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Let $X$ be a smooth projective scheme over a number field $k$, and $V_{p}$ (resp. $V_{\text{dR}}, V_{\text{B}}$) a sub-space of $H_{et,p}^{\ast}(X)$ (resp. $H^{\ast}_{\text{dR}}(X), H^{\ast}_{\text{B} } (X)$) such that $V_{p}$, $V_{\text{dR}}$ and $V_{\text{B}}$ are isomorphic up to tensor and comparison isomorphism. It is also assumed that they are endowed with Galois action (resp. filtration, Hodge structure) which correspond by comparison to isomorphisms.

I would like to consider rational equivalence over cycles, but if you have an answer to the following question for another relation feel free to consider that relation instead.

Question Do we know a condition (necessary? sufficient?) on $V_{p}$, $V_{\text{dR}}$ and $V_{\text{B}}$ such that they are the realization of a pure motive? When they are realisation of a motive $V$, what are its relations with $H^{\ast}(X)$? Does anyone have a reference where this kind of "submotives" is studied?

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    $\begingroup$ Probably, the standard conjectures predict that any Galois submodule of V_p lifts to a pure motif (and then you can envoke some more conjectures to obtain a Chow motif:)). On the other hand, you need the Standard Conjecture C to prove that the whole H^k(X) comes from a pure motif. $\endgroup$
    – Mikhail Bondarko
    Commented Dec 8, 2022 at 20:06
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    $\begingroup$ Yes the $H^{k}$ was a mistake, I'm focusing on cohomology concentrated in one degree for some times so I forgot that projectors are not known to be cycles :) I just make the changes. For the other part of your comment however, I really don't see which part of the standard conjectures are telling us this.. could you be a little bit more precise please? $\endgroup$
    – Marsault Chabat
    Commented Dec 8, 2022 at 20:15
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    $\begingroup$ Actually, the standard conjectures imply that a subspace of V_p lifts to a motif if and only if it is its direct summand (as a Galois representation). Possibly, this assumption is not vacuous (since the whole V_p is not necessarily semi-simple); I don't know. $\endgroup$
    – Mikhail Bondarko
    Commented Dec 8, 2022 at 20:32
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    $\begingroup$ And if you have a subobject in the mixed realization of $X$ (as defined by A. Huber; probably she followed Jannsen and Deligne) then it is a direct summand since one can apply a simple polarization argument. Hence any subobject of this sort should lift to a pure motif. $\endgroup$
    – Mikhail Bondarko
    Commented Dec 9, 2022 at 5:24

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