A Naive Question on Mixed Motives and Mixed Hodge Structures 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)?
 A: 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.


*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.:)
