 '13 at 21:14

2 Answers 2

up vote 7 down vote accepted

The main idea is this: You can always find an $X$ such that $X\supseteq U$ and $X\setminus U\supseteq \Sigma := \mathrm{Sing} X$. Then in characteristic zero apply Hironaka's resolution theorem, which says that there exists a resolution of singularities $\pi:Y\to X$ such that $\pi$ is an isomorphism over $X\setminus \Sigma\supseteq U$. In particular, $\pi^{-1}: U\hookrightarrow Y$ gives an embedding.

In positive characteristic resolution is not known in general and similarly this embedding result is not known either (although I am not saying that knowing this would prove the existence of resolutions).

There is actually a newer, expanded version of Kollár's paper in book form: Lectures on Resolution of Singularities.

And of course it is not unique as long as $U$ itself is not projective, since as you can always blow-up $Y$ outside of $U$.

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A remark: though $U$ doesnotdetermine $Y$, for any cohomology theory that factorizes through Voevodsky′s motives the image $H(Y)\to H(U)$ is canonical and functorial (this is the zeroth level of the weight filtration). One can also look at weight complexes. In characteristic $p$ one can prove a somewhat similar result for any cohomology whose target is a $Z[1/p]$-linear category. –  Mikhail Bondarko Mar 12 '13 at 8:59
    
@Mikhail: I'm sorry, but I don't see your point. Could you explain? –  Sándor Kovács Mar 12 '13 at 16:03
    
So, smooth compactifications are certainly not unique; yet this non-uniqueness can be controlled to a certain extent. –  Mikhail Bondarko Mar 12 '13 at 17:35
    
@Mikhail, I'm sorry, but I still don't understand how this is more control. Perhaps my problem is what Artie mentions that I can't read half of your comment... –  Sándor Kovács Mar 12 '13 at 21:12
    
....or mine, now.... –  Sándor Kovács Mar 12 '13 at 21:13

In characteristic 0, yes, yes, no. Check out Kollar's paper:

http://arxiv.org/abs/math/0508332

In characteristic p, unknown.

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