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Let $X$ be a projective variety over $\mathbb C$ and $T$ an irreducible projective $\mathbb C$-scheme. Let $a,b$ be closed points of $T$.

Suppose we have a flat family $Z\to X\times T\to T$ such that its fibres $A,B$ over $a,b$ are subvarieties of $X$. Is it necessarily true that $A,B$ are rationally equivalent? If not, can we at least make sure that $A,B$ have the same cohomology class (perhaps with coefficients in $\mathbb Q$ or $\mathbb C$)?

I believe this is true if $T$ is linearly connected (in the sense of [1]). It also makes sense since supposedly $A,B$ "deforms" to each other. But is it actually true in general?

[1] Hartshorne, R. Connectedness of the Hilbert scheme. Publications Mathématiques de L’Institut des Hautes Scientifiques 29, 7-48 (1966). https://doi.org/10.1007/BF02684803

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    $\begingroup$ No. All you can say is that $A$ and $B$ are algebraically equivalent — but not rationally. For instance, 2 lines in a (smooth) cubic threefold fit in a flat family as you describe, but they are not rationally equivalent. $\endgroup$
    – abx
    Commented Aug 1, 2022 at 14:53
  • $\begingroup$ @abx Ah yes. I suppose I also could've taken the family of points on an elliptic curve parameterised by itself. How about cohomology class, though? $\endgroup$
    – BAI
    Commented Aug 1, 2022 at 15:01
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    $\begingroup$ The cohomology class is indeed constant. A heuristic reason is that it varies continuously with $t\in T$, but it lives in the discrete space $H^*(X,\Bbb{Z})$. $\endgroup$
    – abx
    Commented Aug 1, 2022 at 16:33
  • $\begingroup$ @abx Thanks so much. Is it easy to turn this heuristic into an actual proof, or is there more machinery needed? $\endgroup$
    – BAI
    Commented Aug 1, 2022 at 16:43
  • $\begingroup$ Well, to define the cohomology class you need already a certain amount of machinery. I don't think much more is needed to ptove the invariance under deformation. $\endgroup$
    – abx
    Commented Aug 1, 2022 at 19:31

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