# Do the local polynomials of Weil representations coincide if they are Artin representations (factor through a finite quotient)?

Let

• $$K$$ be a local field,
• $$G_K$$ its absolute Galois group,
• $$I_K$$ the inertia subgroup of $$G_K$$,
• $$\operatorname{Frob}_K \in G_K$$ be a Frobenius element, i.e. any element of $$G_K$$ acting as $$x \mapsto x^{|k|}$$ on $$\bar{k}$$, the algebraic closure of the residue fiel $$k$$ of $$K$$,
• $$W_K$$ be the Weil group of $$K$$ and
• $$\rho$$ be a Weil representation, i.e. it is a representation $$\rho: W_K \to \operatorname{GL}_n(\mathbb{C})$$ with $$\rho(I_K)$$ being finite.

The local polynomial $$P(\rho,T)$$ is the inverse characteristic polynomial of $$\operatorname{Frob}_K^{-1}$$ on the inertia invariants of $$\rho$$, i.e. $$P(\rho,T) = \det(1-\operatorname{Frob}_K^{-1} T \, | \, \rho^{I_K}).$$

We say that $$\rho$$ factors through a finite quotient if there is a finite Galois extension $$F/K$$ such that $$\operatorname{Gal}(\bar{K}/F) \subset \ker{\rho}$$ ($$\rho$$ is also called an Artin representation then). This means that $$\rho$$ comes from a representation $$\bar{\rho}: \operatorname{Gal}(F/K) \to \operatorname{GL}_n(\mathbb{C})$$.

We can define a local polynomial for $$\bar{\rho}$$ the same way:

$$P(\bar{\rho},T) = \det(1-\operatorname{Frob}_{F/K}^{-1} T \, | \, \bar{\rho}^{I_{F/K}})$$

where $$I_{F/K}$$ is the inertia subgroup of $$\operatorname{Gal}(F/K)$$ and $$\operatorname{Frob}_{F/K} \in \operatorname{Gal}(F/K)$$ is any Frobenius element.

Question Do we have $$P(\rho,T) = P(\bar{\rho},T)$$?

For me, it is especially difficult to understand the first definitions without $$F/K$$ because I am not able to compute them explicitly.

$$\DeclareMathOperator{\Gal}{Gal}\DeclareMathOperator{\Frob}{Frob}$$

This comes down to checking a few things. Let $$\kappa(K), \kappa(F)$$ be the residue fields of $$K$$ and $$F$$.

Let $$\rho: \operatorname{Gal}(\overline{K}/K) \rightarrow \operatorname{GL}(V)$$ be a continuous, finite dimensional representation of the Weil group. Suppose that the kernel of $$\rho$$ contains $$\operatorname{Gal}(\overline{K}/F)$$, so we have a well defined homomorphism $$\overline{\rho}: \Gal(F/K) \rightarrow \operatorname{GL}(V)$$. The inertia group $$I_K$$ is the kernel of the surjective homomorphism

$$\Gal(\overline{K}/K) \rightarrow \Gal(\kappa(K)^{\operatorname{sep}}/\kappa(K))$$

and the inertia group $$I_{F/K}$$ is the kernel of the surjective homomorphism

$$\Gal(F/K) \rightarrow \Gal(\kappa(F)/\kappa(K))$$

First, a given $$\sigma \in \Gal(\overline{K}/K)$$ induces the Frobenius on $$\Gal(\kappa(K)^{\operatorname{sep}}/\kappa(K))$$ if and only if its image in $$\Gal(F/K)$$ induces the Frobenius on $$\Gal(\kappa(F)/\kappa(K))$$.

Second, the image of $$I_K$$ in $$\Gal(F/K)$$ is equal to $$I_{F/K}$$. Thus

$$\{v \in V : \rho(\sigma)v = v \textrm{ for all } \sigma \in I_K\} = \{v \in V : \overline{\rho}(\sigma)v = v \textrm{ for all } \sigma \in I_{F/K}\}$$

or $$\rho^{I_K} = \rho^{I_{F/K}}$$.