Classifications of Hecke Fields and Galois Images of Weight One Exotic Newforms

TL;DR

This paper classifies the Hecke fields and Galois image structures of weight one exotic newforms, providing explicit descriptions based on the nebentypus and group type, advancing understanding of their arithmetic properties.

Contribution

It offers a complete classification of Hecke fields and Galois images for weight one newforms of specific types, linking these to the nebentypus and group structure.

Findings

Hecke fields are explicitly described in terms of nebentypus order.

Galois image classifications are complete for each form type.

Results enhance understanding of the arithmetic of weight one newforms.

Abstract

We determine the Hecke fields associated with weight one newforms of $A_4$-, $S_4$-, and $A_5$-type, expressed in terms of the order of its nebentypus. Furthermore, for each type, we provide a complete classification of the images of the corresponding Galois representations.

Equations

$$f=\sum _{n\ge 1}{a}_n(f)q^n\in S_1(N,\chi )$$

$${\rho }_f:G_{\mathbb{Q}}:=Gal(\overline{\mathbb{Q}}/\mathbb{Q})\to {\mathrm{GL}}_2(\mathbb{C})$$

$$K_f:=\mathbb{Q}(\{a_n(f){\}}_{n\ge 1}).$$

$$d:=\mathrm{ord}(\chi ).$$

$$K_f=\{\begin{array}{ll}\mathbb{Q}({\zeta }_d,\sqrt{-2})\text{ or }\mathbb{Q}({\zeta }_{4d})& \text{ if }{\mathrm{ord}}_2(d)=1,\\ \mathbb{Q}({\zeta }_d)\text{ or }\mathbb{Q}({\zeta }_{2d})& \text{ if }{\mathrm{ord}}_2(d)=2,\\ \mathbb{Q}({\zeta }_{2d})& \text{ if }{\mathrm{ord}}_2(d)\ge 3.\end{array}$$

$$p\le \frac{1}{12}[{\mathrm{SL}}_2(\mathbb{Z}):{\Gamma }_0(N)].$$

$$K_f=\{\begin{array}{ll}\mathbb{Q}({\zeta }_4,\sqrt{5})& \text{ if }d=2,\\ \mathbb{Q}({\zeta }_{12},\sqrt{5})& \text{ if }d=6,\\ \mathbb{Q}({\zeta }_{20})& \text{ if }d=10,\\ \mathbb{Q}({\zeta }_{60})& \text{ if }d=30.\end{array}$$

$$K_f=\{\begin{array}{ll}\mathbb{Q}(\sqrt{-2})\text{ or }\mathbb{Q}({\zeta }_8)& \text{ if }d=2,\\ \mathbb{Q}({\zeta }_4)\text{ or }\mathbb{Q}({\zeta }_8)& \text{ if }d=4,\\ \mathbb{Q}(\sqrt{-2},{\zeta }_3)\text{ or }\mathbb{Q}({\zeta }_{24})& \text{ if }d=6,\\ \mathbb{Q}({\zeta }_{12})\text{ or }\mathbb{Q}({\zeta }_{24})& \text{ if }d=12.\end{array}$$

$$K_f=\{\begin{array}{ll}\mathbb{Q}({\zeta }_{4d})& \text{ if }{\mathrm{ord}}_2(d)=1,\\ \mathbb{Q}({\zeta }_{2d})& \text{ if }{\mathrm{ord}}_2(d)=2.\end{array}$$

$$K_f=\{\begin{array}{ll}\mathbb{Q}({\zeta }_d,\sqrt{-2})& \text{ if }{\mathrm{ord}}_2(d)=1,\\ \mathbb{Q}({\zeta }_d)& \text{ if }{\mathrm{ord}}_2(d)=2.\end{array}$$

$${\mathcal{P}}_f:=\{p\mid \gcd (p,N)=1\text{ and }{K}_f=\mathbb{Q}(a_p(f))\}.$$

$$d({\mathcal{P}}_f)=\{\begin{array}{ll}\frac{3\phi (d)}{4d}& \text{if}\ker (\chi )\not\subset \ker (G_{\mathbb{Q}}\stackrel{{\overline{\rho }}_f}{\twoheadrightarrow }{A}_4\twoheadrightarrow A_4^{\mathrm{ab}}),\\ & \\ \frac{\phi (d)}{d}& \text{if}\ker (\chi )\subset \ker (G_{\mathbb{Q}}\stackrel{{\overline{\rho }}_f}{\twoheadrightarrow }{A}_4\twoheadrightarrow A_4^{\mathrm{ab}}).\end{array}$$

$$d({\mathcal{P}}_f)=\{\begin{array}{ll}\frac{2\phi (d)}{5d}& \text{if}5\nmid d,\\ & \\ \frac{3\phi (d)}{4d}& \text{if}5\mid d.\end{array}$$

$$d({\mathcal{P}}_f)=\{\begin{array}{ll}\frac{3\phi (d)}{8d}& \text{if}{\mathrm{ord}}_2(d)=2,\\ & \\ \frac{5\phi (d)}{8d}& \text{if}{\mathrm{ord}}_2(d)\ge 3.\end{array}$$

$$d({\mathcal{P}}_f)=\{\begin{array}{ll}\frac{\phi (d)}{2d}& \text{if}{\mathrm{ord}}_2(d)\ne 2,\\ & \\ \frac{7\phi (d)}{8d}& \text{if}{\mathrm{ord}}_2(d)=2.\end{array}$$

$$({\mathrm{SL}}_2({\mathbb{F}}_3)\times \mathbb{Z}/2d\mathbb{Z})/⟨(-I,d)⟩.$$

$$({\mathrm{SL}}_2({\mathbb{F}}_3){\times }_{\mathbb{Z}/3\mathbb{Z}}\mathbb{Z}/2d\mathbb{Z})/⟨(-I,d)⟩.$$

$${\mathrm{SL}}_2({\mathbb{F}}_3)\twoheadrightarrow {\mathrm{SL}}_2({\mathbb{F}}_3{)}^{\mathrm{ab}}\simeq \mathbb{Z}/3\mathbb{Z}.$$

$$({\mathrm{SL}}_2({\mathbb{F}}_5)\times \mathbb{Z}/2d\mathbb{Z})/⟨(-I,d)⟩.$$

$$(B{O}_{48}\times \mathbb{Z}/2d\mathbb{Z})/⟨(z,d)⟩.$$

$$(B{O}_{48}{\times }_{\mathbb{Z}/2\mathbb{Z}}\mathbb{Z}/2d\mathbb{Z})/⟨(z,d)⟩.$$

$$B{O}_{48}\twoheadrightarrow B{O}_{48}^{\mathrm{ab}}\simeq \mathbb{Z}/2\mathbb{Z}.$$

$$\frac{\mathrm{trace}(g{)}^2}{\det (g)}=2+{\zeta }_n+{\zeta }_n^{-1}$$

$$\mathrm{ord}(\chi )=\mathrm{lcm}\{|\chi (I_p)|{\}}_{p\mid N}=\mathrm{lcm}\{|\bar{\rho }(I_p)|{\}}_{p\mid N}.$$

$$(\mathbb{Z}/N\mathbb{Z}{)}^{\times }\stackrel{\sim }{\to }\mathrm{Gal}(\mathbb{Q}({\zeta }_N)/\mathbb{Q});p\mapsto ({\mathrm{Frob}}_p:{\zeta }_N\mapsto {\zeta }_N^p),$$

$$Q_m(\chi ):=\left\{p\in {\mathcal{P}}_N\mid \text{ord(χ(p)) is divisible by m}\right\}.$$

$$d(Q_{{\ell }^e}(\chi ))=\frac{({\ell }^e-{\ell }^{e-1})d^{\prime }}{d}=1-\frac{1}{\ell }$$

$${\rho }_f:G_{\mathbb{Q}}\to {\mathrm{GL}}_2(\mathbb{C})$$

$$R_m:=\left\{p\in {\mathcal{P}}_N\mid \text{ρˉf(Frobp) has order m}\right\}.$$

$$\mathbb{Q}(a_p(f))=\{\begin{array}{ll}\mathbb{Q}(\sqrt{\chi (p)})& \text{ if }p\in R_1\bigsqcup R_3,\\ \mathbb{Q}& \text{ if }p\in R_2,\\ \mathbb{Q}(\sqrt{2\chi (p)})& \text{ if }p\in R_4,\\ \mathbb{Q}(\sqrt{5},\sqrt{\chi (p)})& \text{ if }p\in R_5.\end{array}$$

Full text

1. 1 Introduction

2. 2 Preliminaries

3. 2.1 Projective Galois representation

4. 2.2 Density of $Q_{m}(\chi)$

5. 3 Hecke field $K_{f}$

6. 3.1 $A_{4}$-case

7. 3.2 $A_{5}$-case

8. 3.3 $S_{4}$-case

9. 3.4 Refinement of the $S_{4}$-case

10. 4 On a generator of the Hecke field $K_{f}$

11. 5 Strongly minimal newforms

12. 5.1 Local lifting

13. 5.2 Proof of Theorem 5.3

14. 5.3 Strongly minimal newforms with a prescribed order of the nebentypus

15. 6 The group structure of the image of $\rho_{f}$

16. 6.1 Preliminaries

17. 6.2 Main results on the image of $\rho_{f}$

18. 6.3 Criterion for determining whether $\overline{K}=\overline{G}$ or not

19. 6.4 The structure of $G$ when $\overline{K}=\overline{G}$

20. 6.5 The structure of $G$ when $\overline{K}\neq\overline{G}$

1. Introduction

Let

[TABLE]

be a weight one newform of level $N$ and nebentypus $\chi$ with $\chi(-1)=-1$. By the theorem of Deligne and Serre [8, Théorème 4.1], there exists a continuous irreducible odd Galois representation

[TABLE]

associated with $f$, in the sense that $\rho_{f}$ is unramified at each prime $p\nmid N$ and the characteristic polynomial of $\rho_{f}(\mathrm{Frob}_{p})$ is $X^{2}-a_{p}(f)X+\chi(p)$.

It is well-known that the projective image of $\rho_{f}$, namely the image of $\rho_{f}$ in $\mathrm{PGL}_{2}(\mathbb{C}):=\mathrm{GL}_{2}(\mathbb{C})/\mathbb{C}^{\times}$, is isomorphic to one of the finite groups $D_{2n}$, $A_{4}$, $S_{4}$, or $A_{5}$. If the projective image is isomorphic to the dihedral group $D_{2n}$ of order $2n$ for some integer $n$, then we say that $f$ is of dihedral type; otherwise, we say that $f$ is exotic, and we further distinguish the cases of $A_{4}$-type, $S_{4}$-type, or $A_{5}$-type, respectively.

Let $K_{f}$ be the Hecke field associated with $f$; that is,

[TABLE]

Remark 1.1**.**

The Galois representation $\rho_{f}$ is defined over $K_{f}$, that is, it is conjugate to a homomorphism $G_{\mathbb{Q}}\to\operatorname{GL}_{2}(K_{f})$. Although this does not follow directly from the definition of $K_{f}$, it is a consequence of the fact that $\rho_{f}$ is odd (see, for instance, the discussion on MathOverflow [1]).

Since $\rho_{f}$ has finite image, the eigenvalues of $\rho_{f}(\mathrm{Frob}_{p})$ for $p\nmid N$ are roots of unity; hence the Frobenius trace $a_{p}(f)$ lies in a cyclotomic field. Consequently, $K_{f}$ is an abelian extension of $\mathbb{Q}$. In the first part of this paper, we determine the Hecke field $K_{f}$ of a weight one exotic newform $f$ in terms of the order of its nebentypus $\chi$.

In what follows, $d$ will always stand for the order of the nebentypus $\chi$;

[TABLE]

We note that, as $\chi(-1)=-1$, the order $d$ is necessarily even. For any positive integer $n$, let $\zeta_{n}\in\mathbb{C}$ denote a primitive $n$th root of unity. In this setting, the following theorems are among the main results of the present paper.

Theorem 1.2** (Theorem 3.3).**

If $f$ is of $A_{4}$-type, then $K_{f}=\mathbb{Q}(\zeta_{2d})$.

Theorem 1.3** (Theorem 3.5).**

*If $f$ is of $A_{5}$-type, then $K_{f}=\mathbb{Q}(\zeta_{2d},\sqrt{5})$. *

Theorem 1.4** (Theorem 3.7).**

Let $\mathrm{ord}_{2}(d)\geq 1$ denote the $2$-adic valuation of the even integer $d$. If $f$ is of $S_{4}$-type, then

[TABLE]

Remark 1.5**.**

For any exotic type and any even positive integer $d$, there exists a weight one newform of the specified type whose nebentypus has order $d$. This can be shown by twisting with an appropriate Dirichlet character, as follows:

Let $g$ be a weight one exotic newform of the specified type whose nebentypus has order $2$ (the existence of such a newform $g$ can easily be verified, for example, by checking LMFDB [15]). Take a prime $p$ such that $p\equiv 1\pmod{2d}$ and $p$ is coprime to the level of $g$. Then there exists a Dirichlet character $\psi$ of order $2d$ with conductor $p$, and we consider the weight one newform $f$ corresponding to $g\otimes\psi$, that is, $a_{q}(f)=\psi(q)a_{q}(g)$ for almost all primes $q$. Since $\rho_{f}\simeq\rho_{g}\otimes\psi$, the newform $f$ has the same type as $g$, and its nebentypus has order $d$, as follows from $\det(\rho_{f})=\det(\rho_{g})\psi^{2}$.

Remark 1.6**.**

It follows from a result of Hecke and Strum that the Hecke algebra $\mathbb{Z}[T_{n}\mid n\in\mathbb{Z}_{>0}]\subset\mathrm{End}(S_{1}(N,\chi))$ is generated as a $\mathbb{Z}[\chi]$-algebra by $T_{1}$ and $T_{p}$ for all prime $p$ satisfying

[TABLE]

See [5, Proposition 8.2.3] for example. Since the newform $f$ defines a surjective ring homomorphism $\mathbb{Q}[T_{n}\mid n\in\mathbb{Z}_{>0}]\twoheadrightarrow K_{f};T_{n}\mapsto a_{n}(f)$, the Hecke field $K_{f}$ is generated as a $\mathbb{Q}(\chi)$-algebra by $a_{p}(f)$ for all prime $p\leq[\operatorname{SL}_{2}(\mathbb{Z}):\Gamma_{0}(N)]/12$.

Remark 1.7**.**

Throughout this remark, we assume that the level $N$ of $f$ is square-free. Under this assumption, the possible orders of $\chi$ are highly restricted, and this enables us to obtain more explicit results, as follows (see Corollaries 3.4, 3.6, and 3.8).

• (i)

If $f$ is of $A_{4}$-type, then $d=6$ and $K_{f}=\mathbb{Q}(\zeta_{12})$.

• (ii)

If $f$ is of $A_{5}$-type, then $d\in\{2,6,10,30\}$ and

[TABLE]

• (iii)

If $f$ is of $S_{4}$-type, then $d\in\{2,4,6,12\}$ and

[TABLE]

Moreover, a simple check of LMFDB [15] shows that each list contains at least one explicit example.

In Theorem 1.4, when $\mathrm{ord}_{2}(d)\leq 2$, two distinct Hecke fields may occur. A natural question is how one can distinguish between these two possibilities. We provide an answer to this question in the following sense. The details are given in §3.4.

Theorem 1.8** (Theorem 3.9).**

Let $\overline{\rho}_{f}\colon G_{\mathbb{Q}}\to\operatorname{PGL}_{2}(\mathbb{C})$ denote the projective Galois representation associated with $f$. If $f$ is of $S_{4}$-type, i.e, $\overline{\rho}_{f}(G_{\mathbb{Q}})\simeq S_{4}$, then the following hold:

• (i)

If $\chi^{d/2}\neq\operatorname{sgn}\circ\bar{\rho}_{f}$ as $G_{\mathbb{Q}}$-representations, then

[TABLE]

• (ii)

If $\chi^{d/2}=\operatorname{sgn}\circ\bar{\rho}_{f}$ as $G_{\mathbb{Q}}$-representations, then

[TABLE]

Remark 1.9**.**

In his forthcoming work, Yu Miyazawa narrows down the possible Hecke fields for weight one newforms of dihedral type. Combining his results with ours, we conclude that for each positive integer $n$, only finitely many number fields of degree at most $n$ arise as Hecke fields associated with weight one newforms.

As an application of the above results, for each type of the weight one exotic newform $f$, we can compute the Dirichlet density of the set $\mathcal{P}_{f}$ of primes defined by

[TABLE]

In what follows, let $\varphi$ denote the Euler’s totient function.

Theorem 1.10** (Theorem 4.1).**

Suppose that $f$ is of $A_{4}$-type. Then the Dirichlet density $d(\mathcal{P}_{f})$ of the set $\mathcal{P}_{f}$ is given by

[TABLE]

Theorem 1.11** (Theorem 4.2).**

Suppose that $f$ is of $A_{5}$-type. Then the Dirichlet density $d(\mathcal{P}_{f})$ of the set $\mathcal{P}_{f}$ is given by

[TABLE]

Theorem 1.12** (Theorem 4.3).**

Suppose that $f$ is of $S_{4}$-type.

• (i)

If $\mathrm{sgn}\circ\overline{\rho}_{f}\neq\chi^{d/2}$ and $\mathrm{ord}_{2}(d)=1$, then $\mathcal{P}_{f}=\emptyset$, i.e., the Hecke field $K_{f}$ cannot be generated, as a $\mathbb{Q}$-algebra, by a single eigenvalue $a_{p}(f)$.

• (ii)

If $\mathrm{sgn}\circ\overline{\rho}_{f}\neq\chi^{d/2}$ and $\mathrm{ord}_{2}(d)\geq 2$, then the Dirichlet density $d(\mathcal{P}_{f})$ of the set $\mathcal{P}_{f}$ is given by

[TABLE]

• (iii)

If $\mathrm{sgn}\circ\overline{\rho}_{f}=\chi^{d/2}$, then the Dirichlet density $d(\mathcal{P}_{f})$ of the set $\mathcal{P}_{f}$ is given by

[TABLE]

Remark 1.13**.**

By the same method as in the proofs of Theorems 1.10, 1.11, and 1.12, the Dirichlet density of primes $p$ satisfying $K_{f}=\mathbb{Q}(\chi)(a_{p}(f))$ can also be computed (see Remark 4.4).

Remark 1.14**.**

In [12, Corollary 1.1], Koo–Stein–Wiese proved that if a non-CM newform $g\in S_{k}(N,\chi)$ of weight $k\geq 2$ has no nontrivial inner twists, then the Hecke field of $g$ is generated, as a $\mathbb{Q}$-algebra, by a single Hecke eigenvalue $a_{p}(g)$ for a set of primes $p$ of density one. The case where $g$ admits a nontrivial inner twist is treated in [12, Corollary 1.4]. Our results, Theorems 1.10, 1.11, and 1.12, can be viewed as a weight one counterpart of their theorem. See also [7] and [13] for related results.

As noted in Remark 1.5, twisting by Dirichlet characters allows the order of the nebentypus of a newform to vary almost arbitrarily. This motivates a restriction to twist-minimal newforms. However, even among twist-minimal newforms within the same twist class, the orders of the nebentypus need not coincide. Accordingly, in each twist class, we would like to single out a twist-minimal newform whose nebentypus has minimal order. To make this notion explicit, we introduce the notion of a strongly minimal newform in §5 (see Definition 5.2). Roughly speaking, a newform $f$ is strongly minimal if it is twist-minimal and its nebentypus $\chi$ is minimally ramified at every prime among all twist-minimal newforms in the same twist class. In §5, we show that the order of the nebentypus of a strongly minimal newform is subject to restrictions.

Theorem 1.15** (Theorem 5.3).**

Suppose that the weight one exotic newform $f$ is strongly minimal.

• (i)

If $f$ is of $A_{4}$-type, then the order $d$ of the nebentypus takes the form $3\cdot 2^{k}$ with $k\geq 1$.

• (ii)

If $f$ is of $A_{5}$-type, then the order $d$ of the nebentypus takes the form $2^{k}\cdot 3^{\delta_{3}}\cdot 5^{\delta_{5}}$ with $k\geq 1$ and $\delta_{3},\delta_{5}\in\{0,1\}$

• (iii)

If $f$ is of $S_{4}$-type, then the order $d$ of the nebentypus takes the form $2^{k}$ or $3\cdot 2^{k}$ with $k\geq 1$.

Furthermore, we show that, when $f$ is of $A_{4}$-type or $S_{4}$-type, there exists a strongly minimal newform realizing each of the possible orders of the nebentypus described above.

Theorem 1.16** (Propositions 5.14, 5.21, and 5.22).**

For any positive integer $k$, there exist strongly minimal newforms with the following types and orders of nebentypus:

• (i)

$A_{4}$-type with nebentypus of order $3\cdot 2^{k}$,

• (ii)

$S_{4}$-type with nebentypus of order $2^{k}$,

• (iii)

$S_{4}$-type with nebentypus of order $3\cdot 2^{k}$.

Remark 1.17**.**

We expect a similar statement to hold for weight one newforms of $A_{5}$-type; however, it seems difficult to establish this using the same arguments as in the proof of Theorem 1.16.

Finally, in §6, we completely determine the group structure of the Galois image $\rho_{f}(G_{\mathbb{Q}})$.

Theorem 1.18** (Theorem 6.7 and Proposition 6.10).**

Let $I$ denote the $2\times 2$ identity matrix. Suppose that $f$ is of $A_{4}$-type.

• (i)

If $\ker(\chi)\not\subset\ker(G_{\mathbb{Q}}\stackrel{{\scriptstyle\overline{\rho}_{f}}}{{\twoheadrightarrow}}A_{4}\twoheadrightarrow A_{4}^{\rm ab})$, then the Galois image $\rho_{f}(G_{\mathbb{Q}})$ is isomorphic to the group

[TABLE]

• (ii)

If $\ker(\chi)\subset\ker(G_{\mathbb{Q}}\stackrel{{\scriptstyle\overline{\rho}_{f}}}{{\twoheadrightarrow}}A_{4}\twoheadrightarrow A_{4}^{\rm ab})$, then the integer $d$ is divisible by $3$, and the Galois image $\rho_{f}(G_{\mathbb{Q}})$ is isomorphic to the group

[TABLE]

Here, the homomorphism $\operatorname{SL}_{2}(\mathbb{F}_{3})\twoheadrightarrow\mathbb{Z}/3\mathbb{Z}$ used in the fiber product is defined as the composition

[TABLE]

Theorem 1.19** (Theorem 6.8).**

Let $I$ denote the $2\times 2$ identity matrix. If $f$ is of $A_{5}$-type, then the Galois image $\rho_{f}(G_{\mathbb{Q}})$ is isomorphic to the group

[TABLE]

Theorem 1.20** (Theorem 6.9 and Proposition 6.14).**

Let $BO_{48}$ be the binary dihedral group of order $48$ and $z\in BO_{48}$ denotes the unique element of order $2$ (see Remark 6.3). Suppose that $f$ is of $S_{4}$-type.

• (i)

If $\chi^{d/2}\neq\operatorname{sgn}\circ\bar{\rho}_{f}$, then the Galois image $\rho_{f}(G_{\mathbb{Q}})$ is isomorphic to the group

[TABLE]

• (ii)

If $\chi^{d/2}=\operatorname{sgn}\circ\bar{\rho}_{f}$, then the Galois image $\rho_{f}(G_{\mathbb{Q}})$ is isomorphic to the group

[TABLE]

Here, the homomorphism $BO_{48}\twoheadrightarrow\mathbb{Z}/2\mathbb{Z}$ used in the fiber product is defined as the composition

[TABLE]

Acknowledgement

The second author would like to thank Takeshi Ogasawara and Koji Shimuzu for helpful suggestions. RS was supported by JSPS KAKENHI Grant Number 24K16886.

2. Preliminaries

2.1. Projective Galois representation

Lemma 2.1** ([6, Lemma 1]).**

For any matrix $g\in\operatorname{GL}_{2}(\mathbb{C})$, if the image of $g$ in $\operatorname{PGL}_{2}(\mathbb{C})$ has order $n$, then

[TABLE]

for some primitive $n$-th root of unity $\zeta_{n}$.

Proof.

Since $\bar{g}$ has order $n$, the ratio of the two eigenvalues of the matrix $g$ is a primitive $n$-th root of unity. This fact immediately implies the lemma. ∎

For each prime $p$, let $I_{p}\subset G_{\mathbb{Q}_{p}}$ denote the inertia subgroup at $p$.

Lemma 2.2**.**

The order of a continuous character $\mu\colon G_{\mathbb{Q}}\to\mathbb{C}^{\times}$ is equal to the least common multiple of the positive integers $|\mu(I_{p})|$ for all prime $p$.

Proof.

This lemma is a consequence of the fact that the Galois group of a cyclotomic extension over $\mathbb{Q}$ is generated by its inertia groups. ∎

For any complex 2-dimensional Galois representation $\rho\colon G_{\mathbb{Q}}\to\operatorname{GL}_{2}(\mathbb{C})$, we write $\bar{\rho}\colon G_{\mathbb{Q}}\to\operatorname{PGL}_{2}(\mathbb{C})$ for the projective Galois representation attached to $\rho$.

The following result was proved by Nakazato in [16]. However, for the sake of completeness of the present paper, we include a proof. Our proof of the following lemma is based on that of [18, Theorem 7], which treats the case of a prime conductor.

Lemma 2.3** ([16, Lemma 1]).**

Let $V$ be a $2$-dimensional $\mathbb{C}$-vector space, and let $\rho\colon G_{\mathbb{Q}}\to\operatorname{GL}(V)$ be a continuous irreducible odd Galois representation with square-free conductor $N$. Set $\chi:=\det\rho$. Then the following hold.

• (i)

If $\bar{\rho}(G_{\mathbb{Q}})$ is isomorphic to $A_{4}$, then $\chi$ has order $6$.

• (ii)

If $\bar{\rho}(G_{\mathbb{Q}})$ is isomorphic to $S_{4}$, then $\chi$ has order $2$, $4$, $6$, or $12$

• (iii)

If $\bar{\rho}(G_{\mathbb{Q}})$ is isomorphic to $A_{5}$, then $\chi$ has order $2$, $6$, $10$, or $30$.

Proof.

Let $p$ be a prime dividing $N$. Since $N$ is square-free, we have $\dim_{\mathbb{C}}V^{I_{p}}=1$. This in turn implies that $V|_{I_{p}}\simeq 1\oplus\chi|_{I_{p}}$, and hence $\bar{\rho}(I_{p})\simeq\rho(I_{p})\simeq\chi(I_{p})$. Therefore, Lemma 2.2 shows that

[TABLE]

Note that, in this case, the group $\bar{\rho}(I_{p})$ is cyclic for any prime $p$.

• (i)

Suppose that $\bar{\rho}(G_{\mathbb{Q}})$ is isomorphic to $A_{4}$. Since $\bar{\rho}(I_{p})$ injects into $A_{4}$, the group $\bar{\rho}(I_{p})$ is cyclic of order $1$, $2$, or $3$. Also, because $\rho$ is odd, we have $\mathrm{ord}(\chi)\in\{2,6\}$. We claim that $\mathrm{ord}(\chi)=6$. Since $\bar{\rho}(G_{\mathbb{Q}})$ is isomorphic to $A_{4}$, the fixed field of $\ker\bar{\rho}$ contains a cyclic cubic extension $K/\mathbb{Q}$ corresponding to the Klein four-group in $A_{4}$. Let $q$ be a prime ramified in $K$. Then $q\mid N$ and $\bar{\rho}(I_{q})$ surjects onto $\operatorname{Gal}(K/\mathbb{Q})$. Thus $3$ divides $|\bar{\rho}(I_{q})|$ and so the order of $\chi$ is $6$.

• (ii)

Suppose that $\bar{\rho}(G_{\mathbb{Q}})$ is isomorphic to $S_{4}$. Since $\bar{\rho}(I_{p})$ injects into $S_{4}$, the group $\bar{\rho}(I_{p})$ is cyclic of order $1$, $2$, $3$, or $4$. Combined with the fact that $\chi$ has even order, this implies that $\mathrm{ord}(\chi)$ is $2$, $4$, $6$, or $12$.

• (iii)

Suppose that $\bar{\rho}(G_{\mathbb{Q}})$ is isomorphic to $A_{5}$. Since $\bar{\rho}(I_{p})$ injects into $A_{5}$, the group $\bar{\rho}(I_{p})$ is cyclic of order $1$, $2$, $3$, or $5$. Since $\chi$ has even order, it follows that $\mathrm{ord}(\chi)$ is $2$, $6$, $10$, or $30$.

∎

2.2. Density of $Q_{m}(\chi)$

For any positive integer $N$, using the canonical isomorphism

[TABLE]

we identify, as usual, a Dirichlet character $\chi\colon\mathbb{Z}/N\mathbb{Z}\to\mathbb{C}$ of conductor dividing $N$ with a Galois character $\chi\colon\operatorname{Gal}(\mathbb{Q}(\zeta_{N})/\mathbb{Q})\to\mathbb{C}^{\times}$.

Definition 2.4**.**

For any integer $N$, let $\mathcal{P}_{N}$ denote the set of all primes that are coprime to $N$. For any positive integer $m$ and any character $\chi\colon\operatorname{Gal}(\mathbb{Q}(\zeta_{N})/\mathbb{Q})\to\mathbb{C}^{\times}$, we define a set $Q_{m}(\chi)$ of primes by

[TABLE]

Throughout, when $\chi$ is clear from context, we denote $Q_{m}$ simply by omitting $\chi$.

Lemma 2.5**.**

Let $\chi\colon\operatorname{Gal}(\mathbb{Q}(\zeta_{N})/\mathbb{Q})\to\mathbb{C}^{\times}$ be a character of order $d$, and take a prime divisor $\ell$ of $d$. Write $d=\ell^{e}d^{\prime}$ with $(\ell,d^{\prime})=1$. Then, the Dirichlet density $d(Q_{\ell^{e}}(\chi))$ of $Q_{\ell^{e}}(\chi)$ is $1-\frac{1}{\ell}$.

Proof.

Inside $\mathbb{Z}/d\mathbb{Z}\simeq\mathbb{Z}/\ell^{e}\mathbb{Z}\times\mathbb{Z}/d^{\prime}\mathbb{Z}$, there are $(\ell^{e}-\ell^{e-1})d^{\prime}$ elements of order divisible by $\ell^{e}$. By applying the Chebotarev density theorem to the fixed field of $\ker(\chi)$, we obtain

[TABLE]

as desired. ∎

3. Hecke field $K_{f}$

Let $f=\sum_{n\geq 1}a_{n}(f)q^{n}\in S_{1}(N,\chi)$ be a weight one newform of level $N$ and nebentypus $\chi$ with $\chi(-1)=-1$, and

[TABLE]

denotes the continuous irreducible odd Galois representation associated with $f$. Recall that we write $d$ for the order of $\chi$, which is even since $\chi(-1)=-1$.

Definition 3.1**.**

For any positive integer $m$, we define a set $R_{m}$ of primes (depending on $f$) by

[TABLE]

Lemma 3.2**.**

For any prime $p\in\mathcal{P}_{N}$, we have

[TABLE]

Proof.

Lemma 2.1, together with the properties of the Galois representation $\rho_{f}$, shows that

[TABLE]

When $p\in R_{1}\sqcup R_{2}\sqcup R_{3}\sqcup R_{4}$, this lemma is an immediate consequence of this formula. Let us consider the case where $p\in R_{5}$. Since $\sqrt{\frac{3\pm\sqrt{5}}{2}}=\frac{\sqrt{5}\pm 1}{2}$, we obtain

[TABLE]

Moreover,

[TABLE]

so that $\sqrt{5}\in\mathbb{Q}(a_{p}(f))$. It then follows that

[TABLE]

and hence $\mathbb{Q}(a_{p}(f))=\mathbb{Q}(\sqrt{5},\sqrt{\chi(p)})$. ∎

3.1. $A_{4}$-case

In this subsection, we determine the Hecke fields of weight one newforms of $A_{4}$-type.

Theorem 3.3**.**

If $f$ is of $A_{4}$-type, then $K_{f}=\mathbb{Q}(\zeta_{2d})$.

Proof.

Since $f$ is of $A_{4}$-type, for any prime $p\in\mathcal{P}_{N}$, the element $\bar{\rho}_{f}(\mathrm{Frob}_{p})\in\operatorname{PGL}_{2}(\mathbb{C})$ has order $1$, $2$, or $3$. Thus we have $\mathcal{P}_{N}=R_{1}\sqcup R_{2}\sqcup R_{3}$. As $d$ denotes the order of $\chi$, it follows from Lemma 3.2 that $K_{f}\subset\mathbb{Q}(\zeta_{2d})$.

We now prove the converse inclusion. Applying the Chebotarev density theorem to $M/\mathbb{Q}$, where $M$ is the fixed field of $\ker\bar{\rho}_{f}$, we obtain

[TABLE]

where $d(R_{m})$ denotes the Dirichlet density of $R_{m}$. Let $\ell$ be any prime factor of $d$ and denote by $e:=\mathrm{ord}_{\ell}(d)$ the $\ell$-adic order of $d$. By Lemma 2.5, the set $Q_{\ell^{e}}=Q_{\ell^{e}}(\chi)$ of primes has density $1-\frac{1}{\ell}$, which is greater than $d(R_{2})=\frac{3}{12}$. Thus $Q_{\ell^{e}}\not\subset R_{2}$, and so $Q_{\ell^{e}}\cap(R_{1}\sqcup R_{3})\neq\emptyset$. Considering $a_{p}(f)$ for any prime $p\in Q_{\ell^{e}}\cap(R_{1}\sqcup R_{3})$, we have, from Lemma 3.2,

[TABLE]

with $\ell^{e}\mid\operatorname{ord}(\chi(p))$. In particular, $\mathbb{Q}(\zeta_{2\ell^{e}})\subset K_{f}$. Since $\ell$ is an arbitrary prime factor of $d$, it follows that $\mathbb{Q}(\zeta_{2d})\subset K_{f}$. ∎

Corollary 3.4**.**

Suppose that $f$ is of $A_{4}$-type and the level $N$ of $f$ is square-free. Then $d=6$ and $K_{f}=\mathbb{Q}(\zeta_{12})$.

Proof.

By [8, Théorème 4.1], the Galois representation $\rho_{f}$ has conductor $N$, and so Lemma 2.3 implies that $\chi$ has order $6$. Therefore, the assertion follows from Theorem 3.3. ∎

3.2. $A_{5}$-case

In this subsection, we determine the Hecke fields of weight one newforms of $A_{5}$-type.

Theorem 3.5**.**

*If $f$ is of $A_{5}$-type, then $K_{f}=\mathbb{Q}(\zeta_{2d},\sqrt{5})$. *

Proof.

Since $f$ is of $A_{5}$-type, the element $\bar{\rho}_{f}(\mathrm{Frob}_{p})\in\operatorname{PGL}_{2}(\mathbb{C})$ has order $1$, $2$, $3$, or $5$. Hence we have $\mathcal{P}_{N}=R_{1}\sqcup R_{2}\sqcup R_{3}\sqcup R_{5}$, and Lemma 3.2 shows that $K_{f}\subset\mathbb{Q}(\zeta_{2d},\sqrt{5})$.

Let us prove the converse inclusion. The Chebotarev density theorem implies that

[TABLE]

In particular, $R_{5}\neq\emptyset$, and we have $\sqrt{5}\in K_{f}$ by Lemma 3.2. Let $\ell$ be any prime factor of $d$ and denote by $e:=\mathrm{ord}_{\ell}(d)$ the $\ell$-adic order of $d$. By Lemma 2.5, the set $Q_{\ell^{e}}$ of primes has density $1-\frac{1}{\ell}$, which is greater than $d(R_{2})=\frac{15}{60}$. Thus $Q_{\ell^{e}}\cap(R_{1}\sqcup R_{3}\sqcup R_{5})\neq\emptyset$. Considering $a_{p}$ for any prime $p\in Q_{\ell^{e}}\cap(R_{1}\sqcup R_{3}\sqcup R_{5})$, we have, from Lemma 3.2,

[TABLE]

Since $\ell$ is an arbitrary prime factor of $d$, it follows that $\zeta_{2d}\in K_{f}$. Therefore, we conclude that $K_{f}=\mathbb{Q}(\zeta_{2d},\sqrt{5})$. ∎

Corollary 3.6**.**

Suppose that $f$ is of $A_{5}$-type and the level $N$ of $f$ is square-free. Then, $d\in\{2,6,10,30\}$ and

[TABLE]

Proof.

By [8, Théorème 4.1], the Galois representation $\rho_{f}$ has conductor $N$, and so Lemma 2.3 implies that $d\in\{2,6,10,30\}$. Therefore, this corollary follows immediately from Theorem 3.5. Here, note that $\sqrt{5}\in\mathbb{Q}(\zeta_{5})$. ∎

3.3. $S_{4}$-case

In this subsection, we classify the Hecke fields of weight one newforms of $S_{4}$-type. The idea remains the same as in the cases corresponding to $A_{4}$ and $A_{5}$; however, the situation becomes more complicated due to the existence of order-$4$ elements in the projective image of the Galois representation $\rho_{f}$.

Theorem 3.7**.**

Let $k:=\mathrm{ord}_{2}(d)\geq 1$ denote the $2$-adic valuation of the even integer $d$. If $f$ is of $S_{4}$-type, then the following hold:

• (i)

If $k=1$, then $K_{f}=\mathbb{Q}(\zeta_{d},\sqrt{-2})$ or $\mathbb{Q}(\zeta_{4d})$.

• (ii)

If $k=2$, then $K_{f}=\mathbb{Q}(\zeta_{d})$ or $\mathbb{Q}(\zeta_{2d})$.

• (iii)

If $k\geq 3$, then $K_{f}=\mathbb{Q}(\zeta_{2d})$.

Proof.

Since $f$ is of $S_{4}$-type, the element $\bar{\rho}_{f}(\mathrm{Frob}_{p})\in\operatorname{PGL}_{2}(\mathbb{C})$ has order $1$, $2$, $3$, or $4$. Thus $\mathcal{P}_{N}=R_{1}\sqcup R_{2}\sqcup R_{3}\sqcup R_{4}$, and the Chebotarev density theorem implies that the Dirichlet density of $R_{m}$ for each $m\in\{1,2,3,4\}$ is given by

[TABLE]

Let $d^{\prime}:=d/2^{k}\in\mathbb{Z}$. Then, it follows a priori from Lemma 3.2 that

[TABLE]

First, let us show $\zeta_{d^{\prime}}\in K_{f}$. We may assume that $d^{\prime}>1$ since the case $d^{\prime}=1$ is clear. Take any odd prime divisor $\ell$ of $d^{\prime}$ and denote by $e:=\mathrm{ord}_{\ell}(d^{\prime})$ the $\ell$-adic order of $d^{\prime}$. By Lemma 2.5, the set $Q_{\ell^{e}}$ of primes has the density $1-\frac{1}{\ell}$, which is greater than $\frac{15}{24}=d(R_{2}\sqcup R_{4})$. Hence, $Q_{\ell^{e}}\not\subset R_{2}\sqcup R_{4}$, and consequently $Q_{\ell^{e}}\cap(R_{1}\sqcup R_{3})\neq\emptyset$. Taking $a_{p}(f)$ for $p\in Q_{\ell^{e}}\cap(R_{1}\sqcup R_{3})$, we have, from Lemma 3.2, $\mathbb{Q}(a_{p}(f))=\mathbb{Q}(\sqrt{\chi(p)})$ with $\ell^{e}\mid\operatorname{ord}(\chi(p))$. Since $\ell$ is an arbitrary odd prime factor of $d^{\prime}$, it follows that $\zeta_{d^{\prime}}\in K_{f}$.

We now consider the set $Q_{2^{k}}$. Since $d(Q_{2^{k}})=\frac{1}{2}>d(R_{2})=\frac{9}{24}$ by Lemma 2.5, we have $Q_{2^{k}}\cap(R_{1}\sqcup R_{3}\sqcup R_{4})\neq\emptyset$. Lemma 3.2 shows that

[TABLE]

by considering $a_{p}(f)^{d^{\prime}}$ for at least one prime $p$ in $Q_{2^{k}}\cap(R_{1}\sqcup R_{3})$ or $Q_{2^{k}}\cap R_{4}$.

• (i)

Suppose that $k=1$.

• (a)

When $Q_{2}\cap(R_{1}\sqcup R_{3})\neq\emptyset$, we have $\zeta_{4}\in K_{f}$. If $Q_{2}\cap R_{4}\neq\emptyset$, then $\sqrt{-2}\in K_{f}$, and hence $\sqrt{2}\in K_{f}$ since $\zeta_{4}\in K_{f}$. If $Q_{2}\cap R_{4}=\emptyset$, then primes $p\in(\mathcal{P}_{N}\setminus Q_{2})\cap R_{4}$ yield $\sqrt{2\chi(p)}$ with $\chi(p)$ odd order, and hence $\sqrt{2}\in K_{f}$. In any case, we have $K_{f}=\mathbb{Q}(\zeta_{4},\zeta_{d^{\prime}},\sqrt{2})$, which is equal to $\mathbb{Q}(\zeta_{4d})$ since $d=2d^{\prime}$ and $\zeta_{8}=\pm\frac{1\pm\sqrt{-1}}{\sqrt{2}}$.

• (b)

When $Q_{2}\cap(R_{1}\sqcup R_{3})=\emptyset$, we have $Q_{2}\subset R_{2}\sqcup R_{4}$ and so $\sqrt{-2}\in K_{f}$. Hence, $\mathbb{Q}(\zeta_{d^{\prime}},\sqrt{-2})\subset K_{f}$. Since $[\mathbb{Q}(\zeta_{4},\zeta_{d^{\prime}},\sqrt{2})\colon\mathbb{Q}(\zeta_{d^{\prime}},\sqrt{-2})]=2$, we have $K_{f}=\mathbb{Q}(\zeta_{d},\sqrt{-2})$ or $\mathbb{Q}(\zeta_{4},\zeta_{d^{\prime}},\sqrt{2})=\mathbb{Q}(\zeta_{4d})$. Here, we note $\zeta_{d}=-\zeta_{d^{\prime}}$ since $d=2d^{\prime}$ with $d^{\prime}$ odd.

• (ii)

Suppose that $k=2$. Note that $\mathbb{Q}(\zeta_{2d},\sqrt{2})=\mathbb{Q}(\zeta_{2d})$ since $\sqrt{2}=\pm(\zeta_{8}+\zeta_{8}^{-1})\in\mathbb{Q}(\zeta_{8})\subset\mathbb{Q}(\zeta_{2d})$.

• (a)

When $Q_{4}\cap(R_{1}\sqcup R_{3})\neq\emptyset$, we have $\mathbb{Q}(\zeta_{8})\subset K_{f}$, and hence $\mathbb{Q}(\zeta_{2d})\subset K_{f}$, which must in fact be an equality.

• (b)

When $Q_{4}\cap(R_{1}\sqcup R_{3})=\emptyset$, we have $Q_{4}\cap R_{4}\neq\emptyset$, and hence $\mathbb{Q}(\zeta_{4})=\mathbb{Q}(\sqrt{2}\zeta_{8})\subset K_{f}$. In this case, $\mathbb{Q}(\zeta_{d})=\mathbb{Q}(\zeta_{4},\zeta_{d^{\prime}})\subset K_{f}$, and so $K_{f}=\mathbb{Q}(\zeta_{d})$ or $\mathbb{Q}(\zeta_{2d})$.

• (iii)

Suppose that $k\geq 3$. In this case, we observe that $2\zeta_{2^{k}}=(\sqrt{2}\,\zeta_{2^{k+1}})^{2}\in K_{f}$ and that $\sqrt{2}\in\mathbb{Q}(\zeta_{8})\subset\mathbb{Q}(\zeta_{2^{k}})$. It then follows from $Q_{2^{k}}\cap(R_{1}\sqcup R_{3}\sqcup R_{4})\neq\emptyset$ that $\zeta_{2^{k+1}}\in K_{f}$, and therefore $K_{f}=\mathbb{Q}(\zeta_{2^{k+1}},\zeta_{d^{\prime}})=\mathbb{Q}(\zeta_{2d})$.

∎

Corollary 3.8**.**

Suppose that $f$ is of $S_{4}$-type and the level $N$ of $f$ is square-free. Then $d\in\{2,4,6,12\}$ and

[TABLE]

Proof.

By [8, Théorème 4.1] and Lemma 2.3, we have $d\in\{2,4,6,12\}$. Therefore, this corollary follows immediately from Theorem 3.7. ∎

3.4. Refinement of the $S_{4}$-case

Throughout this subsection, we assume that the newform $f$ is of $S_{4}$-type. Since in this case the projective image $\bar{\rho}_{f}(G_{\mathbb{Q}})$ is isomorphic to $S_{4}$, we obtain a quadratic character

[TABLE]

We denote this quadratic character by $\operatorname{sgn}\circ\bar{\rho}$.

Theorem 3.9**.**

Suppose that $f$ is of $S_{4}$-type.

• (i)

If $\chi^{d/2}\neq\operatorname{sgn}\circ\bar{\rho}$ as $G_{\mathbb{Q}}$-representations, then

[TABLE]

• (ii)

If $\chi^{d/2}=\operatorname{sgn}\circ\bar{\rho}$ as $G_{\mathbb{Q}}$-representations, then

[TABLE]

The proof of Theorem 3.9(i) is given in §3.4.2, and that of Theorem 3.9(ii) is given in §3.4.3.

Corollary 3.10**.**

Suppose that $f$ is of $S_{4}$-type and the level $N$ of $f$ is square-free. Then, $d\in\{2,4,6,12\}$ and the following hold.

• (i)

If $\chi^{d/2}\neq\operatorname{sgn}\circ\bar{\rho}$ as $G_{\mathbb{Q}}$-representations, then

[TABLE]

• (ii)

If $\chi^{d/2}=\operatorname{sgn}\circ\bar{\rho}$ as $G_{\mathbb{Q}}$-representations, then

[TABLE]

Proof.

This is an immediate consequence of Lemma 2.3 and Theorem 3.9. ∎

Remark 3.11**.**

Consider here the case of prime conductor. Let $\rho:G_{\mathbb{Q}}\to\operatorname{GL}_{2}(\mathbb{C})$ be a continuous irreducible $2$-dimensional Galois representation with prime conductor $p$ such that $\chi=\det\rho$ is odd. Assume that $\rho$ is not dihedral. It was shown by Serre in [18, Theorem 7] that

• (a)

$p\not\equiv 1\pmod{8}$;

• (b)

if $p\equiv 5\pmod{8}$, then $\rho$ is of type $S_{4}$ (i.e., $\bar{\rho}(G_{\mathbb{Q}})\simeq S_{4}$), and $\chi$ has order $4$ and conductor $p$;

• (c)

if $p\equiv 3\pmod{4}$, then $\rho$ is of type $S_{4}$ or $A_{5}$, and $\chi$ is the Legendre symbol $n\mapsto\left(\dfrac{n}{p}\right)$.

In addition, Serre also proves the following on [18, page 250]: The image $\rho(G_{\mathbb{Q}})$ consists of all elements $s\in\operatorname{GL}_{2}(\mathbb{C})$ whose image $\bar{s}\in\operatorname{PGL}_{2}(\mathbb{C})$ lies in $\bar{\rho}(G_{\mathbb{Q}})$ such that

• •

$\det(s)^{2}=\operatorname{sgn}(\bar{s})$ if $p\equiv 5\pmod{8}$;

• •

$\det(s)=\operatorname{sgn}(\bar{s})$ if $p\equiv 3\pmod{4}$ and $\rho$ is of type $S_{4}$;

• •

$\det(s)=\pm 1$ if $p\equiv 3\pmod{4}$ and $\rho$ is of type $A_{5}$.

Hence if the newform $f$ is of $S_{4}$-type and the level $N=p$ is a prime, then $f$ satisfies the assumption of Theorem 3.9(ii), and we conclude that

[TABLE]

3.4.1. Preliminaries for the proof of Theorem 3.9

Before proving Theorem 3.9, we introduce a bit more notation and make a few observations.

Definition 3.12**.**

For any finite order character $\psi\colon G_{\mathbb{Q}}\to\mathbb{C}^{\times}$ of conductor dividing $N$ and $c\in\mathbb{C}^{\times}$, we define the set $\mathcal{P}_{N}(\psi=c)$ of primes by

[TABLE]

Since $f$ is of $S_{4}$-type, recall that $\mathcal{P}_{N}=R_{1}\sqcup R_{2}\sqcup R_{3}\sqcup R_{4}$, with Dirichlet densities

[TABLE]

The set $R_{2}$ can be further decomposed as $R_{2}=R_{2}^{+}\sqcup R_{2}^{-}$, where

[TABLE]

The corresponding Dirichlet densities are given by

[TABLE]

The following two lemmas follow immediately from the definitions.

Lemma 3.13**.**

We have

[TABLE]

Let $k:=\mathrm{ord}_{2}(d)$ denote the $2$-adic order of $d$.

Lemma 3.14**.**

$Q_{2^{k}}=\mathcal{P}_{N}(\chi^{d/2}=-1)$.

3.4.2. Proof of Theorem 3.9(i)

We assume that $\chi^{d/2}\neq\mathrm{sgn}\circ\bar{\rho}_{f}$. From the proof of Theorem 3.7 (see in particular (i–a) and (ii–a) in the proof), it suffices to show that $Q_{2^{k}}\cap(R_{1}\sqcup R_{3})\neq\emptyset$.

Let $M$ be the fixed field of $\ker(\chi^{d/2})\cap\ker(\mathrm{sgn}\circ\bar{\rho}_{f})$. Since $\chi^{d/2}\neq\mathrm{sgn}\circ\bar{\rho}_{f}$ by assumption, it follows that $M/\mathbb{Q}$ is a Galois extension and

[TABLE]

Hence the Chebotarev density theorem, together with Lemmas 3.13 and 3.14, implies that

[TABLE]

Since $d(R_{2}^{+})=3/24$, we deduce that $d(Q_{2^{k}}\cap(R_{1}\sqcup R_{3}))>0$ and in particular, $Q_{2^{k}}\cap(R_{1}\sqcup R_{3})\neq\emptyset$.

3.4.3. Proof of Theorem 3.9(ii)

We assume that $\chi^{d/2}=\mathrm{sgn}\circ\bar{\rho}_{f}$. From the second paragraph of the proof of Theorem 3.7, we obtain that $\zeta_{d^{\prime}}\in K_{f}$ with $d^{\prime}:=d/2^{k}$.

Since $\chi^{d/2}=\operatorname{sgn}\circ\bar{\rho}$ by assumption, Lemmas 3.13 and 3.14 imply that

[TABLE]

Hence, for any prime $p\in\mathcal{P}_{N}\setminus Q_{2^{k}}$, Lemma 3.2 yields

[TABLE]

Moreover, since $R_{4}\subset Q_{2^{k}}$, Lemma 3.2 once again gives, for any prime $p\in R_{4}$,

[TABLE]

Finally, since $a_{p}(f)=0$ for any prime $p\in R_{2}$ by Lemma 2.1, combining these two facts with the decomposition $\mathcal{P}_{N}=(\mathcal{P}_{N}\setminus Q_{2^{k}})\sqcup R_{2}^{-}\sqcup R_{4}$, we deduce that

[TABLE]

4. On a generator of the Hecke field $K_{f}$

Let $f=\sum_{n\geq 1}a_{n}(f)q^{n}\in S_{1}(N,\chi)$ be a weight one exotic newform of level $N$ and nebentypus $\chi$ with $\chi(-1)=-1$. By the proofs of Theorems 3.3, 3.2, and 3.7, we know that

[TABLE]

In this section, we consider the set of primes

[TABLE]

and compute its Dirichlet density. In what follows, let $\varphi$ denote the Euler’s totient function.

Theorem 4.1**.**

Suppose that $f$ is of $A_{4}$-type. Then the Dirichlet density $d(\mathcal{P}_{f})$ of the set $\mathcal{P}_{f}$ is given by

[TABLE]

Proof.

Let $L_{1}$ denote the field corresponding to the open subgroup $\ker\bigl(G_{\mathbb{Q}}\stackrel{{\scriptstyle\overline{\rho}_{f}}}{{\twoheadrightarrow}}A_{4}\bigr)$, and let $L_{2}$ denote the field corresponding to the open subgroup $\ker\bigl(G_{\mathbb{Q}}\stackrel{{\scriptstyle\chi}}{{\twoheadrightarrow}}\mu_{d}\bigr)$, where $\mu_{d}\subset\mathbb{C}$ denotes the group of $d$-th roots of unity. Then we have a surjective homomorphism

[TABLE]

Since $L_{2}/\mathbb{Q}$ is abelian, the extension $L_{1}\cap L_{2}$ over $\mathbb{Q}$ is also abelian. On the other hand,

[TABLE]

It follows that

[TABLE]

Suppose that $\ker(\chi)\not\subset\ker(G_{\mathbb{Q}}\stackrel{{\scriptstyle\overline{\rho}_{f}}}{{\twoheadrightarrow}}A_{4}\twoheadrightarrow A_{4}^{\rm ab})$. In this case, we have $L_{1}\cap L_{2}=\mathbb{Q}$. Hence it follows from (4.1) that there exists a surjective homomorphism

[TABLE]

whose composition with the first (resp. second) projection agrees with $\overline{\rho}_{f}$ (resp. $\chi$). Since $K_{f}=\mathbb{Q}(\zeta_{2d})$ by Theorem 3.3, it follows from Lemma 3.2 that $p\in\mathcal{P}_{f}$ if and only if $\mathrm{pr}_{1}(\widetilde{\rho}_{f}(\mathrm{Frob}_{p}))\in A_{4}$ has order $1$ or $3$ and $\mathrm{pr}_{2}(\widetilde{\rho}_{f}(\mathrm{Frob}_{p}))\in\mu_{d}$ is a generator. The proportion of such elements is $3\varphi(d)/4d$, and therefore, by the Chebotarev density theorem, we have

[TABLE]

Next, suppose that $\ker(\chi)\subset\ker(G_{\mathbb{Q}}\stackrel{{\scriptstyle\overline{\rho}_{f}}}{{\twoheadrightarrow}}A_{4}\twoheadrightarrow A_{4}^{\rm ab})$. Then we have

[TABLE]

It follows from (6.2) that there exists a surjective homomorphism

[TABLE]

whose composition with the first (resp. second) projection agrees with $\overline{\rho}_{f}$ (resp. $\chi$). In this case, for any prime $p\nmid N$, the element $\mathrm{pr}_{1}\bigl(\widetilde{\rho}_{f}(\mathrm{Frob}_{p})\bigr)\in A_{4}$ has order $3$ whenever $\mathrm{pr}_{2}\bigl(\widetilde{\rho}_{f}(\mathrm{Frob}_{p})\bigr)\in\mu_{d}$ is a generator. Since $K_{f}=\mathbb{Q}(\zeta_{2d})$ by Theorem 3.3, it follows from Lemma 3.2 that $p\in\mathcal{P}_{f}$ if and only if $\mathrm{pr}_{2}\bigl(\widetilde{\rho}_{f}(\mathrm{Frob}_{p})\bigr)\in\mu_{d}$ is a generator. Therefore, the proportion of such elements is $\varphi(d)/d$, and by the Chebotarev density theorem,

[TABLE]

∎

Theorem 4.2**.**

Suppose that $f$ is of $A_{5}$-type. Then the Dirichlet density $d(\mathcal{P}_{f})$ of the set $\mathcal{P}_{f}$ is given by

[TABLE]

Proof.

Since the abelianization of $A_{5}$ is trivial, by the same argument as in the proof of Theorem 4.1, there exists a surjective homomorphism

[TABLE]

whose composition with the first (resp. second) projection agrees with $\overline{\rho}_{f}$ (resp. $\chi$). Since $K_{f}=\mathbb{Q}(\zeta_{2d},\sqrt{5})$ by Theorem 3.3, it follows from Lemma 3.2 that $p\in\mathcal{P}_{f}$ if and only if $\mathrm{pr}_{2}(\widetilde{\rho}_{f}(\mathrm{Frob}_{p}))\in\mu_{d}$ is a generator and

[TABLE]

Since the proportion of such elements is $2\varphi(d)/5d$ if $5\nmid d$, and $3\varphi(d)/4d$ if $5\mid d$, it follows from the Chebotarev density theorem that $d(\mathcal{P}_{f})$ equals this proportion. ∎

Theorem 4.3**.**

Suppose that $f$ is of $S_{4}$-type.

• (i)

If $\mathrm{sgn}\circ\overline{\rho}_{f}\neq\chi^{d/2}$ and $\mathrm{ord}_{2}(d)=1$, then $\mathcal{P}_{f}=\emptyset$, i.e., the Hecke field $K_{f}$ is not generated, as a $\mathbb{Q}$-algebra, by a single $a_{p}(f)$.

• (ii)

If $\mathrm{sgn}\circ\overline{\rho}_{f}\neq\chi^{d/2}$ and $\mathrm{ord}_{2}(d)\geq 2$, then the Dirichlet density $d(\mathcal{P}_{f})$ of the set $\mathcal{P}_{f}$ is given by

[TABLE]

• (iii)

If $\mathrm{sgn}\circ\overline{\rho}_{f}=\chi^{d/2}$, then the Dirichlet density $d(\mathcal{P}_{f})$ of the set $\mathcal{P}_{f}$ is given by

[TABLE]

Proof.

If $\mathrm{sgn}\circ\overline{\rho}_{f}\neq\chi^{d/2}$ and $\mathrm{ord}_{2}(d)=1$, then by Theorem 3.9 we have $K_{f}=\mathbb{Q}(\zeta_{4d})$. However, Lemma 3.2 shows that $\zeta_{8}\notin\mathbb{Q}(a_{p}(f))$ since $\mathrm{ord}_{2}(d)=1$. In particular, $\mathbb{Q}(a_{p}(f))\neq K_{f}$ for any prime $p\nmid N$, which proves claim (i).

Let us prove claim (ii). If $\mathrm{sgn}\circ\overline{\rho}_{f}\neq\chi^{d/2}$ and $\mathrm{ord}_{2}(d)\geq 2$, then the same argument as in the proof of Theorem 4.1, there exists a surjective homomorphism

[TABLE]

whose composition with the first (resp. second) projection agrees with $\overline{\rho}_{f}$ (resp. $\chi$). Take a prime $p\nmid N$. Since $K_{f}=\mathbb{Q}(\zeta_{2d})$ by Theorems 3.7 and 3.9, it follows from Lemma 3.2 that $p\in\mathcal{P}_{f}$ if and only if $\mathrm{pr}_{2}(\widetilde{\rho}_{f}(\mathrm{Frob}_{p}))\in\mu_{d}$ is a generator and

[TABLE]

Since the proportion of such elements is $3\varphi(d)/8d$ if $\mathrm{ord}_{2}(d)=2$, and $5\varphi(d)/8d$ if $\mathrm{ord}_{2}(d)\geq 3$, it follows from the Chebotarev density theorem that $d(\mathcal{P}_{f})$ equals this proportion.

Finally, assume that $\mathrm{sgn}\circ\overline{\rho}_{f}=\chi^{d/2}$. An argument identical to that of the proof of Theorem 4.1 yields a surjective homomorphism

[TABLE]

whose composition with the first (resp. second) projection agrees with $\overline{\rho}_{f}$ (resp. $\chi$). Take a prime $p\nmid N$. Note that if $\mathrm{pr}_{2}(\widetilde{\rho}_{f}(\mathrm{Frob}_{p}))\in\mu_{d}$ is a generator, then $\mathrm{sgn}(\overline{\rho}_{f}(\mathrm{Frob}_{p}))=-1$. This shows that $\overline{\rho}_{f}(\mathrm{Frob}_{p})\in S_{4}$ has order $2$ or $4$. Hence when $\mathrm{ord}_{2}(d)\neq 2$, Theorems 3.7 and 3.9, together with Lemma 3.2, imply that $p\in\mathcal{P}_{f}$ if and only if $\mathrm{pr}_{2}(\widetilde{\rho}_{f}(\mathrm{Frob}_{p}))\in\mathbb{Z}/d\mathbb{Z}$ is a generator and $\mathrm{pr}_{1}(\widetilde{\rho}_{f}(\mathrm{Frob}_{p}))\in S_{4}$ has order $4$. Since the proportion of such elements is $\varphi(d)/2d$, it follows from the Chebotarev density theorem that

[TABLE]

When $\mathrm{ord}_{2}(d)=2$, Theorems 3.7 and 3.9 show that $K_{f}=\mathbb{Q}(\zeta_{d})$. In this case, $p\in\mathcal{P}_{f}$ if and only if

• •

$\mathrm{pr}_{2}(\widetilde{\rho}_{f}(\mathrm{Frob}_{p}))\in\mathbb{Z}/d\mathbb{Z}$ is a generator and $\mathrm{pr}_{1}(\widetilde{\rho}_{f}(\mathrm{Frob}_{p}))\in S_{4}$ has order $4$, or

• •

$\mathrm{pr}_{2}(\widetilde{\rho}_{f}(\mathrm{Frob}_{p}))\in\mathbb{Z}/d\mathbb{Z}$ has order $d/2$ and $\mathrm{pr}_{1}(\widetilde{\rho}_{f}(\mathrm{Frob}_{p}))\in S_{4}$ has order $1$ or $3$.

Since the proportion of such elements is $7\varphi(d)/8d$, it follows from the Chebotarev density theorem that $d(\mathcal{P}_{f})=7\varphi(d)/8d$. ∎

Remark 4.4**.**

Since the newform $f$ has nebentypus $\chi$, one should perhaps consider the field $\mathbb{Q}(\chi)(a_{p}(f))$ instead of $\mathbb{Q}(a_{p}(f))$, where $\mathbb{Q}(\chi):=\mathbb{Q}(\mathrm{Im}(\chi))=\mathbb{Q}(\zeta_{d})$. In other words, we consider the Dirichlet density of the set of primes

[TABLE]

By the same argument as in Theorems 4.1, 4.2, and 4.3, the following result holds.

• (i)

If $f$ is $A_{4}$-type, then

[TABLE]

• (ii)

If $f$ is $A_{5}$-type, then

[TABLE]

• (iii)

If $f$ is $S_{4}$-type, then

[TABLE]

5. Strongly minimal newforms

For any Dirichlet character $\psi$, there exists a unique newform $g:=f\otimes\psi$, called the twist of $f$ by $\psi$, characterized by the relation $a_{p}(g)=\psi(p)a_{p}(f)$ for almost all primes $p$. In this case we say that $f$ and $g$ are twist equivalent.

Definition 5.1**.**

A newform is said to be twist-minimal if its level attains the minimal value in its twist class.

As noted in Remark 1.5, twisting by Dirichlet characters allows one to vary the order of the nebentypus of a newform almost arbitrarily. It is therefore natural to restrict our attention to twist-minimal newforms. However, the orders of the nebentypus of twist-minimal newforms that are twist-equivalent need not coincide. Accordingly, among twist-minimal newforms in a given twist class, we shall focus on one whose nebentypus has minimal order.

Definition 5.2**.**

We say that a newform $f$ with nebentypus $\chi$ is strongly minimal if it satisfies the following conditions:

• (i)

$f$* is twist-minimal;*

• (ii)

for every twist-minimal newform $g$ that is twist-equivalent to $f$, with nebentypus $\chi_{g}$, one has $|\chi(I_{p})|\leq|\chi_{g}(I_{p})|$ for every prime $p$.

Note that each twist class contains a strongly minimal newform. By Lemma 2.2, within any twist class, the nebentypus of a strongly minimal newform has the minimal order among the twist-minimal newforms.

Recall that $d$ denotes the order of the nebentypus $\chi$. The following theorems are the main results of this section.

Theorem 5.3**.**

Suppose that the weight one newform $f$ is strongly minimal.

• (i)

If $f$ is of $A_{4}$-type, then $d$ takes the form $3\cdot 2^{k}$ with $k\geq 1$.

• (ii)

If $f$ is of $A_{5}$-type, then $d$ takes the form $2^{k}\cdot 3^{\delta_{3}}\cdot 5^{\delta_{5}}$ with $k\geq 1$ and $\delta_{3},\delta_{5}\in\{0,1\}$.

• (iii)

If $f$ is of $S_{4}$-type, then $d$ takes the form $2^{k}$ or $3\cdot 2^{k}$ with $k\geq 1$.

The proof of this theorem is given in §5.2.

5.1. Local lifting

In this subsection, we fix a prime $p$ and let

[TABLE]

denote a projective Galois representation.

Definition 5.4**.**

We say that a lifting $\rho\colon G_{\mathbb{Q}_{p}}\to\operatorname{GL}_{2}(\mathbb{C})$ of $\bar{\rho}_{p}$ is minimal if the conductor of $\rho$ is minimal within the set of liftings of $\bar{\rho}$ to $\operatorname{GL}_{2}(\mathbb{C})$.

If we have two lifts of $\bar{\rho}_{p}$ to $\operatorname{GL}_{2}(\mathbb{C})$, then they differ only by a twist by a character on $G_{\mathbb{Q}_{p}}$. This fact will be used frequently below without further mention.

We now consider an explicit minimal lift of $\bar{\rho}_{p}$ to $\mathrm{GL}_{2}(\mathbb{C})$. First, recall the following well-known fact.

Proposition 5.5**.**

Two finite subgroups of $\operatorname{PGL}_{2}(\mathbb{C})$ which are isomorphic are conjugate.

Proof.

See, for example, [4, Proposition 4.1]. ∎

Lemma 5.6**.**

If $\bar{\rho}_{p}(G_{\mathbb{Q}_{p}})$ is cyclic, then there is a minimal lifting $\rho_{p}\colon G_{\mathbb{Q}_{p}}\to\operatorname{GL}_{2}(\mathbb{C})$ of $\bar{\rho}_{p}$ such that $\rho_{p}\simeq\det\rho_{p}\oplus\mathbbm{1}$.

Proof.

Let $\bar{\rho}_{p}(\sigma)\in\bar{\rho}_{p}(G_{\mathbb{Q}_{p}})$ be a generator. By Proposition 5.5, the element $\bar{\rho}_{p}(\sigma)$ is conjugate to $\begin{pmatrix}\zeta&0\\ 0&1\end{pmatrix}\bmod{\mathbb{C}^{\times}}$, where $\zeta$ is a root of unity. Hence, one can choose a lift $g\in\mathrm{GL}_{2}(\mathbb{C})$ of $\bar{\rho}_{p}(\sigma)$ which is conjugate to $\begin{pmatrix}\zeta&0\\ 0&1\end{pmatrix}$. Then $\mathrm{ord}(g)=\mathrm{ord}(\bar{\rho}_{p}(\sigma))$, and the homomorphism

[TABLE]

is a desired lift of $\bar{\rho}_{p}$. The minimality of $\rho_{p}$ follows from the fact that $\rho_{p}\simeq\det\rho_{p}\oplus\mathbbm{1}$. ∎

Lemma 5.7**.**

If $\bar{\rho}_{p}(G_{\mathbb{Q}_{p}})\simeq D_{2n}$ for some odd positive integer $n$, then there is a minimal lifting $\rho_{p}\colon G_{\mathbb{Q}_{p}}\to\operatorname{GL}_{2}(\mathbb{C})$ of $\bar{\rho}_{p}$ such that $\det\rho_{p}$ is a quadratic character.

Proof.

Since $\bar{\rho}_{p}(G_{\mathbb{Q}_{p}})\simeq D_{2n}$, we can define a Galois representation $\tilde{\rho}_{p}$ by

[TABLE]

Since $n$ is odd, the dihedral group $D_{2n}$ has trivial center. It follows that the projective image of $\tilde{\rho}_{p}$ is isomorphic to $D_{2n}$. By Proposition 5.5, this projective image is conjugate to $\bar{\rho}_{p}(G_{\mathbb{Q}_{p}})$, so, if needed, we may conjugate $\tilde{\rho}_{p}$ to obtain the desired lift $\rho_{p}$. By construction, the lift $\rho_{p}$ is minimal. ∎

Lemma 5.8**.**

Suppose that $p$ is an odd prime.

• (i)

If $\bar{\rho}_{p}(G_{\mathbb{Q}_{p}})\simeq D_{8}$, then there is a minimal lifting $\rho_{p}\colon G_{\mathbb{Q}_{p}}\to\operatorname{GL}_{2}(\mathbb{C})$ such that $\det\rho_{p}$ is a (ramified) quadratic character.

• (ii)

Let $k:=\mathrm{ord}_{2}(p-1)$. If $\bar{\rho}_{p}(G_{\mathbb{Q}_{p}})\simeq D_{4}$, then there is a minimal lifting $\rho_{p}\colon G_{\mathbb{Q}_{p}}\to\operatorname{GL}_{2}(\mathbb{C})$ such that $\det\rho_{p}$ is a totally ramified character of order $2^{k}$. Furthermore, for any lifting $\rho_{p}^{\prime}$ of $\bar{\rho}_{p}$, the order of $\det\rho_{p}^{\prime}$ is divisible by $2^{k}$.

Proof.

Let us show claim (i). Since $p\geq 3$, the image $\bar{\rho}_{p}(I_{p})$ is cyclic. Moreover, the quotient $\bar{\rho}_{p}(G_{\mathbb{Q}_{p}})/\bar{\rho}_{p}(I_{p})$ is also cyclic. It then follows from $\bar{\rho}_{p}(G_{\mathbb{Q}_{p}})\simeq D_{8}$ that $|\bar{\rho}_{p}(I_{p})|=4$. Let $K$ be the unramified quadratic extension of $\mathbb{Q}_{p}$. Then there exists a totally ramified character $\mu\colon G_{K}\to\mathbb{C}^{\times}$ of order $4$ such that

[TABLE]

Since $p^{2}\equiv 1\pmod{8}$, we can choose a totally ramified character $\tilde{\mu}\colon G_{K}\to\mathbb{C}^{\times}$ with $\tilde{\mu}^{2}=\mu$ and define

[TABLE]

Because $\mu^{\sigma}=\mu^{-1}$, where $\sigma$ is the generator of $\mathrm{Gal}(K/\mathbb{Q}_{p})$, we have $\tilde{\mu}^{\sigma}=\tilde{\mu}^{b}$ for $b\in\{-1,3\}$ and the image $\tilde{\rho}_{p}(G_{\mathbb{Q}_{p}})$ is (conjugate to) the group

[TABLE]

In particular, the projective image of $\tilde{\rho}_{p}$ is isomorphic to $D_{8}$. Hence, by Proposition 5.5, we may conjugate $\tilde{\rho}_{p}$ if necessary to obtain the desired lift $\rho_{p}$. Since $\tilde{\mu}$ is a tamely ramified character that cannot be extended to $G_{\mathbb{Q}_{p}}$, the lifting $\rho_{p}$ is minimal.

Next, we establish claim (ii). We continue to let $K/\mathbb{Q}_{p}$ be the unramified quadratic extension and $\mathrm{Gal}(K/\mathbb{Q}_{p})=\langle\sigma\rangle$. Since $p^{2}\equiv 1\pmod{2^{k+1}}$, one can take a totally ramified character $\mu\colon G_{K}\to\mathbb{C}^{\times}$ of order $2^{k+1}$. We then define

[TABLE]

Since $p\equiv 1\pmod{2^{k}}$ by the definition of the integer $k$, we see that $\mu^{2}$ is the restriction of a character of $G_{\mathbb{Q}_{p}}$, and hence $(\mu^{\sigma})^{2}=\mu^{2}$. Therefore, we have

[TABLE]

and the image $\tilde{\rho}_{p}(G_{\mathbb{Q}_{p}})$ is (conjugate to) the set

[TABLE]

In particular, the projective image of $\tilde{\rho}_{p}$ is isomorphic to $D_{4}$ and by Proposition 5.5, we may conjugate $\tilde{\rho}_{p}$ if necessary to obtain the desired lift. For the same reason as in claim (i), the lifting $\rho_{p}$ is minimal. Since any lifting $\rho_{p}^{\prime}$ of $\bar{\rho}_{p}$ can be written as $\rho_{p}\otimes\psi$, the remaining assertion follows from the facts that $\mathrm{ord}_{2}(p-1)=k$ and that $\det\rho_{p}$ is a totally ramified character on $G_{\mathbb{Q}_{p}}$ of order $2^{k}$. ∎

5.1.1. Global lifting

First, we recall the result of Tate on a lifting of $\bar{\rho}$ to $\operatorname{GL}_{2}(\mathbb{C})$, as presented in [18].

Theorem 5.9** ([18, Theorem 5]).**

For each prime $p$, let $\rho^{\prime}_{p}\colon G_{\mathbb{Q}_{p}}\to\operatorname{GL}_{2}(\mathbb{C})$ be a lifting of $\bar{\rho}|_{G_{\mathbb{Q}_{p}}}$. Assume that $\rho^{\prime}_{p}$ is unramified for all but finitely many $p$. Then there exists a unique lifting $\rho\colon G_{\mathbb{Q}}\to\operatorname{GL}_{2}(\mathbb{C})$ of $\bar{\rho}$ such that $\rho|_{I_{p}}=\rho^{\prime}_{p}|_{I_{p}}$ for all prime $p$.

Lemma 5.10**.**

For any odd prime $p$, the group $\bar{\rho}(G_{\mathbb{Q}_{p}})$ is either cyclic or dihedral.

Proof.

If $\bar{\rho}$ is at most tamely ramified at $p$, then $\bar{\rho}(G_{\mathbb{Q}_{p}})$ is metacyclic, and the classification of finite subgroups of $\mathrm{PGL}_{2}(\mathbb{C})$ implies that it is cyclic or dihedral. Next, suppose that $\bar{\rho}$ is wildly ramified at $p$. In this case, the non-trivial $p$-Sylow subgroup of $\bar{\rho}(G_{\mathbb{Q}_{p}})$ is normal; hence, again by the classification of finite subgroups of $\mathrm{PGL}_{2}(\mathbb{C})$, either $p=2$, or $\bar{\rho}(G_{\mathbb{Q}_{p}})$ is cyclic or dihedral. ∎

Recall that $\rho_{f}\colon G_{\mathbb{Q}}\to\operatorname{GL}_{2}(\mathbb{C})$ denotes the Galois representation associated with the weight one newform $f$.

Proposition 5.11**.**

Let $S$ (resp. $T$) denote the set of odd primes $p$ for which $\bar{\rho}_{f}(G_{\mathbb{Q}_{p}})$ is cyclic (resp. isomorphic to $D_{4}$). Put $t:=\max_{p\in T}\mathrm{ord}_{2}(p-1)$. If $f$ is strongly minimal, then we have

[TABLE]

Proof.

Note that if $p\not\in S\cup T\cup\{2\}$, then $\bar{\rho}_{f}(G_{\mathbb{Q}_{p}})$ is non-abelian dihedral by Lemma 5.10. Moreover, if $\bar{\rho}_{f}(G_{\mathbb{Q}_{p}})$ is non-abelian, then $\bar{\rho}_{f}$ is ramified at $p$, and hence only finitely many primes lie outside $S\cup T$.

For each odd prime $p\not\in S\cup T$, we denote by $\rho_{p}\colon G_{\mathbb{Q}_{p}}\to\operatorname{GL}_{2}(\mathbb{C})$ the minimal lifting of $\bar{\rho}_{f}|_{G_{\mathbb{Q}_{p}}}$ constructed in Lemmas 5.7 and 5.8. Then there exists a character $\psi_{p}$ on $G_{\mathbb{Q}_{p}}$ such that $\rho_{p}\simeq\rho_{f}|_{G_{\mathbb{Q}_{p}}}\otimes\psi_{p}$. Let $\psi$ be the Dirichlet character satisfying $\psi|_{I_{p}}=\psi_{p}|_{I_{p}}$ for any prime $p\not\in S\cup T\cup\{2\}$ and $\psi|_{I_{p}}=\mathbbm{1}$ for any prime $p\in S\cup T\cup\{2\}$. Then by the definition of $\rho_{p}$, the lifting $(\rho_{f}\otimes\psi)|_{G_{\mathbb{Q}_{p}}}$ is minimal for any odd prime $p$. Moreover, since $f$ is strongly minimal, Lemma 2.2 implies that

[TABLE]

When $p\not\in S\cup T\cup\{2\}$, the construction of $\rho_{p}$ shows that $|(\chi\psi^{2})(I_{p})|$ is $1$ or $2$. When $p\in S$, the strong minimality of $\rho_{f}$ together with Lemma 5.6 implies that $(\rho_{f}\otimes\psi)|_{I_{p}}\simeq\chi\psi^{2}|_{I_{p}}\oplus\mathbbm{1}$. In particular, $|\chi(I_{p})|=|\chi\psi^{2}(I_{p})|=|\bar{\rho}_{f}(I_{p})|$. Thus it follows from Lemmas 2.2 and 5.8 that $\mathrm{ord}(\chi)=\operatorname{lcm}\{2^{t},\,|\chi(I_{2})|,\,|\bar{\rho}_{f}(I_{p})|\ (p\in S)\}$. ∎

5.2. Proof of Theorem 5.3

Note that $\chi(-1)=-1$ implies that the order of $\chi$ is even.

5.2.1. Proof of Theorem 5.3(i): $A_{4}$-type case

Suppose that $f$ is strongly minimal and of $A_{4}$-type. By Proposition 5.11, to prove Theorem 5.3(i), it suffices to show that there exists a prime $p$ such that $\bar{\rho}_{f}(G_{\mathbb{Q}_{p}})$ is cyclic and $|\bar{\rho}_{f}(I_{p})|=3$.

Since the image projective $\bar{\rho}_{f}(G_{\mathbb{Q}})$ is isomorphic to $A_{4}$, the fixed field of $\ker(\bar{\rho}_{f})$ contains a cyclic cubic extension $K/\mathbb{Q}$ corresponding to the Klein four-group in $A_{4}$. Let $q$ be a prime ramified in $K$. Then $\bar{\rho}_{f}(I_{q})$ surjects onto $\mathrm{Gal}(K/\mathbb{Q})$. Hence $q$ is an odd prime and $\bar{\rho}_{f}(G_{\mathbb{Q}_{p}})$ is cyclic or dihedral by Lemma 5.10. However, since the only metacyclic subgroups of $A_{4}$ that admit a surjection onto $\mathbb{Z}/3\mathbb{Z}$ are cyclic of order $3$, the group $\bar{\rho}_{f}(G_{\mathbb{Q}_{p}})$ is cyclic of order $3$.

5.2.2. Proof of Theorem 5.3(ii): $A_{5}$-type case

Since any cyclic subgroup of $A_{5}$ has order $1$, $2$, $3$, or $5$, Theorem 5.3(ii) follows from Proposition 5.11.

5.2.3. Proof of Theorem 5.3(iii): $S_{4}$-type case

Since any cyclic subgroup of $S_{4}$ has order $1$, $2$, $3$, or $4$, Theorem 5.3(iii) follows from Proposition 5.11.

5.3. Strongly minimal newforms with a prescribed order of the nebentypus

In Theorem 5.3, we established the necessary conditions for the order of the nebentypus of a strongly minimal newform. We now turn to the question of whether, for each type, there exists a strongly minimal newform whose nebentypus has the specified order. Here we address this problem in the cases of the $A_{4}$-type and the $S_{4}$-type.

5.3.1. $A_{4}$-type case

Let $K:=\mathbb{Q}(\zeta_{9}+\zeta_{9}^{-1})$ denote the maximal totally real subfield of $\mathbb{Q}(\zeta_{9})$. Note that $K/\mathbb{Q}$ is an abelian extension with Galois group $\mathrm{Gal}(K/\mathbb{Q})\simeq\mathbb{Z}/3\mathbb{Z}$ and that the class number of $K$ is $1$. Hence, every ideal in $K$ is principal.

For any local or global field $L$, let $\mathcal{O}_{L}$ denote the ring of integers in $L$.

Lemma 5.12**.**

For any element $r\in\mathcal{O}_{K}\setminus 2\mathcal{O}_{K}$, at least one of the congruences $X^{2}\equiv\pm r\pmod{4\mathcal{O}_{K}}$ admits a solution in $\mathcal{O}_{K}$.

Proof.

The prime $2$ does not split in $K$. Thus, $K_{2}:=K\otimes_{\mathbb{Q}}\mathbb{Q}_{2}$ is the unramified extension of $\mathbb{Q}_{2}$ of degree $3$. It follows that

[TABLE]

Since the group $(1+2\mathcal{O}{K_{2}})/(1+4\mathcal{O}_{K_{2}})$ is generated by $-1$, we deduce that at least one of the congruences $X^{2}\equiv\pm r\pmod{4\mathcal{O}_{K}}$ has a solution. ∎

Let $\sigma$ denote a generator of the Galois group $\operatorname{Gal}(K/\mathbb{Q})$. By Dirichlet’s unit theorem (see [17, Proposition 8.7.2]), there is an isomorphism of $\mathbb{Q}[\operatorname{Gal}(K/\mathbb{Q})]$-modules

[TABLE]

Hence, one can choose units $\epsilon\in\mathcal{O}_{K}^{\times}$ such that

[TABLE]

Note that $K\simeq\mathbb{Q}[x]/(x^{3}-3x-1)$. The cubic equation $x^{3}-3x-1=0$ has two negative roots and one positive root, and these roots are units of norm $1$. Therefore, the unit $\epsilon$ is neither totally positive nor totally negative.

Lemma 5.13**.**

Let $k$ be a positive ingteger. There exist infinitely many primes $q>3$ with $\mathrm{ord}_{2}(q-1)=k$ for which an odd projective Galois representation $\bar{\rho}\colon G_{\mathbb{Q}}\to\operatorname{PGL}_{2}(\mathbb{C})$ (i.e., $\overline{\rho}(c)\neq 1$) exists satisfying the following properties:

• (a)

$\bar{\rho}(G_{\mathbb{Q}})\simeq A_{4}$,

• (b)

$\bar{\rho}$* is unramified outside $3$, $q$, and $\infty$,*

• (c)

$\bar{\rho}(G_{\mathbb{Q}_{3}})=\bar{\rho}(I_{\mathbb{Q}_{3}})\simeq\mathbb{Z}/3\mathbb{Z}$,

• (d)

$\bar{\rho}(G_{\mathbb{Q}_{q}})\simeq D_{4}$.

Proof.

Consider the field

[TABLE]

Since $\operatorname{Gal}(K(\sqrt{\epsilon},\sqrt{\epsilon^{\sigma}})/\mathbb{Q})\simeq A_{4}$ by the choice of the unit $\epsilon$, we have $K(\sqrt{\epsilon},\sqrt{\epsilon^{\sigma}})\cap K(\zeta_{2^{k+1}})=K$, so that

[TABLE]

Therefore, by the Chebotarev density theorem, there are infinitely many primes $q>3$ such that the Frobenius conjugacy class at $q$ is equal to the set

[TABLE]

under the above isomorphism. By construction, $q\equiv 1+2^{k}\pmod{2^{k+1}}$, which implies that $\mathrm{ord}_{2}(q-1)=k$.

The prime $q$ splits completely in $K$, so there exists a prime element $Q\in\mathcal{O}_{K}$ satisfying $\pm q=QQ^{\sigma}Q^{\sigma^{2}}$. Due to the choice of $q$, the Frobenius automorphism $\mathrm{Frob}_{Q}\in\operatorname{Gal}(K(\sqrt{\epsilon},\sqrt{\epsilon^{\sigma}})/K)$ is non-trivial. It follows that among the following congruences, exactly two have solutions (in $\mathcal{O}_{K}$), while the other two do not:

[TABLE]

Since $\epsilon$ is neither totally positive nor totally negative and

[TABLE]

There exists a unit $\epsilon^{\prime}\in\{1,\epsilon,\epsilon^{\sigma},\epsilon^{\sigma^{2}}\}$ such that $(\epsilon^{\prime}Q)^{\sigma}(\epsilon^{\prime}Q)^{\sigma^{2}}$ is neither totally positive nor totally negative and the congruence

[TABLE]

has no solution. By replacing $Q$ with $\epsilon^{\prime}Q$, we may assume that $\epsilon^{\prime}=1$.

Moreover, if necessary, replacing $Q$ with $-Q$ and applying Lemma 5.12 allows us to assume that the congruence $X^{2}\equiv Q\pmod{4\mathcal{O}_{K}}$ has a solution. Note that $Q^{\sigma}Q^{\sigma^{2}}$ remains unchanged.

Under the above preparations, set $L:=K(\sqrt{QQ^{\sigma}},\sqrt{Q^{\sigma}Q^{\sigma^{2}}})$. Since

[TABLE]

the extension $L/\mathbb{Q}$ is Galois, and we have an isomorphism $\mathrm{Gal}(L/\mathbb{Q})\simeq A_{4}$. Hence we obtain the projective Galois representation

[TABLE]

Since $Q^{\sigma}Q^{\sigma^{2}}$ is neither totally positive nor totally negative, the field $L$ is totally imaginary, and hence $\bar{\rho}$ is odd.

Since the congruence $X^{2}\equiv Q\pmod{4\mathcal{O}_{K}}$ has a solution, the prime $2\mathcal{O}_{K}$ is unramified in $L$. Hence condition (b) follows from the construction of the field $L$. Among the metacyclic subgroups of $A_{4}$, those whose order is a multiple of $3$ are isomorphic to $\mathbb{Z}/3\mathbb{Z}$. From this observation, condition (c) follows since $3$ is ramified in $K$. Since the prime $q$ splits completely in $K$, the subgroup $\bar{\rho}(G_{\mathbb{Q}_{q}})$ is contained in $\operatorname{Gal}(L/K)$. The extension $K(\sqrt{QQ^{\sigma}})/K$ is ramified at $Q$. The prime element $Q$ does not split in $K(\sqrt{Q^{\sigma}Q^{\sigma^{2}}})/K$ since the congruence $X^{2}\equiv Q^{\sigma}Q^{\sigma^{2}}\pmod{Q\mathcal{O}_{K}}$ has no solution. These facts imply that $\bar{\rho}(G_{\mathbb{Q}_{q}})=\operatorname{Gal}(L/K)\simeq D_{4}$. ∎

Proposition 5.14**.**

For any positive integer $k$, there exists a strongly minimal newform of $A_{4}$-type with nebentypus of order $3\times 2^{k}$.

Proof.

By Lemma 5.13, there exists a prime $q>3$ with $\mathrm{ord}_{2}(q-1)=k$ and an odd projective Galois representat $\bar{\rho}\colon G_{\mathbb{Q}}\to\operatorname{PGL}_{2}(\mathbb{C})$ satisfying the following properties:

• (a)

$\bar{\rho}(G_{\mathbb{Q}})\simeq A_{4}$,

• (b)

$\bar{\rho}$ is unramified outside $3$, $q$, and $\infty$.

• (c)

$\bar{\rho}(G_{\mathbb{Q}_{3}})=\bar{\rho}(I_{\mathbb{Q}_{3}})\simeq\mathbb{Z}/3\mathbb{Z}$.

• (d)

$\bar{\rho}(G_{\mathbb{Q}_{q}})\simeq D_{4}$,

Since $\bar{\rho}(G_{\mathbb{Q}})\simeq A_{4}$, it follows from a theorem of Langlands [14] that one can choose a strongly minimal newform $f$ satisfying $\bar{\rho}_{f}\simeq\bar{\rho}$. From Lemma 2.2 and condition (b), we obtain

[TABLE]

Lemma 5.6, combined with condition (c), implies $|\chi(I_{3})|=|\bar{\rho}(I_{3})|=3$. Furthermore, condition (d) and Lemma 5.8(ii) imply that$|\chi(I_{q})|=2^{k}$. Therefore, we have $\mathrm{ord}(\chi)=3\times 2^{k}$. ∎

Remark 5.15**.**

By checking LMFDB [15, Newform orbit 2601.1.x.a], one can find an example of a strongly minimal newform of $A_{4}$-type whose nebentypus has order $48$. This example coincides with the one constructed in Proposition 5.14.

An $A_{4}$-type strongly minimal newform whose nebentypus has order $24$ cannot be found in LMFDB ([15]). However, one can check its existence explicitely as follow: Let us take $q:=89$, which is the second smallest prime satisfying $q\equiv\pm 1\pmod{9}$ and $q\equiv 9\pmod{16}$. Let $a\in\mathbb{R}$ be one of the roots of the equation $X^{3}-3X-1=0$. Then we have $K=\mathbb{Q}(a)$ and $\mathcal{O}_{K}=\mathbb{Z}+\mathbb{Z}a+\mathbb{Z}a^{2}$ (see, for example, [15, Number field 3.3.81.1]). Since

[TABLE]

we have a canonical isomorphism

[TABLE]

Moreover, $a+1\in\mathcal{O}_{K}^{\times}$ and satisfies

[TABLE]

Since $13$ is not a quadratic residue modulo $89$, it follows from the proofs of Lemma 5.13 and Proposition 5.14 that the existence of an $A_{4}$-type strongly minimal newform with nebentypus of order $24$ and level $3^{2}\times 89^{2}=71289$ is guaranteed.

5.3.2. $S_{4}$-type case

Let $K_{1}/\mathbb{Q}$ be the field defined by the polynomial $X^{6}-3X^{4}-5X^{3}+52X^{2}-92X+56$ and let $K_{2}/\mathbb{Q}$ be the field defined by the polynomial $X^{6}-X^{5}-23X^{4}-45X^{3}+241X^{2}+1109X+1177$. First, we list the properties of $K_{1}$ and $K_{2}$ that will be used below (see, for example, [15, Number field 6.0.7880599.1 and Number field 6.0.1172648743.1]:

• (A)

The extension $K_{i}/\mathbb{Q}$ is Galois with $\operatorname{Gal}(K_{i}/\mathbb{Q})\simeq S_{3}$.

• (B)

The class number of $K_{i}$ is equal to $3$.

• (C-1)

The primes ramified in $K_{1}$ are precisely $199$. Moreover, there are exactly three primes of $K_{1}$ lying above $199$.

• (C-2)

The primes ramified in $K_{2}$ are precisely $7$ and $43$. Moreover, there are exactly three primes of $K_{2}$ lying above $7$, and exactly two primes of $K_{2}$ lying above $43$.

• (D)

There are exactly two primes of $K_{i}$ lying above $2$.

Note that, for any group isomorphism $\phi\colon S_{3}\stackrel{{\scriptstyle\sim}}{{\to}}\mathrm{Aut}(V_{4})$, the semi-direct product $V_{4}\rtimes_{\phi}S_{3}$ is isomorphic to $S_{4}$. We will construct an $S_{4}$-extension of $\mathbb{Q}$ by building a $V_{4}$-extension of the $S_{3}$-extension $K_{i}/\mathbb{Q}$. Let $\sigma\in\operatorname{Gal}(K_{i}/\mathbb{Q})\simeq S_{3}$ be an element of order $3$.

Lemma 5.16**.**

Let $i\in\{1,2\}$. For any element $r\in\mathcal{O}_{K_{i}}$ which is coprime to $2\mathcal{O}_{K_{i}}$, the congruence $X^{2}\equiv rr^{\sigma}\pmod{4\mathcal{O}_{K_{i}}}$ admits a solution in $\mathcal{O}_{K_{i}}$.

Proof.

By property (D), there exist two distinct prime ideals $I_{1},I_{2}\subset\mathcal{O}_{K_{i}}$ such that $2\mathcal{O}_{K_{i}}=I_{1}I_{2}$. Since the order of $\sigma$ is $3$, we have $I_{j}^{\sigma}=I_{j}$. Hence the solvability of $X^{2}\equiv r\pmod{I_{j}^{2}}$ is equivalent to that of $X^{2}\equiv r^{\sigma}\pmod{I_{j}^{2}}$. Since $(\mathcal{O}_{K}/I_{j}^{2})^{\times}\otimes_{\mathbb{Z}}\mathbb{F}_{2}\simeq\mathbb{F}_{2}$, the congruence $X^{2}\equiv rr^{\sigma}\pmod{I_{j}^{2}}$ admits a solution. ∎

Lemma 5.17**.**

For each integer $i\in\{1,2\}$, the following exact sequence of $\mathbb{Z}[\operatorname{Gal}(K_{i}/\mathbb{Q})]$-modules is split:

[TABLE]

Proof.

Let $F/\mathbb{Q}$ denote the quadratic extension satisfying $F\subset K_{i}$. Then $F=\mathbb{Q}(\sqrt{-199})$ if $i=1$ and $F=\mathbb{Q}(\sqrt{-7})$ if $i=2$. In particular, $\mathcal{O}_{F}^{\times}=\{\pm 1\}$. Since $[K_{i}\colon F]=3$, the norm map $\mathrm{Norm}_{K_{i}/F}\colon\mathcal{O}_{K_{i}}^{\times}\to\mathcal{O}_{F}^{\times}=\{\pm 1\}$ gives the splliting of the injection $\{\pm 1\}\hookrightarrow\mathcal{O}_{K_{i}}^{\times}$. ∎

Lemma 5.17, together with Dirichlet’s unit theorem, implies that one can choose elements $\epsilon_{1},\epsilon_{2}\in\mathcal{O}_{K_{i}}^{\times}$ such that

[TABLE]

Let $\tau$ denote the non-trivial element in $\operatorname{Gal}(K_{i}/\mathbb{Q})$ satisfying $\epsilon_{1}^{\tau}=\epsilon_{1}$. Note that $\tau$ has order $2$ and satisfies $\tau\sigma=\sigma^{2}\tau$.

Lemma 5.18**.**

Let $i\in\{1,2\}$. For any principal ideal $I\subset\mathcal{O}_{K_{i}}$ with $I^{\tau}=I$, there exists an element $s\in\mathcal{O}_{K_{i}}$ such that $s^{\tau}=\pm s$ and $s\mathcal{O}_{K_{i}}=I$.

Proof.

Dirichlet’s unit theorem (see [17, Proposition 8.7.2]) shows that there is an isomorphism $\mathcal{O}_{K_{i}}^{\times}/\{\pm 1\}\simeq\mathbb{Z}^{2}$ of $\operatorname{Gal}(K_{i}/\mathbb{Q})$-modules such that the automorphism $\tau$ acts on $\mathbb{Z}^{2}$ by $\begin{pmatrix}0&1\\ 1&0\end{pmatrix}$. This fact implies that the first cohomology group $H^{1}(\langle\tau\rangle,\mathcal{O}_{K_{i}}^{\times}/\{\pm 1\})$ vanishes. It therefore follows from Lemma 5.17 that the injection $\{\pm 1\}\hookrightarrow\mathcal{O}_{K_{i}}^{\times}$ induces an isomorphism

[TABLE]

Consequently, the exact sequence of $\operatorname{Gal}(K_{i}/\mathbb{Q})$-modules

[TABLE]

gives rise to an exact sequence

[TABLE]

which proves this lemma. ∎

Lemma 5.19**.**

For any positive integer $k$, there exist infinitely many primes $q\neq 199$ with $\mathrm{ord}_{2}(q-1)=k$ for which an odd projective Galois representation $\bar{\rho}\colon G_{\mathbb{Q}}\to\operatorname{PGL}_{2}(\mathbb{C})$ exists satisfying the following properties:

• (a)

$\mathrm{Gal}(L/\mathbb{Q})\simeq S_{4}$,

• (b)

$\bar{\rho}$* is unramified outside $199$, $q$, and $\infty$,*

• (c)

$\bar{\rho}(G_{\mathbb{Q}_{199}})\simeq\mathbb{Z}/2\mathbb{Z}$* or $D_{4}$,*

• (d)

$\bar{\rho}(G_{\mathbb{Q}_{q}})\simeq D_{4}$.

Proof.

Consider the field

[TABLE]

Since $\operatorname{Gal}(K_{1}(\sqrt{\epsilon_{1}},\sqrt{\epsilon_{2}})/\mathbb{Q})\simeq S_{4}$ by the choice of the units $\epsilon_{1}$ and $\epsilon_{2}$, we have $K_{1}(\sqrt{\epsilon_{1}},\sqrt{\epsilon_{2}})\cap K_{1}(\zeta_{2^{k+1}})=K_{1}$, so that

[TABLE]

Therefore, by the Chebotarev density theorem, there are infinitely many odd primes $q\neq 199$ such that the Frobenius conjugacy class at $q$ contains the element

[TABLE]

under the above isomorphism. By construction, $q\equiv 1+2^{k}\pmod{2^{k+1}}$, which implies that $\mathrm{ord}_{2}(q-1)=k$.

Since $\begin{pmatrix}1&2&3&4\end{pmatrix}\not\in V_{4}$, the choice of the prime $q$ implies that there are exactly three primes $\mathfrak{q}_{1},\mathfrak{q}_{2},\mathfrak{q}_{3}$ of $K_{1}$ lying above $q$. We may assume that $\mathfrak{q}_{1}^{\sigma}=\mathfrak{q}_{2}$ and $\mathfrak{q}_{1}^{\tau}=\mathfrak{q}_{1}$.

Since $\mathfrak{q}_{1}^{\tau}=\mathfrak{q}_{1}$, solvability of $X^{2}\equiv\epsilon_{2}\pmod{\mathfrak{q}_{1}}$ in $\mathcal{O}_{K_{1}}$ is equivalent to that of $X^{2}\equiv\epsilon_{2}^{\tau}\pmod{\mathfrak{q}_{1}}$. Hence, the identity

[TABLE]

implies that the congruence

[TABLE]

admits a solution in $\mathcal{O}_{K_{1}}$. Since, by our choice of the rational prime $q$, the prime $\mathfrak{q}_{1}$ does not split completely in $K_{1}(\sqrt{\epsilon_{1}},\sqrt{\epsilon_{2}})$, the congruence

[TABLE]

has no solution in $\mathcal{O}_{K_{1}}$.

By property (B) and Lemma 5.18, there exists an element $Q\in\mathcal{O}_{K_{1}}$ satisfying $Q\mathcal{O}_{K}=\mathfrak{q}_{1}^{3}$ and $Q^{\tau}=\pm Q$. We then have

[TABLE]

By replacing $Q$ with $\epsilon_{1}Q$, the element $Q^{\sigma}$ is replaced by $\epsilon_{2}Q^{\sigma}$. Hence we may assume that the congruence

[TABLE]

admits a solution in $\mathcal{O}_{K_{1}}$. Using the relation $Q^{\sigma\tau}=Q^{\sigma^{2}}$, we obtain

[TABLE]

which implies that $Q^{\tau}=Q$. It therefore follows from the relation $\mathfrak{q}_{1}^{\tau}=\mathfrak{q}_{1}$ that the congruence

[TABLE]

admits a solution in $\mathcal{O}_{K_{1}}$. Finally, by Lemma 5.16, the congruence $X^{2}\equiv QQ^{\sigma}\pmod{4\mathcal{O}_{K_{1}}}$ also has a solution in $\mathcal{O}_{K_{1}}$.

Under the above preparations, set $L:=K_{1}(\sqrt{QQ^{\sigma}},\sqrt{Q^{\sigma}Q^{\sigma^{2}}})$. Since

[TABLE]

the extension $L/\mathbb{Q}$ is Galois, and there is an isomorphism $\mathrm{Gal}(L/\mathbb{Q})\simeq S_{4}$. Hence we obtain the projective Galois representation

[TABLE]

Since $K_{1}$ is totally imaginary, the projective Galois representation $\bar{\rho}$ is odd.

Since the congruence $X^{2}\equiv QQ^{\sigma}\pmod{4\mathcal{O}_{K_{1}}}$ has a solution, the primes dividing $2\mathcal{O}_{K}$ are unramified in $L$. Hence by construction, condition (b) is satisfied. Condition (c) follows from the property (C-1). Since the congruence $X^{2}\equiv Q^{\sigma}\pmod{\mathfrak{q}_{1}}$ admits a solution, we have

[TABLE]

Therefore,

[TABLE]

Since $K_{1,\mathfrak{q}_{1}}/\mathbb{Q}_{q}$ is the unramified quadratic extension, this fact implies condition (d). ∎

Lemma 5.20**.**

For any positive integer $k$, there exist infinitely many primes $q\neq 7,43$ with $\mathrm{ord}_{2}(q-1)=k$ for which an odd projective Galois representation $\bar{\rho}\colon G_{\mathbb{Q}}\to\operatorname{PGL}_{2}(\mathbb{C})$ exists satisfying the following properties:

• (a)

$\bar{\rho}(G_{\mathbb{Q}})\simeq S_{4}$,

• (b)

$\bar{\rho}$* is unramified outside $7$, $43$, $q$, and $\infty$,*

• (c)

$\bar{\rho}(G_{\mathbb{Q}_{43}})=\bar{\rho}(I_{43})\simeq\mathbb{Z}/3\mathbb{Z}$,

• (d)

$\bar{\rho}(G_{\mathbb{Q}_{q}})\simeq D_{4}$.

Proof.

By applying exactly the same method as in the proof of Lemma 5.19, with $K_{1}$ replaced by $K_{2}$, we obtain an odd projective Galois representation

[TABLE]

satisfying conditions (a), (b), and (d). Among the subgroups of $S_{4}$, only those isomorphic to $\mathbb{Z}/3\mathbb{Z}$ admit a surjection onto $\mathbb{Z}/3\mathbb{Z}$. Therefore, property (C-2) implies condition (c). ∎

Proposition 5.21**.**

For any positive integer $k$, there exists a strongly minimal newform of $S_{4}$-type with nebentypus of order $2^{k}$.

Proof.

By Lemma 5.19, there exists an odd prime $q\neq 199$ with $\mathrm{ord}_{2}(q-1)=k$ and an odd projective Galois representat $\bar{\rho}\colon G_{\mathbb{Q}}\to\operatorname{PGL}_{2}(\mathbb{C})$ satisfying the following properties:

• (a)

$\mathrm{Gal}(L/\mathbb{Q})\simeq S_{4}$,

• (b)

$\bar{\rho}$ is unramified outside $199$, $q$, and $\infty$,

• (c)

$\bar{\rho}(G_{\mathbb{Q}_{199}})\simeq\mathbb{Z}/2\mathbb{Z}$ or $D_{4}$,

• (d)

$\bar{\rho}(G_{\mathbb{Q}_{q}})\simeq D_{4}$.

Since $\bar{\rho}(G_{\mathbb{Q}})\simeq S_{4}$, it follows from the theorem of Langlands–Tunnel [20] that one can choose a strongly minimal newform $f$ satisfying $\bar{\rho}_{f}\simeq\bar{\rho}$. From Lemma 2.2 and condition (b), we obtain

[TABLE]

Lemma 5.6 or Lemma 5.8(ii), combined with condition (c), implies $|\chi(I_{199})|=2$. Furthermore, condition (d) and Lemma 5.8(ii) imply that$|\chi(I_{q})|=2^{k}$. Therefore, we have $\mathrm{ord}(\chi)=2^{k}$. ∎

Proposition 5.22**.**

For any positive integer $k$, there exists a strongly minimal newform of $S_{4}$-type with nebentypus of order $3\times 2^{k}$.

Proof.

By Lemma 5.20, there exists an odd prime $q\neq 7,43$ with $\mathrm{ord}_{2}(q-1)=k$ and an odd projective Galois representat $\bar{\rho}\colon G_{\mathbb{Q}}\to\operatorname{PGL}_{2}(\mathbb{C})$ satisfying the following properties:

• (a)

$\bar{\rho}(G_{\mathbb{Q}})\simeq S_{4}$,

• (b)

$\bar{\rho}$ is unramified outside $7$, $43$, $q$, and $\infty$,

• (c)

$\bar{\rho}(G_{\mathbb{Q}_{43}})=\bar{\rho}(I_{43})\simeq\mathbb{Z}/3\mathbb{Z}$,

• (d)

$\bar{\rho}(G_{\mathbb{Q}_{q}})\simeq D_{4}$.

Since $\bar{\rho}(G_{\mathbb{Q}})\simeq S_{4}$, it follows from the theorem of Langlands–Tunnel [20] that one can choose a strongly minimal newform $f$ satisfying $\bar{\rho}_{f}\simeq\bar{\rho}$. From Lemma 2.2 and condition (b), we obtain

[TABLE]

It follows from condition (c) and Lemma 5.6 that $|\chi(I_{43})|=|\bar{\rho}(I_{43})|=3$. Furthermore, condition (d) and Lemma 5.8(ii) together imply that$|\chi(I_{q})|=2^{k}$. Since $\chi$ is tamely ramified at $7$, the order $|\chi(I_{7})|$ is divisible by $6$. Therefore, we have $\mathrm{ord}(\chi)=3\times 2^{k}$. ∎

Remark 5.23**.**

As shown in Theorem 3.7, when $\mathrm{ord}_{2}(d)\leq 2$, there are two possible Hecke fields for newforms of $S_{4}$-type. In the above construction, it is not clear whether there exist strongly minimal newforms realizing each of these Hecke fields. However, by checking LMFDB ([15]), one can easily see that when $d\in\{2,4,6,12\}$, all of the candidates actually occur as the Hecke fields of strongly minimal newforms.

6. The group structure of the image of $\rho_{f}$

In this section, we continue to let $f$ be a weight one exotic newform with nebentypus $\chi$, and let $d$ denote the order of $\chi$. This section is devoted to determining the image of the Galois representation $\rho_{f}\colon G_{\mathbb{Q}}\to\mathrm{GL}_{2}(\mathbb{C})$ associated with $f$, in terms of the type of $f$ and the order $d$ of $\chi$. In the case of square-free level, an analogous result was obtained by Kida and Sudo [11].

6.1. Preliminaries

6.1.1. Settings

Let $G:=\rho_{f}(G_{\mathbb{Q}})\subset\operatorname{GL}_{2}(\mathbb{C})$ denote the image of $\rho_{f}$. Write $K:=G\cap\operatorname{SL}_{2}(\mathbb{C})=\ker(\det|_{G})$. Also, let $\pi\colon\operatorname{GL}_{2}(\mathbb{C})\twoheadrightarrow\operatorname{PGL}_{2}(\mathbb{C})$ denote the natural projection, and set $\overline{G}:=\pi(G)$ and $\overline{K}:=\pi(K)$; i.e., the projective images of $G$ and $K$, respectively.

We write $I:=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$ for the $2$ by $2$ identity matrix. Since $\ker(\pi|_{G})=G\cap\mathbb{C}^{\times}I$ is a finite subgroup of $\mathbb{C}^{\times}$, it is of the form $\mu_{a}I$ for some integer $a\geq 1$, where $\mu_{a}\subset\mathbb{C}^{\times}$ denotes the group of $a$-th roots of unity. Since $\det\colon G/K\stackrel{{\scriptstyle\sim}}{{\to}}\mu_{d}$ is cyclic, $\overline{G}/\overline{K}$ is also cyclic. Finally, we remark on the following lemma; it follows from the fact that the groups $A_{4}$, $S_{4}$, and $A_{5}$ are centerless.

Lemma 6.1**.**

The center of the group $G$ is $\mu_{a}I$.

6.1.2. The structure of the group $K$

We begin by recalling the classical result of Klein on the classification of finite subgroups of $\operatorname{SL}_{2}(\mathbb{C})$ (see, for example, [9]).

Proposition 6.2**.**

The finite subgroups of $\operatorname{SL}_{2}(\mathbb{C})$ are isomorphic to one of the following (the indices indicate the orders of the groups):

• •

the cyclic group $\mathbb{Z}/n\mathbb{Z}$,

• •

the binary dihedral group $BD_{4n}$,

• •

the binary tetrahedral group $BT_{24}\simeq\operatorname{SL}_{2}(\mathbb{F}_{3})$, which is the double cover of $A_{4}$,

• •

the binary octahedral group $BO_{48}$, which is a double cover of $S_{4}$,

• •

the binary icosahedral group $BI_{120}\simeq\operatorname{SL}_{2}(\mathbb{F}_{5})$, which is the double cover of $A_{5}$.

Remark 6.3**.**

The symmetric group $S_{4}$ of degree four has two double covers: one is $BO_{48}$, and the other is $\operatorname{GL}_{2}(\mathbb{F}_{3})$ (note that $\operatorname{PGL}_{2}(\mathbb{F}_{3})\simeq\mathrm{Aut}(\mathbb{P}^{1}(\mathbb{F}_{3}))\simeq S_{4}$). The binary octahedral group $BO_{48}$ has the presentation

[TABLE]

and $BO_{48}/\langle abc\rangle\simeq S_{4}$. The group $BO_{48}$ has no subgroup of order $2$ other than its center $\langle abc\rangle$, whereas the group $\operatorname{GL}_{2}(\mathbb{F}_{3})$ possesses a non-normal subgroup of order $2$. This property allows one to distinguish between the two groups.

Proposition 6.4**.**

We have the following cases:

[TABLE]

where $V_{4}\subset A_{4}$ denotes the Klein four-group, which is isomorphic to $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$. In particular,

[TABLE]

Proof.

Since $\overline{G}$ is non-commutative and $\overline{G}/\overline{K}$ is cyclic, $\overline{K}$ is a non-trivial normal subgroup of $\overline{G}$, and this fact implies the assertion. ∎

Corollary 6.5**.**

We have the following cases:

[TABLE]

Proof.

Since the binary dihedral $BD_{8}$ of order $8$ is isomorphic to the the quaternion group $Q_{8}$, this corollary follows from Propositions 6.2 and 6.4. ∎

Corollary 6.6**.**

We have $-I\in K$. In particular, $K\cap\mu_{a}I=\{\pm I\}$, the integer $a$ is even, and the group $\overline{G}/\overline{K}$ is cyclic of order $2d/a$.

Proof.

Note that $-I$ is the unique element of order $2$ in $\operatorname{SL}_{2}(\mathbb{C})$. Hence, any finite subgroup of $\operatorname{SL}_{2}(\mathbb{C})$ of even order contains $-I$. The remaining assertions are then clear (see the commutative diagram (6.1) below). ∎

In what follows, for each integer $n\geq 1$, the cyclic group $\mathbb{Z}/n\mathbb{Z}$ is identified with $\mu_{n}\subset\mathbb{C}^{\times}$ via the homomorphism $k\bmod n\mapsto\exp(2\pi\sqrt{-1}k/n)$. Then the situation is summarized as the following commutative diagram, all of whose horizontal and vertical lines are exact:

[TABLE]

6.2. Main results on the image of $\rho_{f}$

In this subsection, we explain the main results of §6. Recall that $\pi\colon\mathrm{GL}_{2}(\mathbb{C})\twoheadrightarrow\mathrm{PGL}_{2}(\mathbb{C})$ is the projection map, and that

[TABLE]

Also recall that $d$ is the order of $\chi$, and that $I=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$ denotes the $2\times 2$ identity matrix.

Theorem 6.7**.**

Suppose that $f$ is of $A_{4}$-type.

• (i)

If $\overline{K}=\overline{G}$, then the Galois image $\rho_{f}(G_{\mathbb{Q}})$ is isomorphic to the group

[TABLE]

• (ii)

If $\overline{K}\neq\overline{G}$, then the integer $d$ is divisible by $3$, and the Galois image $\rho_{f}(G_{\mathbb{Q}})$ is isomorphic to the group

[TABLE]

Here, the homomorphism $\operatorname{SL}_{2}(\mathbb{F}_{3})\twoheadrightarrow\mathbb{Z}/3\mathbb{Z}$ used in the fiber product is defined as the composition

[TABLE]

Proof.

Claim (i) follows from Corollary 6.5 and Proposition 6.15 below. Claim (ii) follows from Proposition 6.20 below. ∎

Theorem 6.8**.**

If $f$ is of $A_{5}$-type, then the Galois image $\rho_{f}(G_{\mathbb{Q}})$ is isomorphic to the group

[TABLE]

Proof.

This theorem follows from Corollary 6.5 and Proposition 6.15 below. ∎

Theorem 6.9**.**

Suppose that $f$ is of $S_{4}$-type. Let $BO_{48}$ be the binary dihedral group of order $48$ and $z\in BO_{48}$ denotes the unique element of order $2$ (see Remark 6.3).

• (i)

If $\overline{K}=\overline{G}$, then the Galois image $\rho_{f}(G_{\mathbb{Q}})$ is isomorphic to the group

[TABLE]

• (ii)

If $\overline{K}\neq\overline{G}$, then the Galois image $\rho_{f}(G_{\mathbb{Q}})$ is isomorphic to the group

[TABLE]

Here, the homomorphism $BO_{48}\twoheadrightarrow\mathbb{Z}/2\mathbb{Z}$ used in the fiber product is defined as the composition

[TABLE]

Proof.

Claim (i) follows from Corollary 6.5 and Proposition 6.15 below. Claim (ii) follows from Proposition 6.26 below. ∎

6.3. Criterion for determining whether $\overline{K}=\overline{G}$ or not

When $f$ is of $A_{4}$- or $S_{4}$-type, it may happen that $\overline{K}\neq\overline{G}$. In this subsection, we give a necessary and sufficient condition for the inequality $\overline{K}\neq\overline{G}$ to occur. We also briefly discuss whether this condition can be effectively checked (under the Generalized Riemann Hypothesis).

6.3.1. $A_{4}$-case

Proposition 6.10**.**

Suppose that $\overline{G}\simeq A_{4}$. Let $\varphi\colon G\twoheadrightarrow\mathbb{Z}/3\mathbb{Z}$ be the composite homomorphism

[TABLE]

Then $\overline{K}\neq\overline{G}$ if and only if $\ker\left(\mathrm{det}\colon G\to\mathbb{Z}/d\mathbb{Z}\right)\subset\ker(\varphi)$. In particular, if $3\nmid d$, then we have $\overline{K}=\overline{G}$ and $a=2d$.

Proof.

Assume first that $\overline{K}\neq\overline{G}$; that is, $\overline{K}\simeq V_{4}$ by Proposition 6.4. In this case, the composite $G\stackrel{{\scriptstyle\pi}}{{\twoheadrightarrow}}\overline{G}\twoheadrightarrow\overline{G}/\overline{K}\simeq\mathbb{Z}/3\mathbb{Z}$ is identified with the homomorphism $\varphi$. Since the homomorphism $G\twoheadrightarrow\overline{G}/\overline{K}$ factors through $G\stackrel{{\scriptstyle\det}}{{\twoheadrightarrow}}G/K$, we obtain $\ker(\mathrm{det})\subset\ker(\varphi)$. When $\overline{K}=\overline{G}$, we have $\varphi(K)=\varphi(G)=\mathbb{Z}/3\mathbb{Z}$, since $\varphi$ factors through $\pi$ by definition. In particular, $\ker(\det)=K\not\subset\ker(\varphi)$. The remaining assertion follows from Corollary 6.6. ∎

Next, we discuss whether the equivalent conditions for $\overline{K}\neq\overline{G}$ established in Proposition 6.10 can be effectively verified.

Suppose that $\overline{G}\simeq A_{4}$ and $3\mid d$. We then have the two Dirichlet characters of order $3$

[TABLE]

Note that Lemma 2.1 shows that $\chi^{\prime}(\mathrm{Frob}_{p})=1$ if and only if $a_{p}(f)=0$ for any prime $p\nmid N$, where $N$ denotes the level of $f$. Since the homomorphism $\rho_{f}\colon G_{\mathbb{Q}}\twoheadrightarrow G$ is surjective by definition, it follows from Proposition 6.10 that $\overline{K}\neq\overline{G}$ if and only if

[TABLE]

Lemma 6.11**.**

Let $L/\mathbb{Q}$ be a Galois extension of degree $9$ unramified outside $N$. Let $\mathrm{rad}(N)$ denote the positive integer obtained as the product of the primes dividing $N$, that is,

[TABLE]

For any element $\sigma\in\mathrm{Gal}(L/\mathbb{Q})$, there exists a prime $p\nmid N$ such that $p\leq(3^{18}\cdot\mathrm{rad}(N)^{8})^{12577}$ and $\mathrm{Frob}_{p}=\sigma$ in $\mathrm{Gal}(L/\mathbb{Q})$. Moreover, if we assume the Generalized Riemann Hypothesis, then one can replace the inequality $p\leq(3^{18}\cdot\mathrm{rad}(N)^{8})^{12577}$ with the inequality

[TABLE]

Proof.

Since $L/\mathbb{Q}$ is a Galois extension of degree $9$ unramified outside $N$, its discriminant satisfies

[TABLE]

Hence, this lemma follows from effective versions of the Chebotarev density theorem proved by Ahn and Kwon [2, Theorem 1.1] and by Bach and Sorenson [3, Theorem 5.1]. ∎

Proposition 6.12**.**

Suppose that $\overline{G}\simeq A_{4}$ and that $3\mid d$. Let

[TABLE]

where $N$ denotes the level of $f$. Then the following are equivalent:

• (a)

$\overline{K}\neq\overline{G}$.

• (b)

$\ker\left(\chi\colon G_{\mathbb{Q}}\twoheadrightarrow\mu_{d}\right)\subset\ker(G_{\mathbb{Q}}\stackrel{{\scriptstyle\bar{\rho}_{f}}}{{\to}}A_{4}\twoheadrightarrow A_{4}^{\rm ab})$.

• (c)

$\{\,p<M\mid(p,N)=1\text{ and }\chi^{d/3}(\mathrm{Frob}_{p})=1\,\}=\{\,p<M\mid(p,N)=1\text{ and }a_{p}(f)=0\,\}$.

Moreover, under the Generalized Riemann Hypothesis, one may replace $M$ by

[TABLE]

Remark 6.13**.**

Given a ($q$-expansion of a) weight one newform $f$ and a prime $p$, the values $a_{p}(f)$ and $\chi(\mathrm{Frob}_{p})=a_{p}(f)^{2}-a_{p^{2}}(f)$ can be checked easily. However, since $3^{18\times 12577}$ has roughly $108000$ digits., it is not realistic to verify the equality in Proposition 6.12(b) by brute-force computation. On the other hand, under the Generalized Riemann Hypothesis, checking the equality in Proposition 6.12(b) is computationally feasible.

Proof.

We have already shown that (a) and (b) are equivalent and that (a) implies (c), so it suffices to prove that (c) implies (a). Assume that $\overline{K}=\overline{G}$. Let $L$ be the number field corresponding to the open subgroup

[TABLE]

By Proposition 6.10, the extension $L/\mathbb{Q}$ is abelian of degree $9$ and

[TABLE]

Hence Lemma 6.11 implies that there exists a prime $p<M$ such that

[TABLE]

Since the condition $\chi^{\prime}(\mathrm{Frob}_{p})=1$ is equivalent to that $a_{p}(f)=0$, this shows that (c) does not hold, and hence completes the proof. ∎

6.3.2. $S_{4}$-case

By the same argument as in Propositions 6.10 and 6.12, a similar result is obtained in the $S_{4}$-case.

Proposition 6.14**.**

Suppose that $\overline{G}\simeq S_{4}$. Let

[TABLE]

where $N$ denotes the level of $f$. Then the following are equivalent:

• (a)

$\overline{K}=\overline{G}$.

• (b)

$\mathrm{sgn}\circ\overline{\rho}_{f}\neq\chi^{d/2}$* as Dirichlet characters.*

• (c)

There exists a prime $p\nmid N$ such that $p<M$ and $a_{p}(f)^{d}=-1$.

Moreover, under the Generalized Riemann Hypothesis, one may replace $M$ by

[TABLE]

Proof.

The equivalence between (a) and (b) follows by the same argument as in the proof of Proposition 6.10. Therefore, it remains to show that (b) and (c) are equivalent. Since $\overline{G}\simeq S_{4}$, it follows from Lemma 2.1 that $|a_{p}(f)|=1$ if and only if $\bar{\rho}_{f}(\mathrm{Frob}_{p})\in S_{4}$ has order $3$, in which case

[TABLE]

Thus, if $\mathrm{sgn}\circ\overline{\rho}_{f}=\chi^{d/2}$, then for any prime $p\nmid N$ such that $\bar{\rho}_{f}(\mathrm{Frob}_{p})\in S_{4}$ has order $3$, it follows that

[TABLE]

This proves that (c) implies (b). We now prove the converse. Suppose that $\mathrm{sgn}\circ\overline{\rho}_{f}\neq\chi^{d/2}$. Let $L_{1}$ denote the $S_{3}$-extension of $\mathbb{Q}$ corresponding to the open subgroup

[TABLE]

and let $L_{2}$ denote the quadratic extension of $\mathbb{Q}$ corresponding to the quadratic character $\chi^{d/2}$. The assumption that $\mathrm{sgn}\circ\overline{\rho}_{f}\neq\chi^{d/2}$ then implies that $L_{1}\cap L_{2}=\mathbb{Q}$, and hence

[TABLE]

The effective version of the Chebotarev density theorem proved by Ahn and Kwon [2, Theorem 1.1] (resp. by Bach and Sorenson [3, Theorem 5.1] under the Generalized Riemann Hypothesis) implies that there exists a prime $p\nmid N$ with $p<M$ (resp. $p<16\cdot(11\cdot\mathrm{log}(\mathrm{rad}(N))+12)^{2}$) such that the Frobenius conjugacy class at $p$ in $\mathrm{Gal}(L_{1}L_{2}/\mathbb{Q})\simeq S_{3}\times\{\pm 1\}$ contains $(\begin{pmatrix}1&2&3\end{pmatrix},-1)$. Then the element $\bar{\rho}_{f}(\mathrm{Frob}_{p})\in S_{4}$ has order $3$, and hence $a_{p}(f)^{d}=\chi^{d/2}(p)=-1$. ∎

6.4. The structure of $G$ when $\overline{K}=\overline{G}$

First, we determine the structure of $G$ under the assumption $\overline{K}=\overline{G}$.

Proposition 6.15**.**

Assume that $\overline{K}=\overline{G}$.

• (i)

We have $a=2d$ and $G=K\cdot\mu_{2d}I$.

• (ii)

The group $G$ is isomorphic to the group

[TABLE]

Proof.

It follows from the assumption $\overline{K}=\overline{G}$ that $G=K\cdot\mu_{a}I$ and that $a=2d$ by Corollary 6.6. Claim (ii) follows from Corollary 6.6 together with claim (i) since $\mu_{a}I$ is the center of $G$. ∎

6.5. The structure of $G$ when $\overline{K}\neq\overline{G}$

Let us consider the case $\overline{K}\neq\overline{G}$. In this case, Proposition 6.4 implies that $f$ is either of $A_{4}$-type or $S_{4}$-type. We shall determine the structure of $G$ in this setting.

Note that, by the commutative diagram (6.1), the group $G$ fits into the central extension

[TABLE]

Such central extensions are classified by the second cohomology group $H^{2}(\overline{G},\mathbb{Z}/a\mathbb{Z})$ (see, for example, [17, Theorem 1.2.4] or [21, Theorem 6.6.3]). We first recall the universal coefficients theorem for group cohomology.

Proposition 6.16**.**

Let $H$ be a finite group and $A$ be a finite abelian group. Then there exists an exact sequence of abelian gruops:

[TABLE]

which splits non-canonically.

Proof.

See [10, VI. Theorem 15.1] or [21, (3.6.5)], for example. ∎

Remark 6.17**.**

The homomorphism $\operatorname{Ext}^{1}_{\mathbb{Z}}(H^{\mathrm{ab}},A)\to H^{2}(H,A)$ is defined by sending each extension of abelian groups

[TABLE]

to its pullback along the surjective homomorphism $H\twoheadrightarrow H^{\mathrm{ab}}$, namely,

[TABLE]

6.5.1. $A_{4}$-case

Suppose here that $(\overline{G},\overline{K})=(A_{4},V_{4})$. In this case, it follows from Corollary 6.6 that

[TABLE]

Since $A_{4}^{\rm ab}\simeq\mathbb{Z}/3\mathbb{Z}$ and $H_{2}(A_{4},\mathbb{Z})\simeq\mathbb{Z}/2\mathbb{Z}$ (see, for example, [19, (2.22), Chapter 3]), the exact sequence given in Proposition 6.16 with $H=A_{4}$ and $A=\mathbb{Z}/a\mathbb{Z}$ yields a split exact sequence of abelian grouops

[TABLE]

Write $a=3^{b}c$ with $3\nmid c$ and $c\in 2\mathbb{Z}$. Then the canonical isomorphism $\mathbb{Z}/a\mathbb{Z}\stackrel{{\scriptstyle\sim}}{{\to}}\mathbb{Z}/3^{b}\mathbb{Z}\times\mathbb{Z}/c\mathbb{Z}$ induces the following identifications:

[TABLE]

Consequently, we obtain the following lemma:

Lemma 6.18**.**

The group $H^{2}(A_{4},\mathbb{Z}/2\mathbb{Z})$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$, and the nontrivial element in $H^{2}(A_{4},\mathbb{Z}/2\mathbb{Z})$ corresponds to the central extension

[TABLE]

Moreover, the unique injective homomorphism $\mathbb{Z}/2\mathbb{Z}\hookrightarrow\mathbb{Z}/c\mathbb{Z}$ induces an isomorphism

[TABLE]

For notational simplicity, we set

[TABLE]

We then obtain two central extensions

[TABLE]

together with the canonical isomorphism

[TABLE]

Lemma 6.19**.**

• (i)

We have an isomorphism of groups

[TABLE]

Moreover, the homomorphism $G_{c}\twoheadrightarrow A_{4}$ corresponds to the first projection.

• (ii)

We have an isomorphism of groups

[TABLE]

Moreover, the homomorphism $G_{3}\twoheadrightarrow A_{4}$ corresponds to the first projection.

Proof.

Let us prove claim (i). Let $p_{c}\colon G\twoheadrightarrow G_{c}$ denote the canonical homomorphism. Since $|K|=8$ (by Corollary 6.5) and $|\ker(p_{c})|=3^{b}$ are coprime, the group $K$ is isomorphic to the image $K_{c}:=p_{c}(K)$, i.e.,

[TABLE]

Also, the homomorphism $\pi_{c}\colon G_{c}\twoheadrightarrow A_{4}$ maps the group $K_{c}$ onto the group $\overline{K}=V_{4}$, i.e.,

[TABLE]

If the central extension

[TABLE]

splits, then $G_{c}\simeq\mathbb{Z}/c\mathbb{Z}\times A_{4}$, and hence the group $K_{c}$ embeds into $\mathbb{Z}/c\mathbb{Z}\times V_{4}$. This contradicts the fact that $K_{c}\simeq Q_{8}$ is non-abelian. Thus, the central extension

[TABLE]

dose not split, and it defines the unique nontrivial element in $H^{2}(A_{4},\mathbb{Z}/c\mathbb{Z})$. Hence the assertion follows from Lemma 6.18.

Next, let us prove claim (ii). Put $p_{3}\colon G\twoheadrightarrow G_{3}$ and $\pi_{3}\colon G_{3}\twoheadrightarrow A_{4}$. Since $|K|=8$ and $|\ker(\pi_{3})|=3^{b}$ are coprime, the homomorphism $\pi_{3}$ maps the group $p_{3}(K)$ isomorphically onto the group $\pi(K)=V_{4}$, i.e.,

[TABLE]

Thus we obtain the following commutative diagram with exact rows:

[TABLE]

This is a pullback diagram, that is,

[TABLE]

Since $G_{3}/p_{3}(K)$ is a quotient of the cyclic group $G/K\stackrel{{\scriptstyle\sim}}{{\to}}\mathbb{Z}/d\mathbb{Z}$, it is itself cyclic. Therefore, $G_{3}/p_{3}(K)\simeq\mathbb{Z}/3^{b+1}\mathbb{Z}$, which completes the proof. ∎

We now determine the structure of $G$ in the case $(\overline{G},\overline{K})=(A_{4},V_{4})$.

Proposition 6.20**.**

Suppose that $(\overline{G},\overline{K})=(A_{4},V_{4})$. Then the group $G$ can be expressed as

[TABLE]

where the fiber product is taken over an arbitrary surjection $\operatorname{SL}_{2}(\mathbb{F}_{3})\twoheadrightarrow\mathbb{Z}/3\mathbb{Z}$ and the natural projection $\mathbb{Z}/2d\mathbb{Z}\twoheadrightarrow\mathbb{Z}/3\mathbb{Z}$.

Proof.

Since $3^{b+1}c=3a=2d$, it follows Lemma 6.19 together with the isomorphism (6.3) that

[TABLE]

∎

6.5.2. $S_{4}$-case.

We now assume that $(\overline{G},\overline{K})=(S_{4},A_{4})$. In this case, it follows from Corollary 6.6 that

[TABLE]

Since $S_{4}^{\mathrm{ab}}\simeq\mathbb{Z}/2\mathbb{Z}$ and $H_{2}(S_{4},\mathbb{Z})\simeq\mathbb{Z}/2\mathbb{Z}$ (see [19, (2.21), Chapter 3], for example), the exact sequence given in Proposition 6.16 with $H=S_{4}$ and $A=\mathbb{Z}/d\mathbb{Z}$ yields a split exact sequence

[TABLE]

Since $d$ is an even integer, we have an isomorphism

[TABLE]

Remark 6.21**.**

The following four groups are pairwise non-isomorphic:

[TABLE]

For instance, this can be seen easily by Remark 6.3 and by comparing their abelianizations. It therefore follows that the four elements of $H^{2}(S_{4},\mathbb{Z}/2\mathbb{Z})\simeq\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ are represented by the central extensions associated with these four groups. For each of the four groups $\mathcal{G}$ above, we let $[\mathcal{G}]\in H^{2}(S_{4},\mathbb{Z}/2\mathbb{Z})$ denote the cohomology class represented by the central extension associated with $\mathcal{G}$. Note that

[TABLE]

We denote by $[S_{4}\times_{\mathbb{Z}/2\mathbb{Z}}\mathbb{Z}/2d\mathbb{Z}]$ the nontrivial element of $\operatorname{Ext}^{1}_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/d\mathbb{Z})\subset H^{2}(S_{4},\mathbb{Z}/d\mathbb{Z})$ determined by the central extension

[TABLE]

and by $[G]$ the element of $H^{2}(S_{4},\mathbb{Z}/d\mathbb{Z})$ determined by the central extension (6.2).

We consider the homomorphism

[TABLE]

induced by the injective homomorphism $j\colon\mathbb{Z}/2\mathbb{Z}\hookrightarrow\mathbb{Z}/d\mathbb{Z}$.

Lemma 6.22**.**

The cohomology group $H^{2}(S_{4},\mathbb{Z}/d\mathbb{Z})$ is generated by the classes $[S_{4}\times_{\mathbb{Z}/2\mathbb{Z}}\mathbb{Z}/2d\mathbb{Z}]$ and $j_{*}[BO_{48}]$.

Proof.

The injective homomorphism $j\colon\mathbb{Z}/2\mathbb{Z}\hookrightarrow\mathbb{Z}/d\mathbb{Z}$ induces a commutative diagram with exact rows:

[TABLE]

It follows from Remark 6.21 that

[TABLE]

Since the (image of the) subgroup $\operatorname{Ext}^{1}_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/d\mathbb{Z})$ is generated by the class $[S_{4}\times_{\mathbb{Z}/2\mathbb{Z}}\mathbb{Z}/2d\mathbb{Z}]$, the commutative diagram (6.4) shows that $H^{2}(S_{4},\mathbb{Z}/2d\mathbb{Z})\simeq\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ is generated by the classes $[S_{4}\times_{\mathbb{Z}/2\mathbb{Z}}\mathbb{Z}/2d\mathbb{Z}]$ and $j_{*}[BO_{48}]$. ∎

Remark 6.23**.**

The commutative diagram (6.4) shows the following:

• (i)

If $d/2$ is odd, then the homomorphism $j_{*}$ is an isomorphism.

• (ii)

If $d/2$ is even, then $j_{*}[BO_{48}]=j_{*}[\operatorname{GL}_{2}(\mathbb{F}_{3})]$.

Lemma 6.24**.**

$[G]\neq j_{*}[BO_{48}]$*. *

Proof.

Let $z\in BO_{48}$ denotes the unique element of order $2$ (see Remark 6.3). Suppose, to the contrary, that $[G]=j_{*}[BO_{48}]$, which implies that

[TABLE]

Note that, in this case, the center of $G$ is $\mathbb{Z}/d\mathbb{Z}$ and the subgroup $K\subset G$ corresponds to the subgroup $\operatorname{SL}_{2}(\mathbb{F}_{3})\subset BO_{48}$ by Lemma 6.5. Let

[TABLE]

be the fixed field corresponding to the open subgroup $\ker(\rho_{f})$. Then the homomorphism $\rho_{f}$ induces an isomorphism

[TABLE]

Let $G^{\prime}\subset G$ be the subgroup corresponding to $BO_{48}$. Set $M:=L^{K}$ and $M^{\prime}:=L^{G^{\prime}}$. Since $K\subset G^{\prime}$ and $G^{\prime}/K\simeq\mathbb{Z}/2\mathbb{Z}$, we have

[TABLE]

Moreover, since $\operatorname{Gal}(M/\mathbb{Q})\simeq G/K\simeq\mathbb{Z}/d\mathbb{Z}$, the extension $M/\mathbb{Q}$ is cyclic. Hence $M^{\prime}/\mathbb{Q}$ is the unique subextension of $M/\mathbb{Q}$ of index $2$, and in particular $M^{\prime}$ is totally real. Since $L$ is totally imaginary as $\rho_{f}$ is odd, it follows that for every complex conjugation $c\in G_{\mathbb{Q}}$, the element $\rho_{f}(c)$ belongs to $G^{\prime}\simeq BO_{48}$, i.e.,

[TABLE]

On the other hand, since $\rho_{f}$ is odd, the element $\rho_{f}(c)\in\operatorname{GL}_{2}(\mathbb{C})$ has eigenvalues $1$ and $-1$. In particular, $\rho_{f}(c)$ does not lie in the center of $G^{\prime}$. This contradicts the fact that the center of $BO_{48}$ is the only subgroup of order 2 (see Remark 6.3). ∎

Proposition 6.25**.**

$[G]=j_{*}[BO_{48}]+[S_{4}\times_{\mathbb{Z}/2\mathbb{Z}}\mathbb{Z}/2d\mathbb{Z}]$.

Proof.

Since $\operatorname{SL}_{2}(\mathbb{F}_{3})\simeq K\subset G$ by Lemma 6.5, the group $G$ cannot be isomorphic to either $S_{4}\times\mathbb{Z}/d\mathbb{Z}$ or $S_{4}\times_{\mathbb{Z}/2\mathbb{Z}}\mathbb{Z}/2d\mathbb{Z}$. In particular, $[G]\neq 0$ and $[G]\neq[S_{4}\times_{\mathbb{Z}/2\mathbb{Z}}\mathbb{Z}/2d\mathbb{Z}]$. Since $H^{2}(S_{4},\mathbb{Z}/d\mathbb{Z})\simeq\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$, it follows from Lemmas 6.22 and 6.24 that $[G]=j_{*}[BO_{48}]+[S_{4}\times_{\mathbb{Z}/2\mathbb{Z}}\mathbb{Z}/2d\mathbb{Z}]$. ∎

Proposition 6.26**.**

Suppose that $(\overline{G},\overline{K})=(S_{4},A_{4})$ Then the group $G$ is isomorphic to the group

[TABLE]

Here, $z\in BO_{48}$ denotes the unique element of order $2$ and the homomorphism $BO_{48}\twoheadrightarrow\mathbb{Z}/2\mathbb{Z}$ used in the fiber product is defined as the composition

[TABLE]

Proof.

This proposition follows from Proposition 6.25 (note that the sum of central extensions can be computed using the Baer sum; see [21, Definition 3.4.4] for the definition of the Baer sum). ∎

Remark 6.27**.**

The group $G$ admits the following description in terms of $\operatorname{GL}_{2}(\mathbb{F}_{3})$.

• (i)

If $d/2$ is odd, then Proposition 6.25, together with Remark 6.23(i), implies that

[TABLE]

It follows that

[TABLE]

• (ii)

If $d/2$ is even, then Proposition 6.25, together with Remark 6.23(ii), implies that

[TABLE]

It follows that

[TABLE]