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Let $F$ be a number field and $E/F$ a Galois extension. Suppose we have a representation $\rho_E : Gal(\overline{F}/E) \rightarrow GL_n(\overline{Q}_p)$. My question is : what are sufficiant conditions so that $\rho_E$ can be extended to a representation $ Gal(\overline{F}/F) \rightarrow GL_n(\overline{Q}_p)$ ?

A necessary condition is that $\rho_E$ is invariant under $Gal(E/F)$. This paper (last line of page 1)

claims that such an extension exists if moreover $\rho_E$ is irreducible and $E/F$ is cyclic of prime ordre, but I don't know why it is true.

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What do you mean by "$\rho_E$ is invariant under $Gal(E/F)$" ? – A M May 19 '11 at 10:16
It means that : for all $\sigma \in G(\overline{{F})/F$, the representation $\rho_E^\sigma := \rho_E(\sigma \cdot \sigma^{-1})$ is isomorphic to $\rho_E$. Note that the isomorphic class of $\rho_E^\sigma$ does not depend on $\sigma \in G(E/F)$. – Auguste Hoang Duc May 19 '11 at 10:23
If $n=1$, this follows from the "Inflation-Restriction" sequence (see applied to $A=\overline{Q}_p^*$. For $n \geq 2$, the argument is quite likely of the same nature. – Laurent Berger May 19 '11 at 15:08
up vote 6 down vote accepted

Pick a $\sigma \in \Gamma_F$ such that $\sigma |_E$ generates $\mathrm{Gal}(E/F)$. There is an $A \in \mathrm{GL}_n$ such that for all $x \in \Gamma_E$, $\rho(\sigma x \sigma^{-1}) = A \rho(x) A^{-1}$, and if $q$ denotes the order of $\mathrm{Gal}(E/F)$, we get that $\rho(\sigma^q x \sigma^{-q}) = A^q \rho(x) A^{-q}$, which by irreducibility and Schur's lemma implies that $\rho(\sigma^q) A^{-q} = \lambda \mathrm{Id}$. Choose any $\mu$ such that $\mu^q = \lambda$. Then setting $\rho(\sigma)=\mu A$ is one of the $q$ ways to extend $\rho$.

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