Local polynomials of Frobenius-semisimple Weil representations which are tensor products of an Artin representation and an unramified character

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.

• We don't need that the $\lambda_i^{-1}$ are roots of $P(\rho|_K, T)$ - instead we need the converse that the roots of $P(\rho|_K, T)$ are among the $\lambda_i^{-1}$. This is easy because $\rho^{I_K}$ is a subspace of $V$. – Will Sawin Dec 17 '18 at 20:08
• @WillSawin: Thank you for your response! But don't we need to assume that $\rho^{I_K}$ is a $W_K$-invariant subspace? (otherwise, $\rho(\operatorname{Frob}_K)$ is not diagonalizable on the subspace $\rho^{I_K}$) – Diglett Dec 17 '18 at 22:06
• But this is clear because $I_K$ is a normal subgroup of $W_K$. – Will Sawin Dec 17 '18 at 22:08