Is there a "universal" cohomology theory for varieties over p-adic fields? Let $K$ be a $p$-adic field, $X$ a smooth proper algebraic variety over $K$, and $0 \le i \le 2 \dim X$. For a prime $\ell \ne p$ one can consider the $\ell$-adic cohomology $H^i(\overline{X}, \mathbb{Q}_\ell)$, and massage this in the usual way (via Grothendieck's abstract monodromy theorem) to get a Weil–Deligne representation of $K$ with coefficients in $\mathbb{Q}_\ell$. For $p$-adic étale cohomology, there is a more complicated construction starting from $H^i(\overline{X}, \mathbb{Q}_p)$ going via Fontaine's $D_{\text{pst}}$ functor. I gather it is conjectured that all of these Weil–Deligne representations are in fact definable over $\mathbb{Q}$, and they should have the same character and thus be isomorphic up to semisimplification; and this is known in some cases.
Is it expected that there should be a "universal" cohomology theory taking values in the category of Weil–Deligne representations over $\mathbb{Q}$, from which all of the above can be obtained by extending scalars? If so, have there been any attempts to construct such a cohomology theory?
 A: The following is an adaptation of an argument of Serre, explaining why there shouldn't be a universal cohomology theory for $\mathbb{F}_p$ varieties taking values in $\mathbb{Q}$ vector spaces. Be warned that this is not my field, so I may be missing something basic.
Let $p$ be a prime which is $3 \bmod 4$, let $X$ be the elliptic curve $y^2 = x^3-x$ over $\mathbb{Q}_p$ and let $Y$ be the base change of $X$ to $\mathbb{Q}_p(i)$. In any of the cohomology theories you describe, $H^1(X)$ and $H^1(Y)$ are two dimensional. In etale cohomology, $H^1(X) \cong H^1(Y)$; in $p$-adic cohomologies I believe you usually have $H^1(X) \otimes_{\mathbb{Q}_p} \mathbb{Q}_p(i) \cong H^1(Y)$. I assume in your hypothetical $\mathbb{Q}$-valued theory, you would have $H^1(X) \cong H^1(Y)$.
Let $F$ be the Frobenius automorphism of $H^1(X)$; let $J$ be the automorphism of $H^1(Y)$ induced by $(x,y) \mapsto (-x, iy)$. Identifying $H^1(X)$ and $H^1(Y)$, these maps should obey the relations
$$F^2 = -p \quad J^2 = -1 \quad FJ=-JF$$
These equations are not solvable in $2 \times 2$ matrices over $\mathbb{Q}$ (or even over $\mathbb{R}$).
So any theory would have to be "unnatural" enough that this is not an obstacle.
The category of motives is designed to be the recipient of a universal cohomology theory. It gets around this issue by being $\mathbb{Q}$-linear, meaning that $\mathrm{Hom}(U,V)$ is a $\mathbb{Q}$-vector space for any motives $U$ and $V$, but not having a natural functor to $\mathbb{Q}$-vector spaces, so the motives themselves cannot be thought of as $\mathbb{Q}$-vector spaces.
I don't know if there is any way in which motives over $p$-adic fields are better than motives over general fields.
A: I'm seeing this old question just now, and simply wanted to remark that the situation may be slightly better.
Namely, enlarging the category of $\mathbb Q$-vector spaces into the larger semisimple $\mathbb Q$-linear tensor category $\operatorname{Rep}(\mathrm{Kt}_{\mathbb Q})$ defined by Kottwitz (see for example the article Representations of the Kottwitz gerbes of Iakovenko for a definition), one expects that there is a (canonical) cohomology theory valued in Weil–Deligne representations with coefficients in $\operatorname{Rep}(\mathrm{Kt}_{\mathbb Q})$. This should recover the other constructions via natural functors to $\operatorname{Rep}(\mathrm{Kt}_{\mathbb Q_\ell})$ for all primes $\ell$. This is a variant of Conjecture 9.5 in my 2018 ICM paper. (Warning: In that paper, I write $\mathrm{Kt}_{\mathbb Q}$ for what I denote $\operatorname{Rep}(\mathrm{Kt}_{\mathbb Q})$ in this answer.)
While the definition of $\operatorname{Rep}(\mathrm{Kt}_{\mathbb Q})$ is as representations of a gerbe $\mathrm{Kt}_{\mathbb Q}$ that is more-or-less explicitly constructed by class field theory — quite similar to the global Weil group —, we still lack a direct linear-algebraic description of $\operatorname{Rep}(\mathrm{Kt}_{\mathbb Q})$, unfortunately.
