# On tetrahedral Artin representation

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.

• Do you mean show $Ad(\rho)$ is induced from $\xi$? $Ind_E^F \xi$ will be 3 dimensional. – 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.