Definition: Let $K$ be a (non-Archimedean) local field and $k$ its residue field.
- A Frobenius element of the absolute Galois group $G_K$ is any element of $G_K$ which is a lift of the Frobenius automorphism $x \mapsto x^{|k|}$ in $G_k \simeq G_K/I_K$.
- A Weil representation $\rho$ is called Frobenius-semisimple if $\rho(\Phi)$ is diagonalizable for some Frobenius element $\Phi$ in $G_K$.
Question: If $\rho$ is a Frobenius-semisimple Weil representation why is $\rho(\Phi')$ diagonalizable for any Frobenius element $\Phi'$ in $G_K$?
Attempts:
- I know that the Frobenius element in $G_K$ is unique up to an element in $I_K$, i.e. if $\Phi, \Phi'$ are two Frobenius elements in $G_K$, there exists an element $\sigma \in I_K$ with $\Phi' = \Phi \circ \sigma$.
- Let us say that $\rho(\Phi')$ is diagonalizable and $(v_1,\dots,v_n)$ is a basis of the representation space consisting of eigenvectors of $\rho(\Phi')$. The only thing I was able to show with that $(v_1,\dots,v_n)$ is also a basis of eigenvectors of $\rho(\sigma^{-1}\circ \Phi \circ \sigma)$. But this is not helpful since I want to find a basis of eigenvectors for $\Phi$.
Could you please help me with this problem? Thank you!