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I've been learning about p-adic Hodge theory recently (I'm a beginner), and I've been wondering about the following question the past couple of weeks. Sorry for the long setup, it's mainly background; the question is down towards the bottom.

Background: For a field $k$, and a second field $K$ of characteristic 0, we can consider the $K$-linear category $Mot(k)_{K}$ of motives defined by smooth projective varieties over $k$ (pure motives) up to homological or numerical equivalence (let's assume homological equivalence equals numerical equivalence for each Weil cohomology we consider). Then this is a semi-simple abelian category (due to Jannsen), and there are realization functors to various categories of vector spaces extending the functor sending a variety $X$ to its cohomology groups (e.g. its etale cohomology groups $H^i(X,\mathbb{Q}_\ell)$ for $\ell \not= p$ in which case $K = \mathbb{Q}_\ell$).

In the case when $k = \mathbb{C}$ and we take $K = \mathbb{Q}$, we can consider the singular cohomology realization $Mot(\mathbb{C})_\mathbb{Q} \rightarrow {Vect}_{\mathbb{Q}}$ sending $X$ to $H_{sing}^*(X,\mathbb{Q})$. One way of stating the Hodge conjecture is that when we consider the extra structure of the Hodge decomposition on $H^*_{sing}(X,\mathbb{Q}) \otimes \mathbb{C}$ and therefore consider the enriched realization $Mot(\mathbb{C})_{\mathbb{Q}} \rightarrow {Hdge}_{\mathbb{Q}}$ to pure $\mathbb{Q}$-Hodge structures, this functor is fully faithful.

Similarly, if $k$ is a finite field or a number field, we can take $K = \mathbb{Q}_\ell$, we can consider the functor $Mot(k)_{\mathbb{Q}_{\ell}} \rightarrow GalRep(\mathbb{Q}_\ell)$ which sends $X$ to its $\ell$-adic cohomology $H^i(X_{\overline{k}},\mathbb{Q}_\ell)$ as an $\ell$-adic Galois representation. Then the Tate conjecture is that this functor is fully faithful. In the number field case, there are some other conjecturally fully faithful enriched realization functors discussed in Ch. 7 of Andre's book "Une Introduction Aux Motifs".

Motivation: Now in $p$-adic Hodge theory, we deal with varieties over a complete discrete valuation field (for example $k = \mathbb{Q}_p$), and study the extra structures which one obtains on the de Rham cohomology of such a variety from Hodge theory, crystalline cohomology, and etale cohomology. To me it seems that one is trying to endow the de Rham cohomology of a variety $X$ with as much "natural" structure as possible, at least enough to determine the Galois representation $H^i(X_{\overline{k}},\mathbb{Q}_p)$, but possibly one can add more. So I wonder the following:

Let $k$ be a finite extension of $\mathbb{Q}_p$. Is it reasonable to believe that there exist a characteristic-0 field $K$ and a fully faithful functor $Mot(k)_K \rightarrow \mathcal{C}$, where $\mathcal{C}$ is the category of $K$-vector spaces with some "natural" extra structure, for example that considered in $p$-adic Hodge theory? If not, why not? If so, is there a candidate for $\mathcal{C}$?

EDIT: I want to add an additional condition: that when one takes the forgetful functor $\mathcal{C} \rightarrow Vect(K)$, the composition $Mot(k)_K \rightarrow Vect(K)$ is the functor defined by some Weil cohomology (e.g., de Rham cohomology or $p$-adic etale cohomology).

My impression from the example given by user ulrich in the comment to

Should the etale cohomology of a smooth projective variety (over rationals) be semi-simple; why?

is that we CANNOT take $K = \mathbb{Q}_p$ and take the functor sending $X$ to $H^*(X_{\overline{k}},\mathbb{Q}_p)$ , because the the $p$-adic cohomology of $X$ might not be semi-simple, for example when $X$ is an elliptic curve with non-split multiplicative reduction. But perhaps there's another possible $\mathcal{C}$?

EDIT: As wccanard explained in a comment (now deleted), the example above doesn't actually show that the $p$-adic etale cohomology functor isn't fully faithful, since the subobject of $H^1(X,\mathbb{Q}_p)$ which is not split is not "motivic". However, there are other reasons why $p$-adic etale cohomology can't work as explained in wccarnard's answer below.

share|improve this question
Thanks wccanard, your comments were definitely helpful! One thing which I also wondered was whether one could also keep track of the comparison isomorphism between etale and de Rham cohomology via B_{pst}, like in the de Rham-Betti realization functor in Andre's Ch. 7. Potentially keeping track of this isomorphism could yield more information than just looking at D_{pst}, even if it determines de Rham cohomology. (Maybe that fails as well though?) –  Peter Mar 19 '13 at 22:41
Right; I think your idea has far more going for it than any of mine. I think that the general philosophy is that for two varieties with the same reduction, the cohomology groups themselves should be the same, but what distinguishes the varieties is the Hodge filtration. However I'm now a bit confused because for these ell curves with ss reduction all the filtrations seem to be isomorphic. –  user30035 Mar 19 '13 at 22:52
Oh, somehow the way it works is that for two varieties with the same reduction, the crystalline cohomologies are canonically isomorphic, so one can see the different varieties giving different Hodge filtrations. But this doesn't help you here, because $\mathcal{C}$ can't tell the canonical isomorphism from all of the non-canonical ones :-/ –  user30035 Mar 19 '13 at 22:54
Since the category of motives (under your assumptions) is a Tannakian category and any Weil cohomology theory is a fibre functor, it follows tautologically that your question has a positive answer. Of course, this doesn't give you an explicit way of determining the "extra structure" that is needed. –  ulrich Mar 20 '13 at 6:10
Right, I definitely don't want to just take $\mathcal{C}$ to be the essential image of some Weil cohomology functor. I purposely left the question vague because I was curious if there are other possibilities besides $D_{pst}$ and $p$-adic realization, but hopefully it's clear that $\mathcal{C}$ should be the category of $K$-vector spaces with some extra structure like a filtration or a Galois representation or... –  Peter Mar 20 '13 at 6:45
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1 Answer

This is not an answer, it's just a note that some of the more standard examples of functors don't seem to work because they're not fully faithful. This was originally in the comments but it clogged them up so I deleted them and moved them here.

The observation I want to make is simply that if $k=\mathbb{Q}_p$ then I think that none of the following work in general: $K=\mathbb{Q}_\ell$ and the functor is $\ell$-adic etale cohomology; $K=\mathbb{Q}_p$ and the functor is $p$-adic etale cohomology; and finally $K=\mathbb{Q}_p$, the target category is $K$-vector spaces equipped with a filtration, a Frobenius and a monodromy operator in the usual $p$-adic Hodge theory waym and the functor is $D_{pst}$.

The reason none of these work is the following. Fix an elliptic curve over $\mathbb{F}_p$ with $a_p=0$ (assume such a curve exists). Then there are uncountably many lifts of this curve to $\mathbb{Q}_p$, so there are two non-isogenous lifts. The motives attached to the $h^1$ of these curves will, I imagine, have no non-zero maps between them. However the functors described above are all controlled by the special fibre so produce isomorphic linear algebra objects. In the $D_{pst}$ case what is happening is that the Frobenius has trace zero and the $Fil^1$ can't be either of the eigenspaces so it must be another line; however all other lines give isomorphic filtered $\phi$-modules.

The reason this doesn't contradict the Hodge conjecture is that you can take coefficient field equal to $\mathbb{Q}$ in that setting, which is much smaller, and then the filtration (which is defined once you tensor up to $\mathbb{C}$ gives you far more information.

share|improve this answer
[PS the content of the above non-answer was originally comments to the q] –  user30035 Mar 19 '13 at 22:52
I guess instead of taking $D_{pst}$ of the etale cohomology one should just take the crystalline cohomology, which gives an isomorphic object (and then the comments still apply). –  user30035 Mar 19 '13 at 22:55
This might be as good an answer as I can hope for, since there aren't really any other obvious candidates for a category $\mathcal{C}$. I'll see if other people want to chime in though. –  Peter Mar 19 '13 at 22:59
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