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.
$\def\R{\mathbf{R}}$
$\def\Z{\mathbf{Z}}$
$\def\Q{\mathbf{Q}}$
$\def\Qbar{\overline{\Q}}$
$\def\F{\mathbf{F}}$
$\def\GL{\mathrm{GL}}$
$\def\Gal{\mathrm{Gal}}$
Here are some thoughts on your question from a number theorist. First of all, I think the only sensible
answer to your question is that your naive guess is the right one, namely, that a tamely ramified Weil-Deligne
representation is one for which the wild inertia group $I_{>0}$ acts trivially. But let me take the opportunity
to give a more expansive answer which touches on a few other concepts. This discussion will all be at the cartoon
level, but that may still be useful to some. Let's fix ideas and always talk about the
case of $G = \GL_n(\Q_p)$. Let $\rho$ be a Weil-Deligne representation and let $\pi$ be the corresponding
representation associated to $\rho$ by local Langlands. The following three things are then equivalent:
- The Weil-Deligne representation is trivial on the entire inertia group.
- $\pi$ has an Iwahori fixed vector.
- The corresponding $l$-adic representations for $l \ne p$ are unipotent on inertia.
With that out of the way:
I would certainly only ever use "tamely-ramified" to refer to the weaker condition that the image of wild inertia is trivial, exactly as you hypothesized.
A really basic example to keep in mind is the case of characters. We have $\GL_1(\Q_p) = \Q^{\times}_p$.
Let $\chi$ be a (continuous) character of $\Q^{\times}_p$. Let's specialize the conditions above in this case. We get
- The Weil-Deligne representation is trivial on the entire inertia group.
- $\pi$ has a fixed vector under $\Z^{\times}_p \subset \Q^{\times}_p$.
- The corresponding $l$-adic representations for $l \ne p$ are trivial (equivalently in dimension one unipotent) on inertia.
The nice thing about the case $n = 1$ is that the local Langlands correspondence is very transparent, namely, it is given by local class field theory. To recall (briefly) what class field theory says (in one formulation), it gives a canonical map
$$\Q^{\times}_p \hookrightarrow \Gal(\Qbar_p/\Q_p)^{\mathrm{ab}}$$
with dense image. Indeed, the image is precisely the image of the Weil group, and the image of $\Z^{\times}_p$ is the image of the inertial
part of the Weil group. It follows that the wild inertia group maps to the pro-$p$ part of $\Z^{\times}_p$ which is $1 + p \Z_p$.
A (Galois) character of $\Gal(\Qbar_p/\Q_p)$ canonically gives (by restriction) a character of $\Q^{\times}_p$.
Certainly one would want to allow a "tamely-ramified" Galois character to be tamely ramified, so (in dimension one) we get the following equivalences:
- The Weil-Deligne representation is trivial on $1 + p \Z_p \subset \Q^{\times}_p$,
- $\pi$ has a fixed vector under $1 + p \Z_p$.
- The corresponding $l$-adic representations are tamely ramified on inertia.
I honestly only found very few papers in the literature in which "tamely ramified" was implied to have the meaning (1), (2), or (3). I think they were just in error.
Here are a few speculations on how they could get confused. First of all, the maximal tamely ramified extension of a local field has a particularly simple Galois group.
Namely, it is given by (the profinite completion of) the group
$$\Gamma: \langle \tau, \sigma | \sigma \tau \sigma^{-1} = \tau^q \rangle,$$
where $\sigma$ is Frobenius, $\tau$ is a (pro-finite) generator of tame inertia, and $q$ is the order of the residue field (So $q = p$ for $\Q_p$).
In characteristics $l \ne p$, unipotent implies pro-$l$ implies tame, and thus unipotent implies tamely ramified.
Let us now compare condition 3 above with a new condition 4.
The corresponding $l$-adic representations for $l \ne p$ are unipotent on inertia, that is, they are representations of $\Gal(\Qbar_p/\Q_p)$ which factor through $\Gamma$ and which therefore give rise to matrices $T$ and $F$ satisfying $FTF^{-1} = T^q$, and where $T$ is unipotent.
The corresponding $l$-adic representations for $l \ne p$ are tamely ramified, that is, they are representations of $\Gal(\Qbar_p/\Q_p)$ which factor through $\Gamma$ and which give rise to
matrices $T$ and $F$ satisfying $FTF^{-1} = T^q$.
One reason these look quite similar is that the condition that $T$ is conjugate to $T^q$ almost looks as though it should imply that $T$ is unipotent, or equivalently that the (generalized) eigenvalues of $T$ are all $1$. But this is not quite true, it doesn't force the eigenvalues to satisfy $\lambda = 1$ but only $\lambda^{q-1}= 1$ and so
only $q-1$th roots of unity. (Correction: as Will Sawin points out, this should be $\lambda^{q^k - 1} = 1$ for some $k \le n$. Indeed, in the supercuspidal case you can get primitive $(q^n-1)$th roots of unity.) So 3 and 4 are quite similar, but 4 gives a more relaxed condition
and is the "correct" definition of tamely ramified.
So now, let us return to your next question (slightly modified):
Why should [...] translate into having an Iwahori fixed vector? What do tamely ramified representations correspond do?
OK, so answering "why" to questions in Langlands is always tricky. (What is Langlands? A vast generalization of class field theory. What is class field theory? A vast generalization of quadratic reciprocity. What is quadratic reciprocity? NOBODY KNOWS.) But let me give a little attempt to answer your question by a somewhat circuitous route. I want to start by talking about conductors. (I'm not actually going to say very much interesting about conductors, but I'm going to start out this way.) If you are a graduate student in number theory, sooner or later you are going to read
about the modularity of elliptic curves. And to understand the statement, there is some mysterious invariant $N$ associated to an elliptic curve $E/\Q$.
At first glance, it does something reasonable to measure the bad reduction of $E$: if $p |N$, then $E$ (or a suitably good model) has bad reduction at $p$.
And there's some sense that the power of $p$ dividing $N$ measures "how bad" the reduction really is in some controllable way for $p > 3$. And then for $p = 3$ or $p = 2$ everything goes crazy, but it's OK because some guy called Tate came up with an algorithm and $\texttt{gp}$ can compute it for you. But if you try to think about it, it makes bugger all sense. Then you have to believe the same $N$ turns up as the level of the modular form, and this is even more confusing.
In order to even begin to understand it, it's really key to once again step back and think about the case of $\GL(1)$.
In this case, we have a character $\chi$ which we can think of as a character of $\mathbf{Q}^{\times}_p$.
And what class field theory tells us is that the ramification behavior of $\chi$ is all wrapped up in the restriction of $\chi$ to $\Z^{\times}_p$.
The "automorphic" and "Galois" sides are almost literally the same here, but let me still try to distinguish them. We can now define the conductor as follows:
The automorphic conductor of $\chi$ is the smallest $p^n$ such that $\chi$ has an invariant under the subgroup $1 + p^n \Z_p$.
The Galois conductor of $\chi$ is the smallest $p^n$ such that $\chi$ is trivial on $1 + p^n \Z_p$.
Of course these are clearly the same. But even though I haven't really said it here, what is important in class field theory is that the filtration of $\Z^{\times}_p$
by $1+p^n \Z_p$ is intimately related to the inertia filtrations on the Galois groups. The special case that $1+p \Z_p$ corresponds to $I_{>0}$ is clear,
but the others are more subtle. So here the "equality" of conductors coming from "local Langlands for $\GL(1)$" is expressing something rather
deep in class field theory. To give at least one concrete example, there is Hasse's Führerdiskriminantenproduktformel which
expresses the discriminant of the corresponding
cyclic fixed field in terms of the conductors of the non-trivial powers of $\chi$. The study of relations of this flavour were actually key to Artin's formulation
of his various reciprocity laws and recognizing the correct way to define $L$-functions. I think Noah Snyder wrote an interesting undergraduate thesis
about this once. Anyway, I'm drifting slightly from my main topic. To get back on point, the key thing about this example which generalizes is that the following
two things are intimately related:
- The restriction of the Weil--Deligne representation $\rho$ to inertia.
- The restriction of $\pi$ as a $\GL_n(\Q_p)$ representation to $\GL_n(\Z_p)$.
There is (some sort of) converse to this, in that when $\pi$ is spherical (unramified), then it is determined
by the Hecke operators at $p$, and the classical Hecke operators at $p$ come from the appropriate actions of diagonal matrices with powers
of $p$, which are (in a non-technical sense) ``orthogonal'' to the copy of $\GL_n(\Z_p)$. (Warning, this is a very vague remark.)
Of course, this connection is much deeper when $n > 1$ than when $n = 1$. For example, when $n > 1$, then $\pi$
is infinite dimensional, and so its restriction to $\GL_n(\Z_p)$ decomposes into infinitely many different representations.
(each with finite multiplicity by admissibility). Here, I think, is one reasonable answer to your question:
Claim: $\rho$ is tamely ramified if and only if $\pi$ has an invariant vector under the $p$-congruence subgroup $K(p)$
of $\GL_n(\Z_p)$. The broader takeaway is that the image of inertia in $\rho$ is determined by the restriction of $\pi$ to $K = \GL_n(\Z_p)$.
Of course I can't really prove this because it requires (at least) the local Langlands correspondence itself. But let's talk about some examples,
even for the case when $n = 2$ because you see most of the phenomena in this case. In fact, for many people, the first thing to do is
just to understand the statement of local Langlands for $n = 2$ and $p > 2$ by just understanding the objects of both sides.
Kevin Buzzard wrote a great note about this
here: http://wwwf.imperial.ac.uk/~buzzard/maths/research/notes/old_introductory_notes_on_local_langlands.pdf
and then he even gave a 20 lecture course as well:
https://www.youtube.com/watch?v=Rv59aRUMfio
In order to talk about it, I guess one has to say a little bit about what the representations $\pi$ really are.
For $n = 2$, there are two types of basic construction: (parabolic) induction from the Borel $B$ of $\GL_2(\Q_p)$,
and (compact) induction from representations of the maximal compact $K = \GL_2(\Z_p)$. (I won't even try to keep track of normalizations which I always find confusing). Let's also denote by $I$ the Iwahori subgroup of $K$ and by $P$ the pro-$p$ Iwahori (the inverse image in $K$ of the unipotent subgroup the Borel of $\GL_2(\F_p)$), and by $K(p)$ the full congruence subgroup of $K$.
Tamely Ramified Principal Series: Let $\chi_1$, $\chi_2$ be a pair of tamely ramified admissible characters of $\Q^{\times}_p$,
which one can think of as a character of the torus $T$ and thus the Borel $B$, and let $\pi$ be
the corresponding principal series representation, which is roughly the induction from $B$ to $G$ of the corresponding character of $B$.
Suitably defined, $\pi$ has invariants under the pro-$p$-Iwahori subgroup $P$ of $K$. Note that $P$ is normalized by $I$, and thus $\pi^{P}$ has an action of $I/P$.
But $I/P$ is nothing but $(\F^{\times}_p)^2$, and this acts on $\pi^{P}$.
The assumption that $\chi_i$ are tame means that their restriction to $\Z^{\times}_p$ give characters on $\F^{\times}_p$,
and hence the pair naturally give a character of $I/P$.
If $\pi$ is irreducible (as it will be most of the time) then $\pi^{P}$ is $1$-dimensional and $I/P$ acts exactly by this character.
So $\pi$ never has invariants under the Iwahori unless the $\chi_i$ are both unramified. In this case, one is either just talking about a unramified principal series, or a (possibly unramified twist of) the Steinberg representation, which have invariants under $K$ and $I$ respectively.
(tamely ramified) Supercuspidals: If we want something with $K(p)$-invariants, we can take a representation $\tau$ of $K/K(p) = \GL_2(\F_p)$
and take the compact induction from $K$ to $G$. If you take $\tau$ to be induced from a Borel then you again get a $\pi$ which has invariants under
the pro-p Iwahori. But if you take $\tau$ to be one of the "exotic" representations of dimension $p - 1$, then $\tau$ turns out to have no $P$-invariant vectors.
The compact inductions in these cases now turn out to be irreducible. Let's compare $\pi$ on the automorphic side with the Weil-Deligne
side. On the automorphic side, I always think of $\tau$ as "almost" constructed as follows: take a character $\psi$ of the
non-split Cartan (which is isomorphic to $\F^{\times}_{p^2}$), and take the induction to $\GL_2(\F_p)$.
Of course, this is not quite correct, because inducing something from the non-split Cartan is not irreducible, but it at least
sees $\tau$ and other similar representations. At any rate, this automorphic side is related to the induction of a character $\psi$ of $\F^{\times}_{p^2}$. But now the Weil-Deligne side is very similar --- the representation is exactly induced from a tamely ramified
character ($\psi$) of the quadratic unramified extension of $\Q_p$. But, by class field theory, on inertia $\psi$ is tamely ramified and hence the same as a character of
the units in the residue field $k^{\times} = \F^{\times}_{p^2}$.
These examples actually exhaust all the $\pi$ which are tamely ramified. Even for $\mathrm{GL}_n(\Q_p)$, the situation is similar. There are representations built up from smaller Levi subgroups, and there are the supercuspidals. But at least the tame supercuspidals are "easy" in that we know on the Weil-Deligne side they should be irreducible (and tamely ramified) and hence induced from an unramified cyclic degree $n$ extension of $\mathbf{Q}_p$. (The supercuspidals were constructed by Howe - the construction requires at least one understands the irreducibles of $\mathrm{GL}_n(\mathbf{F}_p)$.) And on the automorphic side they can be constructed from compact inductions from "known" representations of $\mathrm{GL}_n(\mathbf{F}_p)$.
Back to conductors:
You might guess that the conductor of $\pi$ is given by the minimal $p^n$ such that $\pi$ has an invariant vector under $\Gamma(p^n)$.
But in fact this is not quite correct --- the automorphic conductor is the minimal $p^n$ such that $\pi$ has an invariant vector under
the group $\Gamma_1(p^n)$. In the case of the tamely ramified principal series representation the conductor is thus $p^2$ if $\chi_i$
are both ramified and $p$ if exactly one of $\chi$ is ramified. In particular, $\pi$ will have a twist of conductor $p$.
Similarly, taking ramified principal series which may now be wildly ramified, their explicit construction shows that the conductor
in the automorphic sense is the product (or sum if you take the exponent) of the conductors of $\chi_1$ and $\chi_2$.
And similarly, the conductor on the "Galois" side is the same.
On the other hand,
the tamely ramified supercuspidals have conductor $p^2$, and all their twists have conductor $p^2$. In particular, they have no
invariant vectors under the pro-$p$ Iwahori subgroup $P$ (this can be checked more directly). On the Galois side,
the representation is tamely ramified but irreducible so also has conductor $p^2$.
Best Answer
It's true that if $\rho$ is tamely ramified, then $\rho$ is de Rham. In fact, it's even potentially crystalline with all Hodge-Tate weights equal to 0.
First, note that $\rho(I_{\mathbb Q_p})$ is finite. The reason is that the image of $\rho$ lands in $GL_n(\mathbb Z_p)$, which has a pro-$p$ subgroup of finite index, namely the principal congruence subgroup $1+pM_n(\mathbb Z_p)$. Since $I_{\mathbb Q_p} \to GL_n(\mathbb Z_p)$ factors through a prime-to-$p$ group (tame inertia) by assumption, $\rho(I_{\mathbb Q_p})$ injects into $GL_n(\mathbb F_p)$, hence it is finite.
It follows that there is a finite extension $K/\mathbb Q_p$ such that $\rho|_{I_K}$ is trivial. (The kernel of ($\rho$ restricted to $I_{{\mathbb Q}_p}$) corresponds to a finite (tame) extension of $\mathbb Q_p^{nr}$ and we can choose $K$ such that that extension is contained in $\mathbb Q_p^{nr} \cdot K$.)
It's a general fact that $\rho|_{I_K}$ crystalline implies $\rho|_{G_K}$ crystalline, hence $\rho$ is potentially crystalline. (Added: this follows from Hilbert's theorem 90. If $L/K$ is a Galois extension and $Gal(L/K)$ acts semilinearly and continuously on a finite-dimensional $L$-vector space $V$, then $V$ has a basis that is $Gal(L/K)$-invariant.)
In particular, you only get WD representations with $N = 0$.