7
$\begingroup$

If we have a fibration of smooth projective complex varieties $F\to E\to B$, which is locally trivial in the analytic topology, and the global monodromy is trivial. Then is it true that the Chow motives satisfy $h(E)\cong h(F)\otimes h(B)$? I am pretty sure this should fail, but cannot think of a counterexample.

$\endgroup$
0

3 Answers 3

6
$\begingroup$

This is an expansion of my comment to Victor Petrov's answer.

First, it is possible to reformulate Victor Petrov's answer so as to make the question slightly more precise: we can consider De Rham cohomology $H^\bullet_{DR}$ as a functor on the category of Chow motives. A decomposition as in the question implies that the Leray-Serre spectral sequence for De Rham cohomology degenerates, so that De Rham cohomology of $E$ has a tensor-product decomposition as in the Künneth formula - not just as vector spaces, but as Hodge structures. Making the question slightly more precise, we want to find fibrations $F\to E\to B$ for which the Leray-Serre spectral sequence for De Rham cohomology degenerates and yields a product decomposition as Hodge structures. (I think that the conditions in the question - locally analytically trivial with trivial global monodromy - imply this, but I am not sure about all the higher differentials. If this is not the case, then we may detect counterexamples on De Rham cohomology...)

Second, I think that there can be no counterexample with $F$ a curve (at least in my sharper reformulation above). For $F=\mathbb{P}^1$, local triviality in the analytic topology implies that the fibration is the projectivization of a rank two vector bundle, and then the projective bundle formula shows that the motive is additively split, so the question has a positive answer. (The ring structure on cohomology can also be incorporated into the motivic vs. analytic comparison, because in both worlds it is explicitly given via the Chern polynomial for the rank two vector bundle.)

Now, for $F$ a curve of genus $\geq 1$, there should be no counterexamples because of the Torelli theorem: if the family is such that the Leray-Serre spectral sequence degenerates and gives a product decomposition of Hodge structures, then all the curves have isomorphic Jacobians. The Torelli theorem then implies that the family is isotrivial. From isotriviality and triviality of the global monodromy we should get global triviality of the family. If this latter claim can be made rigorous, it would imply that any family of genus $\geq 1$ curves for which the condition of the question is true is actually trivial and hence does not serve as a counterexample.

More generally - extracting from the above discussion:

  1. No counterexamples should exist whenever $F$ has a Tate motive (like projective bundles, Grassmannian bundles and stuff like that).

  2. No counterexamples should exist whenever $F$ lies in a family for which a suitable period map is injective (like the general type curves above).

  3. The question (with $\mathbb{Q}$-coefficients) should be a question of comparison of moduli spaces of motives vs. moduli spaces of Hodge structures.

Ok, this was for motives with rational coefficients. With finite coefficients, one can play the same game: a decomposition of motives with finite coefficients implies a decomposition of étale cohomology (or equivalently singular cohomology) and so we are trying to find examples of fibrations where the Leray-Serre spectral sequence for étale cohomology has a product decomposition. I am not so sure about this case. At least we could use the comparison results of Suslin-Voevodsky (comparing Suslin homology with finite coefficients to singular homology of the associated complex manifold) to show that the Chow groups with finite coefficients look like a splitting exists.

$\endgroup$
4
$\begingroup$

I know very little about motives, but conjecturally you should expect that $h(E) \cong h(F) \otimes h(B)$.

Your hypotheses imply that $E$ and $F \times B$ have isomorphic cohomology in the category of $\mathbf Q$-Hodge structures. In fact if $E \to B$ is smooth and proper, and $B$ is smooth and connected, then Saito's theory of mixed Hodge modules implies degeneration of the Leray spectral sequence and that the Leray filtration splits in the category of mixed Hodge structures. The additional assumptions of locally trivial (I'm assuming this means in the category of complex analytic spaces) and trivial monodromy imply also that $H^p(B,R^q f_\ast \mathbf Q) \cong H^p(B) \otimes H^q(F)$. (It also implies more directly that Leray spectral sequence degenerates.)

In any case, the conjectural functor (pure/mixed motives) $\to$ (pure/mixed $\mathbf Q$-Hodge structures) is conjectured to be fully faithful. This statement contains in particular the Hodge conjecture. So you should expect $E$ and $F\times B$ to have isomorphic motives, although proving this may be hard.

On a sidenote, do you have any nontrivial examples of such fibrations? I have a hard time imagining one.

$\endgroup$
8
  • $\begingroup$ Using the conservativity conjecture maybe a bit more subtle. It is surely ok if you have a morphism $E\to F\times B$ or the other way - conservativity then tells you that if this morphism induces an isomorphism of Hodge structures, it induces an isomorphism of motives. But you will probably not have such a morphism on the scheme level... $\endgroup$ Commented Jun 30, 2014 at 9:02
  • $\begingroup$ @Matthias Wendt I don't understand your comment. Do you disagree that the realization functor to Hodge structures is expected to be fully faithful? $\endgroup$ Commented Jun 30, 2014 at 9:28
  • $\begingroup$ No, I don't. I just know of the conjecture that a morphism of motives which induces an isomorphism of Hodge structures is an isomorphism. Could you explain how the Hodge conjecture implies that there is a splitting of $F\to E$ on the level of motives? I certainly agree with you that examples as required in the question are unlikely to exist, I am just unsure which conjectures to use to deduce that... $\endgroup$ Commented Jun 30, 2014 at 9:44
  • 1
    $\begingroup$ Suppose X and Y are smooth projective and the Hodge strictures H(X) and H(Y) are isomorphic. This isomorphism is a class in $H(X) \otimes H(Y)^\ast = H(X \times Y)$ which is pure of Tate type, by Hodge conjecture it is class of algebraic cycle and gives morphism of motives. $\endgroup$ Commented Jun 30, 2014 at 10:10
  • 2
    $\begingroup$ It may be helpful (or pedantic) to note that the realisation functor is expected to be fully faithful from homological motives to $\mathbb{Q}$-HS. This question is about Chow motives, for which this is not true: all rational cycles that are homologically trivial are killed by the realisation functor (so it is only conjectured to be full). $\endgroup$
    – jmc
    Commented Jun 30, 2014 at 10:26
1
$\begingroup$

I don't know an explicit counter-example, but in general you have only a spectral sequence similar to Leray-Serre. See Markus Rost's paper http://www.math.uiuc.edu/documenta/vol-01/16.html

$\endgroup$
1
  • $\begingroup$ True, but for Hodge theory the global monodromy being trivial (as required in the question) would imply that the Leray-Serre spectral sequence degenerates. I think that there are two questions here: 1) is it possible to deform smooth projective varieties such that Chow groups change but the deformation is holomorphically trivial? 2) does trivial global monodromy in the analytic world imply trivial global monodromy in the motivic world? Both are interesting questions, but they sound like they are close to conservativity of deRham cohomology on motives... $\endgroup$ Commented Jun 29, 2014 at 14:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .