Let $K$ be a local field and $\rho: W_K \to \operatorname{GL}(V)$ be a Weil representation. The for any finite extension $F/K$, we define the local polynomial $$ P(\rho|_F,T) = \det{(1 - \operatorname{Frob}^{-1}_F T \, | \, \rho^{I_F})} $$ where
- $\rho^{I_F} = \{ x \in V \, | \, \rho_\sigma(x) = x \text{ for all } \sigma \in I_F \}$ is the subspace of inertia invariants,
- $\operatorname{Frob}_F$ is a Frobenius element,
- $I_F$ denotes the inertia subgroup of the absolute Galois group $G_F$.
Let us say that $\rho$ is a twist, meaning $\rho = A \otimes \psi$ where $A$ is an Artin representation and $\psi$ is an unramified character.
One can show that if $\rho$ is Frobenius-semisimple, i.e. $\rho(\operatorname{Frob}_K)$ is diagonalizable, and a finite extension $F/K$ such that $\rho|_F$ is unramified, we obtain $$ P(\rho|_F,T) = (1-\mu^{-f}T)^n $$ where
- $n = \dim{\rho}$,
- $\mu = \psi(\operatorname{Frob}_K)$ and
- $f = f(F/K)$, the inertial degree of $F/K$.
Problem: I read in "Euler Factors determine local Weil Representations" by Tim and Vladimir Dokchitser (more specifically, in the Proof of Theorem 1 (Step 4)) that the roots of $P(\rho|_L, T)$ lie in the equivalence class of $\psi(\operatorname{Frob}_K)^{f(L/K)}$ in $\mathbb{C}^*/\mu_\infty$ ($\mu_\infty$ denotes the set of complex roots of unity) for any finite extension $L/K$. I don't really understand why this is true.
Since $\rho$ is Frobenius-semisimple, $\rho(\operatorname{Frob}_K)$ is diagonalizable. Let $\lambda_1,\dots, \lambda_{\dim{\rho}}$ be its eigenvalues. But since $\rho^{I_K}$ is not $V$ in general, we can not directly say yet that the $\lambda^{-1}_i$ are roots of $P(\rho|_K,T)$. I think that there might be a relationship between these roots and the $\mu$ from before. I think it should be $\lambda_i^{-f} = \mu^{-f}$ for all $i$. But I am not sure how I can show a more general statement.
Could you please help me to understand this result? Thank you!
