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As a physicist, I have some naive questions about mixed motives and its mixed Hodge structure (MHS) realization. Any references, comments, answers will be appreciated!

The category of mixed motives over $\mathbb{Q}$ has not been constructed, but anyway let us suppose it exists and denote it by $\mathcal{MM}_{\mathbb{Q}}$, and I want to know some expected properties of it. Every mixed motive $M$ then has a Hodge realisation, denoted by $H(M)$, which is a MHS (an obeject in the abelian category of \mathbb{Q}-MHS), i.e. a functor \begin{equation} H:\mathcal{MM}_{\mathbb{Q}} \rightarrow \mathbb{Q}\,MHS \end{equation} First, suppose $H(M)$ is the direct sum of $S_1$ and $S_2$ in the category $\mathbb{Q}$-MHS, do we expect there exist $M_1$ and $M_2$ in $\mathcal{MM}_{\mathbb{Q}}$ such that $H(M_i)=S_i$ and $M=M_1 \oplus M_2$? If yes, does this property have anything to do with Hodge conjecture?

Second, suppose there is a sequence in $\mathcal{MM}_{\mathbb{Q}}$, \begin{equation} 0\rightarrow M_1 \rightarrow M \rightarrow M_2 \rightarrow 0 \end{equation} which we do not require to be exact. If its Hodge realization is exact in the category $\mathbb{Q}$-MHS, i.e. the following sequence is exact, \begin{equation} 0\rightarrow H(M_1) \rightarrow H(M) \rightarrow H(M_2) \rightarrow 0 \end{equation} Do we expect the sequence upstairs is exact!

Third, Veovodsky has constructed a triangulated category which is candidate for the derived category of the assumed category $\mathcal{MM}_{\mathbb{Q}}$, denote it by $\mathcal{DMM}_{\mathbb{Q}}$, does there exist a functor which looks like a Hodge realization functor? i.e. a functor \begin{equation} \widetilde{H}:\mathcal{DMM}_{\mathbb{Q}} \rightarrow \mathbb{Q}\,MHS \end{equation}

Fourth, if third is true, $\widetilde{H}$ exists, is a similar property like (First) true when we replace $\mathcal{MM}_{\mathbb{Q}}$ by $\mathbb{DMM}$? Similarly, is there a similar property like (second) when we replace $\mathcal{MM}_{\mathbb{Q}}$ by $\mathbb{DMM}$? (need to replace SES by a distinguished triangle in the question)?

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    $\begingroup$ Questions 1 and 2 boil down to the statement that the Hodge realisation functor is full and faithful, respectively. For pure motives with homological equivalence, this is the Hodge conjecture. For mixed motives, it does not follow from the Hodge conjecture, but is sometimes stated as a separate conjecture generalising the Hodge conjecture. See for example Conjecture 3.22 in Mark Levine's chapter of the $K$-theory handbook. $\endgroup$
    – R. van Dobben de Bruyn
    Commented May 30, 2017 at 22:19
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    $\begingroup$ There should be a special tag for questions titled "A silly/naive/stupid question about [insert fancy technical topic]" which are then followed by a perfectly intelligent and well-formed question. $\endgroup$
    – Marty
    Commented May 31, 2017 at 17:02
  • $\begingroup$ @Marty, hahaha! I am doing string theory and know very little about this abstract theories, hence I guess my question is very naive for an expert! $\endgroup$
    – Wenzhe
    Commented May 31, 2017 at 17:16

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I will try to answer.

1) I suspect that the answer is "no", but justifying this should be difficult. The Hodge conjecture gives a positive answer under the assumption that $M$ splits as the direct sum of its weight factors (one may say that $M$ is semi-pure).

2) The answer to this question as well as to its "triangulated version 4.2" should be positive. Indeed, these statements follow from the following (widely believed to be true) conservativity conjecture: the singular realization of a non-zero geometric motif (with rational coefficients) is non-zero. Note here that the exactness of an exact sequence of MHS is equivalent to that of the underlying $\mathbb{Q}$-vector spaces.

  1. Yes, there is a Hodge realization. The first construction was described by Huber (see https://pdfs.semanticscholar.org/2b04/2f81bc16df356e7efb35ac2504ef0aadd5ff.pdf and the erratum to this text); there is also a paper by Lecomte and Wach and by Ivorra on this subject.

4.1. This question seems to be easier that question 1; still I don't know how to answer it.:)

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  • $\begingroup$ For part (1), When $M$ is semi-pure, do you know any reference that implies Hodge conjecture gives a positive answer? (Sorry about this question) I haven't seen the definition of what does it mean that $M$ splits as the direct sum of its weight factors! Could you describe it briefly or any reference? Thank you very much for your answer! $\endgroup$
    – Wenzhe
    Commented May 30, 2017 at 22:21
  • $\begingroup$ Actually, semi-purity should be equivalent to the semi-simplicity of $M$.:) Note that simple mixed motives should be pure. i.e., of a single weight (this should be equivalent to their singular realization being pure). To obtain the statement in question one should apply the Krull-Schmidt theorem for semi-simple mixed motives (that follows from the finite dimensionality of endomorphism rings of pure motives); some work appears to be necessary to pass from cycles over $\mathbb{C}$ provided by the Hodge conjecture to that over $\mathbb{Q}$. $\endgroup$
    – Mikhail Bondarko
    Commented May 30, 2017 at 22:44
  • $\begingroup$ Thank you. Are there any conjectures about whether the Hodge realisation functor of Voevodsky's category is full ( faithful)? $\endgroup$
    – Wenzhe
    Commented May 31, 2017 at 12:53
  • $\begingroup$ If you want to make the realization functor to be "closer to a full embedding" then I suggest you to take the triangulated category of graded polarizable Hodge structures for the target. However, I suspect that neither of these properties is valid even after this modification. $\endgroup$
    – Mikhail Bondarko
    Commented May 31, 2017 at 13:13

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