# Variety without a compactification whose complement is smooth

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?

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 $$4$$.
• @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. Commented Jul 22 at 23:55