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Let $B$ be a smooth quasi-projective variety over a field of characteristic zero. Let $\pi\colon \mathcal{X} \rightarrow B$ be a smooth and projective morphism with geometrically integral fibres. Let $Z$ be a cycle on $\mathcal{X}$, i.e. a $\mathbb{Z}$-linear combination of closed integral subvarieties of $\mathcal{X}$. Suppose in addition that $Z$ is flat over $B$, i.e. each component of the support of $Z$ is flat over $B$.

Question: is the locus $B_{\text{trivial}} = \{ b\in B \mid \text{ the cycle }Z_b \text{ of }\mathcal{X}_b \text{ is rationally trivial}\}$ a closed subset of $B$?

(Recall that a cycle on a variety $X$ is called rationally trivial if it lies in the subgroup of cycles of the form $W_0-W_\infty$ where $W$ is a cycle $X\times \mathbb{P}^1$ flat over $\mathbb{P}^1$.)

I can show that $B_{\text{trivial}}$ is closed under specialization, since by smoothness it suffices to consider the case of a DVR and in that case there is a well-defined specialization morphism on the level of Chow groups, see the last chapter of Fulton's intersection theory book. So it suffices to prove that $B_{\text{trivial}}$ is a constructible set. Any reference/help would be appreciated!

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    $\begingroup$ No that is not true. Let $X$ be a very general (projective) K3 surface, let $B$ be $X$, let $\mathcal{X}$ be $X\times X$ with its first projection, and let $Z_1$ be the diagonal copy of $X$ in $X\times X$. Let $Z_0$ be a cross-section $X\times\{x_0\}$, where $x_0$ is a rational point contained in a rational curve on $X$. Let $Z$ be $1Z_1 - 1Z_0$. The locus in $B$ where this cycle is rationally trivial is the (countable) union of all rational curves in $X$. $\endgroup$ Commented Dec 1, 2023 at 11:30
  • $\begingroup$ Thanks a lot for this great example! $\endgroup$
    – Jef
    Commented Dec 4, 2023 at 20:27
  • $\begingroup$ @JasonStarr if you have any thoughts on my followup question I would be very grateful if you would like to share them: mathoverflow.net/questions/462776/… $\endgroup$
    – Jef
    Commented Jan 24 at 17:34

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