# Is the Weil–Deligne representations coming from $\ell$-adic cohomology independent of $\ell$?

Let $F$ be a $p$-adic field. Let $(G_{F}, W_{F}, I_{F})$ denote the (absolute Galois group, Weil group, inertia group) of $F$. Let $X/F$ be a proper smooth variety. Let $\ell$ be a prime number $\ne p$. The $\ell$-adic cohomology $H_{\ell}^{i} = H_{\text{ét}}^{i}(X_{\bar{F}}, \mathbb{Q}_{\ell})$ is naturally endowed with a continuous Galois representation $\rho$.

# The Weil–Deligne representation

By the $\ell$-adic monodromy theorem of Grothendieck, one associates a Weil–Deligne representation with $H_{\ell}^{i}$: It gives a nilpotent “monodromy” operator $N$. One restricts the representation $\rho$ to $W_{F}$, and changes it to $$\sigma (\Phi^{a}x) = \rho(\Phi^{a}x) \exp(-t(x)N), \qquad a \in \mathbb{Z}, x \in I_{F}$$ where $\Phi \in W_{F}$ is a Frobenius element, and $t \colon I_{F} \to \mathbb{Z}_{\ell}$ a projection onto the $\ell$-adic component of $I_{F}$.

The Weil–Deligne representation associated with $H_{\ell}^{i}$ is $(\sigma, H_{\ell}^{i}, N)$. It is an object in $\mathrm{WDRep}_{\mathbb{Q}_{\ell}}(W_{F})$.

# Question

Choose an (non-canonical, non-continuous!) embedding $i_{\ell} \colon \mathbb{Q}_{\ell} \to \mathbb{C}$. By extending scalars, we obtain an object $(\sigma, H_{\ell}^{i}, N) \otimes_{i_{\ell}} \mathbb{C}$ in $\mathrm{WDRep}_{\mathbb{C}}(W_{F})$.

Let $\ell' \ne p$ be another prime. Let $i_{\ell'} \colon \mathbb{Q}_{\ell'} \to \mathbb{C}$ be an embedding. We can repeat the entire process to obtain an object $(\sigma, H_{\ell'}^{i}, N) \otimes_{i_{\ell'}} \mathbb{C}$ in $\mathrm{WDRep}_{\mathbb{C}}(W_{F})$.

Q. Are $(\sigma, H_{\ell}^{i}, N) \otimes_{i_{\ell}} \mathbb{C}$ and $(\sigma, H_{\ell'}^{i}, N) \otimes_{i_{\ell'}} \mathbb{C}$ isomorphic in $\mathrm{WDRep}_{\mathbb{C}}(W_{F})$?

From a motivic viewpoint this should certainly be true, but on the other hand maybe this is one of the strongest forms of $\ell$ independence that one can ask for.

If there is no answer in general, I would be very happy to learn about partial cases, where this is known.

• Setting aside the isomorphisms with $\mathbf{C}$ for a moment, consider the more basic question of identifying the finite extension of $F$ corresponding to $\ker(t)$. Is it independent of $\ell$? The mere fact that $\ker(t)$ is at least bounded independently of $\ell$ (i.e., contained in a finite extension of $F$ independent of $\ell$) rests on deJong's alterations theorem, as far as I know. Is there a proof of the finer result about the actual field for $\ker(t)$? That is much weaker than what you ask for, so is a proof known? Dec 9, 2014 at 2:59
• @user74230 — Good remark! I do not know of such a proof. Do you have a reference for the weaker result, that you mentioned?
– jmc
Dec 9, 2014 at 8:43
• I garbled what I meant to say, but I think you read into my comment what I had intended: literally $t$ itself cuts out the $\ell$-part of the maximal tame extension of $F^{\rm{un}}$, so what I meant was to focus on the unique minimal finite extension of $F^{\rm{un}}$ over which $\rho|_{I_F}$ becomes unipotent. To prove this is bounded independently of $\ell$ one uses cohomological descent applied to a proper hypercovering built from semistable models (provided by dJ). So it is an "exercise" with cohomological descent, but I don't know a reference (any local expert could explain it to you). Dec 9, 2014 at 8:51