I am trying to understand the proof of existence of $\pi(\rho)$ for tetrahedral representation (Galois representation of dim 2 having image $A_4$ in $PGL_2(\mathbb{C})$) explained by Rogawski, Functoriality and the Artin conjecture.

The first step is to show that the representation $\rho$ is induced from a character $\xi$ of an extension $E/F$ of degree 3.

At this point, I guess Rogawski mimics Langlands in showing that $\pi(Ad(\rho))$ and $\pi(\rho_E)$ exist to deduce the existence of $\pi(\rho)$.

But a few paragraphs before, Rogawski introduces the concept of automorphic induction and the result of Arthur-Clozel showing that automorphic induction exists for cyclic extension.

What I naturally want to do is: show that $\rho$ is induced from $\xi$ and then use automorphic induction to prove the existence of $\pi(\rho)$.

Is there a reason for Rogawski to not use automorphic induction ? I guess Langlands didn't have the result of Arthur-Clozel that explains why he goes around.

  • $\begingroup$ Do you mean show $Ad(\rho)$ is induced from $\xi$? $Ind_E^F \xi$ will be 3 dimensional. $\endgroup$ – Kimball Jun 7 '17 at 7:01

Say $\rho$ is a tetrahedral representation of $G_F$, i.e., it is an irreducible 2-dimensional representation and the projection $\bar \rho$ to $PGL_2(\mathbb C)$ has image $A_4$. Then you can take $E/F$ to be a normal cubic extension such that $\bar \rho$ has image $V_4$ in $A_4$.

I guess you want to do the following: take a character $\xi$ of $G_E$ and induce to $G_F$ and say $Ind_E^F(\xi) \simeq \rho \oplus \psi$ for some character $\psi$. You cannot have such a decomposition because $E/F$ is normal, so this argument will not work.

To take an explicit example, suppose $\rho$ factors through an SL(2,3) extension, so you can view $\rho$ as an irreducible 2-dimensional representation of SL(2,3) (the double-cover of $A_4$). Then SL(2,3) has no index 2 subgroups so $\rho$ is not induced, and there is only one subgroup of index 3, a normal $Q_8$. The restriction of $\rho$ to this is irreducible, so you cannot construct $\rho$ as a (component of) induction of a character along a degree 2 or degree 3 extension. This is why Langlands' argument is so beautiful--it gives modularity for representations that are not easily constructed from characters.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.