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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.

Could you please help me with this question? Thank you in advance!

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1 Answer 1

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

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