6
$\begingroup$

Let $X$ be a smooth, separated complex algebraic variety. By Hironaka, there exists a compactification $j : X \to \bar{X}$ of $X$ so that $\bar{X} \setminus X$ is a simple normal crossings divisor.

Is there some example of a variety $X$ where you cannot choose the compactification such that $\bar{X} \setminus X$ is smooth?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
14
$\begingroup$

Presumably, you want $\bar X$ to be smooth as well. Then there are many examples. Here is a simple one. Let $\bar Y$ be smooth projective curve of positive genus. Now remove at least two points to get $Y$. Then

$X= Y\times Y$ does not have a smooth compactification with a smooth complement.

The proof requires some basic facts from mixed Hodge theory [Deligne, Théorie de Hodge II]. If $X$ possessed such a compactifcation, then the MHS on $H^2(X)$ would have at most weights $2$ and $3$. But by Künneth, $H^2(X)$ has weights $2,3$ and $4$.

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ Thank you! Actually, a question about history: does happen to know if there any counterexample that Hironaka would have known at the time he proved his theorem? $\endgroup$
    – Michael Barz
    Commented Jul 22 at 21:50
  • 3
    $\begingroup$ This I don't know, but it seems possible. There are other examples which checked using different methods which certainly would have been known at the time. $\endgroup$
    – Donu Arapura
    Commented Jul 22 at 22:03
  • 3
    $\begingroup$ @MichaelBarz Surely the following argument would have been more or less obvious to Hironaka: Let $Y$ be a minimal surface of Kodaira dimension $0$. Then $Y$ is the unique minimal surface birational to $Y$, so any smooth proper surface birational to $Y$ is obtained by blowing up points on $Y$. Thus if we take $X$ to be $Y$ minus any curve with a node, every smooth compactification is obtained by blowing up $Y$ and therefore the complement has at least as many nodes. $\endgroup$
    – Will Sawin
    Commented Jul 22 at 23:55
  • 1
    $\begingroup$ (Hironaka was spending a lot of time thinking about much more complicated questions about what happens when you blow up certain kinds of singularities, of course.) $\endgroup$
    – Will Sawin
    Commented Jul 22 at 23:56
  • $\begingroup$ @WillSawin Ah, this is a very nice construction -- thank you!! $\endgroup$
    – Michael Barz
    Commented Jul 23 at 0:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .