## Hodge numbers of compactifications

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Let $X$ be a smooth complex quasi-projective variety. We can find good compactification: a smooth proper variety $\bar{X}$ such that ${\bar X} \setminus X$ is a divisor with normal crossing. The variety $\bar{X}$ is then stratified by the singulartities of the divisor. And one can compute the mixed Hodge structure on $H^{\bullet}(X)$ in terms of the pure Hodge structures $H^{\bullet}(S_\alpha)$ of the smooth closed strata using a spectral sequence.

Let's say a variety $Y$ is Hodge-Tate if $h^{p,q}(Y) = 0$ for $p\neq q$.

If all the closed strata of $\bar{X}$ are Hodge-Tate then $X$ is Hodge-Tate.

Question: Let $X$ be a smooth complex quasi-projective variety. Assume $X$ is Hodge-Tate.

1. Can one find a good compactification $\bar{X}$ with Hodge-Tate strata?
2. Are all good compactifications of $X$ of this type? (Edit: Answer is no, see Torsten's elementary example).
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## 2 Answers

I asked the question to Claire Voisin and she immediatly gave me the following counter example:

Consider Fermat's cubic of dimension 3 $X$. There are 5 cones on elliptic curves $E_i$ inside $X$ and one can verify that $Y=X\setminus U_iE_i$ is Hodge-Tate. But $Y$ doesn't admit a Hodge-Tate compactification as that would imply that $X$ is birational to a Tate variety and this would contradict Clemens-Griffiths' theorem saying that the intermediate jacobian of $X$ is not a direct sum of jacobians of curves as a polarized variety.

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Let $E$ and $F$ be two elliptic curves and let the involution $\sigma$ act on $E\times F$ by $\sigma(e,f)=(-e,f+\alpha)$, $\alpha$ is an element of order two of $F$. Finally let $\overline{X}=(E\times F)/\sigma$ (this is a so called hyperelliptic surface). We have an inclusion $F':=0\times F/\langle\alpha\rangle\subseteq S$ and put $X:=\overline{X}\setminus F'$. Then $X$ is Hodge-Tate but all other good compactifications of $X$ are obtained by blowing ups and downs of $\overline{X}$ which means that you can never get rid of $F'$ (alternatively any good compactification $X'$ has $H^1(X')=H^1(X)$ and you need something non-Hodge-Tate at the boundary to kill that off).

Addendum: This example is all wrong it took care of $H^3(X)$ but not (the more interesting) $H^1(X)$. At the moment I am less sure than I was that the answer to 1) is no.

As for 2) you can just look at $\mathbb A^3\subseteq\mathbb P^3$ which is a good Hodge-Tate compactification with $\mathbb P^2$ as divisor at infinity and then blow up something non-Hodge-Tate in $\mathbb P^2$. This gives a good compactification with two components one of which (the exceptional divisor for the blowing up) is non-Hodge-Tate (as is the intersection of these two divisors).

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