2
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.