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Let $\mathsf{MM}(\mathbf C)$ be the hypothetical category of mixed motives over the complex numbers, and consider the realization functor $\Phi : \mathsf{MM}( \mathbf C) \to \mathsf{MHS}$ to integral mixed Hodge structures.

Is $\Phi$ expected to be faithful? If not, what kind of information does one expect to lose?

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I don't think the question has much meaning. Even in the pure setting, the answer depends on which equivalence relation you use to define the category of motives. In any case, you should ask the question on the level of triangulated categories. Then a number of candidates for the triangulated category of motivic complexes have been defined, and your question becomes more precise. –  abz Aug 5 '13 at 21:58
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I do not agree with Anon's comment, at least not fully. In the setting of pure motives, the question makes perfect sense, independently on the equivalence relation used, the key word being "expected" : indeed the different equivalence relations are expected to yield the same categories of pure motives. Moreover the answer (in the pure setting) is expected to be yes, by the Hodge conjecture. –  Joël Aug 5 '13 at 22:10
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As for mixed motives, I am much less sure, and perhaps what I am going to say is utterly naive, but the category of mixed motives should be a nice abelian categories, with a realization functor to MHS as the OP says, and it makes perfect sense to ask if such functor is expected be faithful. +1 for the question. –  Joël Aug 5 '13 at 22:15
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The question also seems to presume some kind of integral theory of motives. Certainly, if we are considering rational Hodge structures and pure motives, then, as Joel points out, this comes down to the Hodge conjecture + a Standard conjecture. EDIT: Actually, if we only require faithfulness, then we don't even need the Hodge conjecture, just the standard conjecture on num. eq.=hom. eq. –  Keerthi Madapusi Pera Aug 5 '13 at 22:35
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@Joel: If you take rational equivalence, the answer is no; if you take homological equivalence, the answer is yes. But, as I said, this should be asked for triangulated categories of motivic categories, where at least there are candidates for the motivic category. –  abz Aug 5 '13 at 23:23

1 Answer 1

up vote 8 down vote accepted

Yes, this functor is expected to be faithful, but this has very little to do with Hodge theory. The reason for this is that, if $\Psi$ denotes the forgetful functor from integral mixed Hodge structures to abelian groups, then the composed functor $\Psi\circ\Phi$ simply is the Betti realization functor. As $\Psi$ itself is faithful, it is thus sufficient to prove that the Betti realization functor is faithful. The reason why the Betti realization functor is expected to be faithful is the following. The abelian category $\mathsf{MM}(\mathbf{C})$ is expected to be the heart of the motivic $t$-structure on the triangulated category $\mathsf{DM}_{et}(\mathbf{C})$ of Voevodsky's constructible (or geometric) étale motives (the one obtained from étale sheaves with transfers; the triangulated category of motives obtained from Nisnevich sheaves with transfers (which is known to coincide with $\mathsf{DM}_{et}(\mathbf{C})$ with rational coefficients) is known not to have a motivic $t$-structure with integral coefficients anyway). The faithfulness of the Betti realization functor as above boils down to the following two properties of the triangulated version of the Betti realization functor $$R:\mathsf{DM}_{et}(\mathbf{C})\to\mathsf{D}^b(\mathit{Ab})$$ 1) it must be conservative;

2) it must be $t$-exact.

Note that the functor $R$ is actually well defined and that property 1) is equivalent to its version with rational coefficients (because, for torsion coefficients, the functor $R$ is known to be an equivalence, by the rigidity theorem of Suslin and Voevodsky). Property 2) is required in the very definition of a motivic $t$-structure. This essentially means that, for this problem, the main difficulty is about rational coefficients. This is then very related with the standard conjectures, as explained in this paper of Beilinson: arXiv:1006.1116.

Finally, it might be worth to mention that there is an actual candidate for $\mathsf{MM}(\mathbf{C})$, constructed by Madhav Nori, which, by definition, comes with an exact and faithful functor to $\mathsf{MHS}$.

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Thank you for your detailed answer. I guess I was not really asking the "right" question. –  Dan Petersen Aug 6 '13 at 7:32

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