This post concerns the proof of the Artin conjecture in the case of degree 2 representations.

It is known that the finite subgroups of $\textrm{GL}(2,\mathbb{C})$ can be classified according to their images in $\textrm{PGL}(2,\mathbb{C})$, being isomorphic to one of: $C_n$, $D_n$, $A_4$, $S_4$, $A_5$, in which case they are called cyclic, dihedral, tetrahedral, octahedral, and icosahedral respectively.

Any degree 2 representation $(V,\rho)$ of a finite group $G$ can hence be classified by the image of $\rho(G) < \textrm{GL}(2,\mathbb{C})$ under the canonical projection $\pi: \textrm{GL}(2,\mathbb{C}) \to \textrm{PGL}(2,\mathbb{C})$. We say that $(V,\rho)$ is cyclic if $\pi(\rho(G)) \cong C_n$, dihedral if $\pi(\rho(G)) \cong D_n$, et cetera.

The work of Langlands and Tunnell resolved the Artin conjecture for tetra- and octahedral representations. I will not be going into that here. But in the cases of cyclic and dihedral representations, it is said that the following two are equivalent:

(1) $(V,\rho)$ is a cyclic or dihedral degree 2 representation.

(2) $(V,\rho)$ is a monomial representation.

This would resolve the Artin conjecture for cyclic and dihedral representations.

I have been trying to prove the equivalence of (1) and (2), but all my attempts so far have only led to a cul-de-sac.

All help or input would be highly appreciated.


2 Answers 2


This comes down to just group theory. Let $Z(G)$ denote the centre of $\rho(G)$.

If $(V, \rho)$ is cyclic, then $\rho(G)/Z(G)$ is cyclic, from which it follows that $\rho(G)$ is abelian. Hence, by Schur's lemma, $(V, \rho)$ is reducible, so a direct sum of one dimensional representations, and Artin's conjecture follows from class field theory.

If $(V, \rho)$ is dihedral, then $\rho(G)/Z(G)$ has an index two subgroup isomorphic to $C_n$. Pulling back this subgroup to $G$, we see that $G$ has an index $2$ subgroup $H$ such that $\rho(H)/Z(G)$ is cyclic. Hence, $\rho(H)$ is abelian.

By Schur's lemma, $\rho|_H$ is therefore a sum of two characters. If $\chi$ is one of these characters, by Frobenius reciprocity $$(\rho|H, \chi) = (\rho, \mathrm{Ind}_H^G(\chi)).$$

Since $\rho$ is irreducible (e.g. it does not have abelian image), it follows that $\rho\simeq\mathrm{Ind}_H^G(\chi)$.

Conversely, if $\rho \simeq \mathrm{Ind}_H^G(\chi)$ is irreducible, where $H$ is an index two subgroup of $G$, then $H/Z(G)$ is an index two subgroup of $G/Z(G)$. Since $A_4, S_4$ and $A_5$ have no index two subgroups, $G/Z(G)$ must be $D_n$.

You should take a look at Section 4 of this paper, which gives an accessible account of the theory of dihedral Galois representations.

  • $\begingroup$ I am having some trouble understanding the "pulling back" part. Are you assuming that $\rho(G) \cong G$ (that $\rho$ is faithful)? $\endgroup$ Jan 21, 2020 at 15:59
  • 1
    $\begingroup$ I'm not, but making that assumption would be fine! You can pull back the subgroup to a subgroup $H\le \rho(G)\cong G/\ker(\rho)$ and then to $G$. $\endgroup$
    – Mathmo123
    Jan 22, 2020 at 8:21
  • $\begingroup$ By $Z(G)$, do you mean $Z(\rho(G))$? Also, is it true that $\pi(\rho(G)) \cong \rho(G)/Z(\rho(G))$, when $\pi: \textrm{GL}(V) \to \textrm{PGL}(V) \cong \textrm{GL}(V) / Z(V)$ is the quotient map, and $Z(V)$ is the group of homotheties on $V$? $\endgroup$ Jan 30, 2020 at 15:31
  • 1
    $\begingroup$ @HeinrichWagner Yes that's what I mean. And yes that's true - by Schur's Lemma, $Z(\rho(G))$ can only consist of scalar matrices. $\endgroup$
    – Mathmo123
    Jan 30, 2020 at 20:27

About the $(1)\implies (2)$ direction

Find by hand the irreducible representations of $D_3$ (of dimension $\ge 2$) and show they are of dimension $2$, induced from a character $\phi:C_3\to \Bbb{C}^*$.

With $L/E$ Galois $Gal(L/E)=D_3$, the corresponding Artin L-function $L(s,\rho,L/E)$ is thus $=L(s,\phi,F/E)$.

By class field theory $L(s,\phi,F/E)=L(s,\psi,F/E)$ for some (non-trivial, finite order) Hecke character of $F$, Neukrich's book proves it is entire.

When $E=\Bbb{Q}$, the odd representation stuff in Langlands is to ensure $F$ is an imaginary quadratic field, so that $L(s,\psi,F/Q)=L(s,f)$ where $f$ is a weight 1 modular form (when $F$ is a real quadratic field, $f$ is a Maass form).

Note that when $F/\Bbb{Q}$ is imaginary quadratic then class field theory has a more explicit version : the theory of elliptic curves with complex multiplication by $O_F$, whose $j$-invariant and torsion points generate all the abelian extensions of $F$.

Mathmo123's answer shows it works the same way for every $D_n$.

  • $\begingroup$ "When $F$ is a real quadratic field, $f$ is a Maass form". That isn't quite right. If the projective image is $D_2\cong C_2\times C_2$, then $f$ is induced from $3$ distinct extensions, at least one of which must be real. This case can occur when $f$ is a weight one modular form. $\endgroup$
    – Mathmo123
    Jan 20, 2020 at 13:55
  • $\begingroup$ Rereading, what you say is correct, as you are only talking about $D_3$, but it is nevertheless slightly misleading. Another misleading thing: $\rho(G)$ will not be dihedral in general (it has dihedral projective image not dihedral image), so $\rho$ will not in general be induced from a character of $C_3$. $\endgroup$
    – Mathmo123
    Jan 20, 2020 at 14:01

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