Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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).
share|improve this question

2 Answers 2

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.

share|improve this answer

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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.