1
$\begingroup$

Let $f : X \to Y$ be a flat proper morphism of complex varieties whose fibers are normal varieties. Is it true that $\mathrm{dim}_{\mathbb{Q}} H^i(X_t, \mathbb{Q})$ is constant?

For non-normal fibers, the cohomology rank can jump down. Can it also jump up?

If not, what is a counterexample? Is there any sort of semicontinuity (upper or lower) that does hold? I am particularly interested in the case $i = 1$.

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ The cohomology can jump down for a family of quadric hypersurfaces that acquires an ordinary double point. This is the model for “vanishing cycles”. This case is analyzed in detail in SGA 7 (as well as other sources, going back to “Analysis Situs” by Lefschetz). $\endgroup$
    – Jason Starr
    Commented Sep 16, 2023 at 21:19
  • 1
    $\begingroup$ Or even more simply, consider a family of smooth quadrics in $\mathbb{P^3}$ degenerating to a cone. The second Betti number drops from $2$ to $1$. However, a long time I convinced myself the $i=1$ case is OK for a $1$-dim family of normal projective varieties. I can try to reconstruct the argument when I have time. $\endgroup$
    – Donu Arapura
    Commented Sep 16, 2023 at 21:41
  • $\begingroup$ @DonuArapura thank you. If I understand correctly, vanishing cycles theory should give an exact triangle $\underline{\mathbb{Q}}|_{X_0} \to \psi \underline{\mathbb{Q}} \to \varphi \underline{\mathbb{Q}} \to +1$ where $\psi$ is the nearby cycles and $\varphi$ the vanishing cycles. If the singularities $X_0$ are isolated then the Milnor fiber being $(n-1)$-connected implies that $\varphi \underline{\mathbb{Q}}$ is $(n-1)$-truncated so the LES shows that the map $H^i(X_0, \mathbb{Q}) \to H^i(X_s, \mathbb{Q})$ is an isomorphism for $i \le n - 1$ and injective for $i = n$. $\endgroup$
    – Ben C
    Commented Sep 16, 2023 at 22:51
  • $\begingroup$ This should say no jumping for $i \le n-1$ and only jumping down for $i = n$. Does this sound right? If so, do you know if this is optimal? Can I get jumping up for $i = n + 1$? $\endgroup$
    – Ben C
    Commented Sep 16, 2023 at 22:55
  • $\begingroup$ @BenC I'm not sure if the Milnor fibre is known to be $(n-1)$ connected in general, but it is for hypersurfaces. So your argument seems fine in this case. $\endgroup$
    – Donu Arapura
    Commented Sep 17, 2023 at 13:16

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .