2
$\begingroup$

Let $K$ be an extension of $\mathbb{Q}_p$. Consider an Galois representation $\rho: G_K \to GL_2(\mathbb{C})$ and for any finite extension $M/K$ we call

$$P(\rho|_M,T) = \det(1-\operatorname{Frob}_{M}^{-1}T \, | \rho^{I_M})$$

the local polynomial of $\rho$ over $M$. Here, $\operatorname{Frob}_{M}$ is an Frobenius element of $G_M$, i.e. a lift of the automorphism $x \mapsto x^{q_M} \in G_{\mathbb{F}_{q_M}}$ (where $q_M$ denotes the cardinality of the residue field of $M$). Note that the local polynomial is independent of the choice of the Frobenius element.

Now we make the following assumptions:

  • $L/K$ is a cyclic and finite extension,
  • $M$ is a subextension of $L/K$ such that $L/M$ is unramified,
  • $\rho = A \otimes \psi$ where $A: G_K \to GL_{2}(\mathbb{C})$ factors through $\operatorname{Gal}(L/K)$ (i.e. is an Artin representation) and $\psi: G_K \to \mathbb{C}^*$ is an unramified character (i.e. $\psi(I_K) = 1$),
  • $P(\rho|_L,T) = (1-\mu^{-f}T)^2$ where $\mu = \psi(\operatorname{Frob}_K) \in \mathbb{C}^*$ for a fixed Frobenius element $\operatorname{Frob}_K \in G_K$ and $f$ is the inertial degree of $L/K$.

Question: Can we compute $P(\rho|_M,T)$ by using the result $P(\rho|_L,T)$?

Because $L/M$ is unramified, a Frobenius element in $G_M$ is given by $\operatorname{Frob}_L^\ell$ where $\ell$ is the degree of $L/M$. Now I am not sure how to proceed - a next step would be to consider the vector spaces $\rho^{I_L}$ and $\rho^{I_M}$ in the definition of the local polynomials.

Could you please help me here? Any help is really appreciated!

$\endgroup$
2
  • $\begingroup$ The notation $\operatorname{Frob}_M^{-1} T | \rho^{I_M}$ does not make sense to me. I assume $M/K$ is Galois, and $\rho$ factors through $\operatorname{Gal}(M/K)$? In that case, is $I_M$ the inertia group of $\operatorname{Gal}(M/K)$? $\endgroup$
    – D_S
    Commented Apr 19, 2021 at 1:59
  • $\begingroup$ $I_M=Gal(\overline{M}/M^{ur})$ and $\rho^{I_M}$ is the subspace of $\Bbb{C}^2$ where $I_M$ acts trivially, so that $\rho(Frob_M)$ doesn't depend on a choice of $Frob_M$ @D_S $\endgroup$
    – reuns
    Commented Apr 19, 2021 at 2:01

1 Answer 1

2
$\begingroup$

This is what I understand of your "setting".

Optional:

  • $P(\rho|_L,T) = (1-\mu^{-f}T)^2$ gives that $P(A|_L,T) = (1-\psi(Frob|_L)\mu^{-f} T)^2$.

    Since $A$ has finite image it means that on $A$, $Frob_L$ is the multiplication by $\psi(Frob_L)^{-1}\mu^f$ and thus on $\rho|_L$ it is the multiplication by $\mu^f$.

The main point:

  • Since $L/M$ is unramified of degree $\ell$ then for any $Frob_M$ you'll get that $Frob_M^\ell$ is a $Frob_L$. Also $M^{ur}=L^{ur}\implies Gal(\overline{L}/L^{ur})=I_M=I_L$, therefore $$P(\rho|_M,T) = (1- \zeta_{\ell}^a\mu^{-f/\ell}T)(1- \zeta_{\ell}^b\mu^{-f/\ell}T)$$ for some integers $a,b$, with in general no restriction on $a,b$.
$\endgroup$
2
  • $\begingroup$ Thank you for your response! Could you explain to me why these $\zeta_\ell^a$ and $\zeta_\ell^b$ occur in $P(\rho_M,T)$ in the first place? This is the only part which is unclear to me. $\endgroup$
    – Diglett
    Commented Apr 19, 2021 at 12:10
  • $\begingroup$ Eigenvalues of a matrix whose $\ell$ power is $\mu^{-f}$ (or if you skip the first part so you don't know that it is diagonalizable, the diagonal entries of its jordan normal form whose $\ell$ power has $\mu^{-f}$ diagonal entries) $\endgroup$
    – reuns
    Commented Apr 19, 2021 at 16:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .