On the Experimental verification of the Artin conjecture for 2-dimensional odd Galois representations over Q

Liftings of 2-dimensional projective Galois representations over Q

Ian Kiming

0. Introduction

0.0. Consider equivalence classes of 2-dimensional, irreducible, continuous, odd Galois representations over ℚ:

ρ : Gal (\overline{ℚ}/ℚ) → GL_{2}(ℂ)

with Artin conductor N ∈ ℕ and determinant character det ρ = ε . Here, ℂ has the discrete topology and Gal (\overline{ℚ}/ℚ) has the natural topology as a profinite group so that “continuous” means “having finite image”. The determinant character ε is the character on Gal (\overline{ℚ}/ℚ) obtained by composing ρ with the determinant homomorphism:

det : GL_{2}(ℂ) → ℂ^{x} .

Then ε is a character on Gal (\overline{ℚ}/ℚ) with class field theoretic conductor dividing N . By class field theory we may identify ε with a Dirichlet character modulo N ; that ρ is \underline{odd} then means that ε(−1) = −1 .

It is conjectured that these equivalence classes are in 1-1 correspondence with the normalized newforms f(z) of weight 1 and nebentype ε on the congruence subgroup Γ_{0}(N) of SL_{2}(ℤ):

Γ_{0}(N) = \{( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ SL_{2}(ℤ) | c ≡ 0 (N)\}

(see for example [10] for definitions). More explicitly, one expects that if the Artin L-series of ρ (which depends only on the equivalence class of ρ ) is:

(∗)  L(s, ρ) = ∑_{n=1}^{∞} a_{n}n^{−s} , Re(s) > 1 ,

then:

(∗∗)  f(z) = ∑_{n=1}^{∞} a_{n}e^{2πinz} , Im(z) > 0 ,

should be a (normalized) newform of weight 1 on Γ_{0}(N) with nebentype ε. This can also be expressed in this way: The Artin L-series of ρ should be the Hecke L-series of the corresponding newform f(z).

Now, a theorem of Deligne and Serre (cf. [4]) states that if f(z) is a normalized newform on Γ_{0}(N) of weight 1 and nebentype ε with Fourier expansion at ∞ as in (**), then there is a Galois representation ρ as above with Artin L-series as in (*). A classical theorem of Hecke (cf. for example [10], chap. 4) then implies that this L-series, enlarged by the usual Γ-factor, has a holomorphic continuation to the whole complex plane, i.e. it satisfies the Artin conjecture. With this, a theorem of Weil (cf. [16], [9] or [10]) shows that the above conjecture conjunctively for all Galois representations of the above type is equivalent to the Artin conjecture for these representations. We also see that the existence, for given N and ε, of the abovementioned conjectural (set-theoretic) 1-1 correspondence implies the existence of a 1-1 correspondence with “preservation of L-functions” as in (*), (**).

0.1. If ρ is a Galois representation of the above type, we may consider its projectivisation:

\overline{ρ} : Gal (\overline{ℚ}/ℚ) → P GL_{2}(ℂ)

obtained by composing ρ with the canonical homomorphism GL_{2}(ℂ) → P GL_{2}(ℂ). As a finite subgroup of P GL_{2}(ℂ) the image of \overline{ρ} is a priori either isomorphic to a cyclic group, a dihedral group, to the alternating group A_{4}, to the symmetric group S_{4}, or to the alternating group A_{5}, where however the cyclic case is excluded since ρ was required to be irreducible. We shall distinguish these cases by saying that ρ is of dihedral, A_{4}- , S_{4}- or A_{5}-type, respectively. Now, if ρ is of dihedral type, then one may by class field theory show that ρ is associated to a newform of weight 1 as in 0.0., i.e. its L-series is the L-series of such a modular form. By results of Langlands and Tunnell (cf. [8] and [15]) this is also the case if ρ is of A_{4}- or S_{4}-type. Hence the question of whether ρ is in this way associated to a newform of weight 1 is only interesting in case that ρ is of A_{5}-type. If this is the case, the above-mentioned methods of associating to ρ a newform of weight 1 fail completely, and so the question of displaying at least \underline{examples} of representations of A_{5}-type whose Artin L-series are L-series of newforms of weight 1 arises. Up till now there was in the literature only 1 such example, namely the example of J.P. Buhler (cf. [3]). How could we produce such examples in a systematic way? Conceptually, there is a simple way of doing that:

Suppose that we are given a natural number N and a Dirichlet character ε with ε(−1) = −1. Denote by S_{1}^{new}(N, ε) the complex vectorspace generated by the newforms of weight 1 and nebentype ε on Γ_{0}(N), and by d(N, ε) the number of equivalence classes of 2-dimensional, complex, irreducible, continuous, odd representations of Gal (\overline{ℚ}/ℚ) with Artin conductor N and determinant character ε. The above-mentioned theorem of Deligne and Serre then tells us that:

dim S_{1}^{new}(N, ε) ≤ d(N, ε) ,

and that if we have equality here, all of these representations are associated with newforms of weight 1 as in 0.0. (and so the enlarged L-series of these representations as well of their “twists” by 1-dimensional characters of Gal (\overline{ℚ}/ℚ) will have holomorphic continuations to the whole complex plane, i.e. they will satisfy the Artin conjecture). If we do not have equality, the Artin conjecture is false.

Hence we should ask ourselves how one can determine the number dim S_{1}^{new}(N, ε) and d(N, ε) for given N and ε. The problem of determining d(N, ε) will be considered in sections 1, 2 and 3 below. For the rest of this section, we shall briefly show how the problem of computing dim S_{1}^{new}(N, ε) can be reduced to analogous “higher weight” problems.

Let k and N be natural numbers and let ε be a Dirichlet character mod N with ε(−1) = (−1)^{k}. The complex vectorspace S_{k}(N, ε) of cusp forms of weight k and nebentype ε on Γ_{0}(N) has a basis consisting of forms all of whose Fourier coefficients (at ∞) are integers; for k ≥ 2 this follows from [14], Th. 3.52, and for k = 1 it follows from the considerations below. Call any such basis an “integral basis”.

Clearly, if one can find an integral basis for S_{1}(N, ε), then one can also determine the subspace generated by the oldforms in it, and hence one can compute dim S_{1}^{new}(N, ε). In order to find such a basis we can, as a first shot, use the well-known trick (cf. [13] or [3], chap. 6) of multiplying with modular forms without a common zero (neither in the upper half plane nor at any cusp):

Let p be a prime number such that N is divisible by 2p, if p ≡ 3 (4), or divisible by 4p, if p ≡ 1 (4); if no such prime number exists, replace N by a suitable multiple for which there does. Consider the Eisenstein series:

E_{p}(z) = −\frac{1}{2}h + ∑_{n=1}^{∞}(∑_{d|n} χ_{−p}(d))e^{2πinz} , Im (z) > 0 ,

where χ_{−p} is the character of the quadratic field ℚ(\sqrt{−p}) and h its class number. The function E_{p}(z) is a modular form of weight 1 and nebentype χ_{−p} on Γ_{0}(M), where:

M = \{ \begin{matrix}p , & if p ≡ 3 (4) \\ 4p , & if p ≡ 1 (4) \end{matrix} .

Consider furthermore the theta-series:

θ_{2}(z) = ∞ ; ∑ ; \begin{aligned}m & = −∞ \\ m & ≡ 1 (2) \end{aligned} e^{πim^{2}z/4} , Im (z) > 0 ;

it is a modular form of weight \frac{1}{2} on Γ_{0}(2) with the multiplier-system:

(+)  v_{2}(( \begin{matrix}a & b \\ c & d \end{matrix} )) = (\frac{c}{d})_{∗}e^{πi(d−1+bd)/4} , for ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ Γ_{0}(2) ,

where:

(\frac{c}{d})_{∗} = (\frac{c}{|d|})σ(c, d) ,

with:

σ(c, d) = \{ \begin{matrix}−1 , & if c < 0 and d < 0 \\ 1 & otherwise \end{matrix} .

Cf. [11], p. 269. We conclude that θ_{2}^{8} is a modular form of weight 4 on Γ_{0}(2) with trivial nebentype. Now, θ_{2} vanishes only at the cusp ∞ (cf. [11], p. 23) and E_{p} is non-zero at ∞. Consequently, the map:

S_{1}(N, ε) → S_{2}(N, εχ_{−p}) × S_{5}(N, ε)

given by:

f ↦ (f · 2E_{p}, fθ_{2}^{8}) ,

maps S_{1}(N, ε) isomorphically onto the subspace:

V ≤ S_{2}(N, εχ_{−p}) × S_{5}(N, ε)

consisting of pairs (f_{1}, f_{2}) with:

(++)  f_{1}θ_{2}^{8} = f_{2} · 2E_{p} in S_{6}(N, εχ_{−p}) .

Since the equation (++) is equivalent to a linear system of equations involving (finitely many) Fourier coefficients of the forms f_{1} and f_{2}, it is clear that if we know how to obtain integral bases for the spaces S_{2}(N, εχ_{−p}) and S_{5}(N, ε), then the problem of computing dim S_{1}(N, ε) or even obtaining an integral basis for the space S_{1}(N, ε) is reduced to a question in linear algebra. An algorithm for obtaining an integral basis for a space S_{k}(·, ·), where k ≥ 2, will be described in IV and V.

0.3. The algorithm described in 0.2. is of course general but has the disadvantage of requiring the construction of an integral basis for a space of cusp forms of weight 5, which may be computationally unmanagable for interesting cases. Here we shall show that in some favorable cases there is an algorithm which only involves computation in a space of cusp forms of weight 2. In this subsection we shall assume that N is a natural number of the form:

N = 2^{s+1}mn ,

where m and n are coprime natural numbers which are both odd and squarefree, and

s ≥ \{ \begin{matrix}1 , & if n ≡ 3 (4) \\ 2 , & if n ≡ 1 (4) \end{matrix} .

Define the function g_{s,n} by:

g_{s,n}(z) = θ_{2}(2^{s}z)θ_{2}(2^{s}nz) , Im (z) > 0 ,

where θ_{2} is the theta-function considered in 0.2. Recall that if f is a non-zero holomorphic modular form of some weight and with a multiplier-system on some subgroup Γ in SL_{2}(ℤ) of finite index and ζ is a cusp of Γ , then f has a well-defined \underline{order} at ζ with respect to Γ (cf. [11], p.12 ). This order depends only on the equivalence class of ζ with respect to Γ .

Lemma: The function g_{s,n} is a modular form on Γ_{0}(2^{s+1}mn) of weight 1 and nebentype χ_{−n}, where χ_{−n} is the character of the quadratic field ℚ(\sqrt{−n}). It vanishes only at cusps (equivalent to a cusp) of the form:

−(2^{s+1}kl)^{−1} , with k|m , l|n ,

and its order at such a cusp −(2^{s+1}kl)^{−1} is:

2^{s−3} · \frac{m}{k} · (\frac{n}{l} + l) .

Proof: For information (equivalence classes, widths) on the cusps of groups Γ_{0}(N) we refer to [11], pp. 241-251.

Let us start by noting that the cusps −(2^{s+1}kl)^{−1} , where k and l run independently through the divisors of m and n respectively, are mutually inequivalent under Γ_{0}(2^{s+1}mn). Here we have used the fact that m and n are coprime.

We shall use the usual notation Γ(μ, ν) for the congruence subgroup:

Γ(μ, ν) = \{( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ SL_{2}(ℤ) | c ≡ 0 (μ) , b ≡ 0 (ν)\} .

Recall that θ_{2} is a holomorphic modular form of weight \frac{1}{2} on Γ_{0}(2) with the multiplier system v_{2} given in 0.2. by (+).

Let us now fix k and l with k|m, l|n . The cusp −(2k)^{−1} is equivalent to ∞ under Γ_{0}(2) , and with respect to Γ(2m, 2^{s}) it has width 2^{s} · \frac{m}{k} . Since θ_{2} has order \frac{1}{8} with respect to Γ_{0}(2) at ∞ , we conclude that θ_{2} has order 2^{s−3} · \frac{m}{k} at −(2k)^{−1} with respect to Γ(2m, 2^{s}). Acting with the matrix:

( \begin{matrix}2^{s} & 0 \\ 0 & 1 \end{matrix} )

on θ_{2}(z) , we find that θ_{2}(2^{s}z) is a form on Γ_{0}(2^{s+1}m) which has order 2^{s−3} · \frac{m}{k} at the cusp −(2^{s+1}k)^{−1} . (Use for example a reasoning along the lines of the argument on p.248 in [11] ). The cusps −(2^{s+1}k)^{−1} and −(2^{s+1}kl)^{−1} are equivalent under Γ_{0}(2^{s+1}m) with width \frac{m}{k} . Since −(2^{s+1}kl)^{−1} has width \frac{mn}{kl} with respect to Γ_{0}(2^{s+1}mn) , we conclude that θ_{2}(2^{s}z) has order 2^{s−3} · \frac{mn}{kl} at the cusp −(2^{s+1}kl)^{−1} with respect to Γ_{0}(2^{s+1}mn) .

Using the fact that m and n are coprime and m is squarefree, we find that −(2^{s+1}k)^{−1} and −\frac{n}{l} · (2^{s+1}k)^{−1} are equivalent under Γ_{0}(2^{s+1}m):

Put α = \frac{n}{l}, β = 2^{s+1}k and M = 2^{s+1}m . Choose u, v ∈ ℤ such that:

−βu − αv = 1 ,

and μ, ν ∈ ℤ such that:

νβ + μ · \frac{M}{β} = β − 1 − v .

Then the matrix:

( \begin{matrix}−α & u \\ β & v \end{matrix} ) ( \begin{matrix}1 & ν \\ 0 & 1 \end{matrix} ) ( \begin{matrix}β − 1 & 1 \\ −β & −1 \end{matrix} )

belongs to Γ_{0}(M) and it maps −\frac{1}{β} to −\frac{α}{β} .

Now, the cusp −(2^{s+1}k)^{−1} has width \frac{m}{k} with respect to Γ_{0}(2^{s+1}m) and the cusp −\frac{n}{l} · (2^{s+1}k)^{−1} has width \frac{m}{k} · l with respect to Γ(2^{s+1}m, n). Since θ_{2}(2^{s}z) has order 2^{s−3} · \frac{m}{k} at −(2^{s+1}k)^{−1} with respect to Γ_{0}(2^{s+1}m) , we find that θ_{2}(2^{s}z) has order 2^{s−3} · \frac{m}{k} · l at −\frac{n}{l}(2^{s+1}k)^{−1} with respect to Γ(2^{s+1}m, n) . Acting on θ_{2}(2^{s}z) with the matrix:

( \begin{matrix}n & 0 \\ 0 & 1 \end{matrix} ) ,

we then find that θ_{2}(2^{s}nz) is a form on Γ_{0}(2^{s+1}mn) which has order 2^{s−3} · \frac{m}{k} · l at −(2^{s+1}kl)^{−1} .

Consequently, g_{s,n} is a modular form of weight 1 on Γ_{0}(2^{s+1}mn). Given the multiplier v_{2} of θ_{2} , one easily finds that the multiplier of g_{s,n} is:

v ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) = (\frac{n}{|δ|}) · (−1)^{\frac{δ−1}{2} + \frac{n+1}{2} ·2^{s−1}βδ} ,

for ( \begin{matrix}α & β \\ γ & δ \end{matrix} ) ∈ Γ_{0}(2^{s+1}mn) . Using the assumption that s ≥ 2 if n ≡ 1 (4) , we find

that this value is:

(\frac{n}{|δ|}) · (−1)^{\frac{δ−1}{2}} ,

and this is precisely χ_{−n}(δ) . Hence g_{s,n} is a form of weight 1 and nebentype χ_{−n} on Γ_{0}(2^{s+1}mn) .

Let us compute the sum of the orders of g_{s,n} with respect to Γ_{0}(2^{s+1}mn) at the cusps (2^{s+1}kl)^{−1} , where k and l run independently through the divisors of m and n respectively. Using the fact that m and n are coprime and both squarefree, we find that this sum is:

\begin{aligned}∑_{k|m} ∑_{l|n} 2^{s−3} · \frac{m}{k} · (\frac{n}{l} + l) & = 2^{s−2} ∑_{k|m} ∑_{l|n} \frac{mn}{kl} \\ & = 2^{s−2}mn ∏_{p|mn}(1 + \frac{1}{p}) , \end{aligned}

where the product is over all prime divisors of mn . Since mn is odd, we see that this sum is precisely \frac{1}{12}[SL_{2}(ℤ) : Γ_{0}(2^{s+1}mn)] . Hence we may conclude that g_{s,n} vanishes \underline{only} at the above cusps.

Q.E.D.

If now ε is a Dirichlet character modulo N with ε(−1) = −1 , the above lemma tells us that multiplication by g_{s,n} maps S_{1}(N, ε) isomorphically onto the subspace of S_{2}(N, εχ_{−n}) consisting of forms which vanish at any cusp −(2^{s+1}kl)^{−1} , where k|m , l|n , with order greater than:

2^{s−3} · \frac{m}{k} · (\frac{n}{l} + l) .

Let us now fix divisors k|m , l|n , and consider the “Atkin-Lehner operator” (cf. [1]):

W_{N,kl} = ( \begin{matrix}\frac{mn}{kl} & b \\ −N & \frac{mn}{kl} d \end{matrix} ) ,

where b and d are such that:

\frac{mn}{kl} d + 2^{s+1}klb = 1 .

(Such numbers b and d exist since mn is an odd, squarefree number). Now, εχ_{−n} is a Dirichlet character mod N , and so we have εχ_{−n} = χ_{1}χ_{2} , where χ_{1} and χ_{2} are Dirichlet charcters mod \frac{mn}{kl} and mod 2^{s+1}kl respectively. Acting with W_{N,kl} on modular forms gives us an isomorphism:

S_{2}(N, εχ_{−n}) ˜ ⟶ S_{2}(N, \overline{χ_{1}}χ_{2}) .

Since we have:

W_{N,kl} = ( \begin{matrix}1 & b \\ −2^{s+1}kl & \frac{mn}{kl}d \end{matrix} ) ( \begin{matrix}\frac{mn}{kl} & 0 \\ 0 & 1 \end{matrix} ) ,

and since det W_{N,kl} = \frac{mn}{kl} , which is precisely the width of the cusp −(2^{s+1}kl)^{−1} with respect to Γ_{0}(N) , we find that a form f ∈ S_{2}(N, εχ_{−n}) has order greater than:

2^{s−3} · \frac{m}{k} · (\frac{n}{l} + l) at − (2^{s+1}kl)^{−1}

if and only if the first 2^{s−3} · \frac{m}{k} · (\frac{n}{l} + l) Fourier coefficients at ∞ (counting from the

coefficient of e^{2πiz}) of f|W_{N,kl} are 0 .

Thus it is clear that if we have an algorithm for finding integral bases for spaces of cusp forms of weight 2, and as we noted above such an algorithm will be given in IV and V, together with an algorithm for determining the action of the above “Aktin-Lehner operators”, then we have in certain favorable cases (and in fact in all cases to be considered in VI) an algorithm for computing dim S_{1}(N, ε) (and in fact for determining an integral basis for S_{1}(N, ε)). The question of determining the action of the “Atkin-Lehner operators” will be considered in IV (and in VI).

1. Liftings of 2-dimensional projective Galois representations over Q

We return to the discussion of 0.1. and 0.2. and in particular to the question of determining the number d(N, ε) , i.e. the number of equivalence classes of irreducible representations:

ρ : Gal (\overline{ℚ}/ℚ) → GL_{2}(ℂ)

with Artin conductor N and determinant character ε . In the following we do not assume that the representations are odd, i.e. we do not require ε(−1) = −1 . Given the representation ρ , we may consider its projectivisation:

\overline{ρ} : Gal (\overline{ℚ}/ℚ) → P GL_{2}(ℂ) .

We now want to reverse this situation, i.e. to consider \overline{ρ} as being given and ask for “liftings” of \overline{ρ} , where by a lifting of \overline{ρ} we shall understand a representation ρ as above whose projectivisation is \overline{ρ} . According to a theorem of Tate, such liftings always exist. Our prime concern is now of course to obtain information on the Artin conductors and determinant characters of such liftings. We want to ask the following question: Given \overline{ρ} as above, what are the possible pairs (N, ε), where N ∈ ℕ and ε is a Dirichlet character modulo N, such that \overline{ρ} has a lifting with Artin conductor N and determinant (character) ε? For each occuring pair (N, ε) one also wants to know its “multiplicity”, i.e. the number of inequivalent liftings of \overline{ρ} with Artin conductor N and determinant ε.

Given an answer to this question, we can reduce the problem of enumerating all (irreducible) Galois representations:

ρ : Gal (\overline{ℚ}/ℚ) → GL_{2}(ℂ)

with given Artin conductor, N, and determinant, to a question in geometry of numbers: For if ρ has Artin conductor N, then the \underline{minimal} Artin conductor of a lifting of the associated projective representation \overline{ρ} will certainly be ≤ N, and this gives, as will become clear from the following, an explicit bound for the discriminant D(K/ℚ), where K is the fixed field of the kernel of \overline{ρ}. The finitely many possibilities for K can thus, at least in principle, be found by geometry of numbers.

Let us now return to the situation where the projective representation \overline{ρ} is given. Now, if ρ is any lifting of \overline{ρ}, then the other liftings of \overline{ρ} are ρ ⊗ χ, where χ runs through the characters of Gal (\overline{ℚ}/ℚ). The determinant of ρ ⊗ χ is:

det (ρ ⊗ χ) = det(ρ) · χ^{2},

hence it is clear that we can answer the above question, if we can point to \underline{one} lifting ρ, with such precision that we may determine det (ρ) and the Artin conductor of every “twist” ρ ⊗ χ. Let us now localize the question by choosing for each prime number p a place of \overline{ℚ} over p; let D_{p} resp. I_{p} be the associated decomposition resp. inertia group. The restriction \overline{ρ}_{p} of \overline{ρ} to D_{p} can be viewed as a projective representation of Gal (\overline{ℚ}_{p}/ℚ_{p}). The following theorem of Tate is now helpful.

Theorem: (Tate, cf. [13]) Let \overline{ρ} be a projective representation of Gal (\overline{ℚ}/ℚ). Assume that for each prime number p there is a given lifting r_{p} of \overline{ρ}_{p}. Assume further that r_{p} is unramified (i.e. r_{p}(I_{p}) = 1) for all but finitely many p. Then there is a lifting ρ of \overline{ρ} such that:

ρ|I_{p} = r_{p}|I_{p} for all p,

and ρ is unique.

Given \overline{ρ}, the restriction \overline{ρ}_{p} is unramified for almost all p, and one knows that there is always a system (r_{p}) of liftings of \overline{ρ}_{p} satisfying the requirements of the theorem, cf. [13]. In the situation of the theorem the determinant of ρ is given, once one knows its restriction to I_{p} for all p, and this restriction is det (r_{p})|I_{p}. Viewing via local classfield theory the character det (r_{p}) as a character of ℚ_{p}^{x}, this restriction is simply the restriction of det (r_{p}) to the group of units of ℤ_{p}. Furthermore, if χ is a character of Gal (\overline{ℚ}/ℚ), then we may by global classfield theory view χ as an idele class character and consider its restriction χ_{p} to ℚ_{p}^{x} for every p. The Artin conductor of ρ ⊗ χ is the product of the Artin conductors of r_{p} ⊗ χ_{p} for all p. (Note that these latter conductors depend only on the restriction of r_{p} to I_{p}.)

Concerning the question of equivalence of twists ρ ⊗ χ in case ρ is 2-dimensional, one must know for what characters χ the representations ρ and ρ ⊗ χ are equivalent. If χ is non-trivial this can only happen, if Im (\overline{ρ}) is a dihedral group, and this case can be completely analysed, as will become clear from the following, by use of the well-known theorem of Mackey concerning induced representations. Thus, we shall not pursue this question further.

It is now clear that we can answer the above question once we have solved the following problem.

Problem: Let p be a prime number and let \overline{ρ} : Gal (\overline{ℚ}_{p}/ℚ_{p}) → P GL_{2}(ℂ) be a (continuous) representation. Determine for some lifting ρ of \overline{ρ} the following:

(1) the restriction of det (ρ) to the group of units of ℤ_{p}, viewing det (ρ) as a
character of ℚ_{p}^{x},
(2) the Artin conductor of ρ ⊗ χ, where χ runs through all characters of ℚ_{p}^{x}.

(ρ has to be chosen to be unramified, if \overline{ρ} is unramified.)

This problem will be solved in the following two sections.

Given \overline{ρ} : Gal (\overline{ℚ}_{p}/ℚ_{p}) → P GL_{2}(ℂ), let us consider the finite extension M/ℚ_{p} which is cut out by \overline{ρ}, i.e. M is the fixed field of the kernel of \overline{ρ}. For the Galois group G = Gal (M/ℚ_{p}) we have a priori the following possibilites:

(a) G is a cyclic group,
(b) G is a dihedral group,
(c) G is isomorphic to A_{4} or S_{4},

since G is a finite, solvable subgroup of P GL_{2}(ℂ).

Here, we may dispose of case (a) immediately: If G is a cyclic group, then \overline{ρ} is given by a character χ_{0} of Gal (\overline{ℚ}_{p}/ℚ_{p}), and the liftings of \overline{ρ} are the representations:

ρ(χ) : g ↦ ( \begin{matrix}χ_{0}(g)χ(g) & 0 \\ 0 & χ(g) \end{matrix} ) ,

where χ runs through all characters of Gal (\overline{ℚ}_{p}/ℚ_{p}). The determinant of ρ(χ) is χ_{0}χ^{2} and its Artin conductor is the product of the conductors of χ_{0}χ and χ.

The cases (b) and (c) will be considered in sections 2 and 3 respectively. As the proofs are somewhat technical we shall merely give the main points; details will appear elsewhere. For case (c) there is already essential information available: Building upon [17], the \underline{minimal} conductor of a lifting of \overline{ρ} was determined by Buhler and Zink, cf. [3] and [18]. In fact, the conductors of twists ρ ⊗ χ, where ρ is a lifting of \overline{ρ} with minimal conductor, were determined in [18]. Hence, in this case our problem is to complement these works by discussing the associated determinant characters.

Let us now introduce the following notation. If p is a prime number and M/ℚ_{p} a finite extension, let O_{M} denote the ring of integers in M, ℘_{M} its prime ideal, π_{M} a prime element of ℘_{M}, U_{M}^{0} = U_{M} the group of units of O_{M} and for i ∈ ℕ let U_{M}^{i} denote the group of 1-units of level ≥ i. Let E_{M} denote the group of roots of unity in M^{x} of order prime to p, and let for l a prime number μ_{l^{∞}}(M) be the group of roots of unity in M^{x} of l-power order. The extension of M obtained by adjoining the p′th roots of unity will be denoted by M(μ_{p}). Finally, denote by ℘_{M}^{c_{M} (χ)} the (classfield theoretic) conductor of χ, if χ is a character of M^{x}; for convenience, we shall refer to c_{M}(χ) as the conductor of χ.

2. The dihedral case

Consider a projective representation:

\overline{ρ} : Gal (\overline{ℚ}_{p}/ℚ_{p}) → P GL_{2}(ℂ)

of dihedral type, i.e. the extension M/ℚ_{p} cut out by \overline{ρ} has Galois group isomorphic to:

D_{n} = ⟨σ, τ|σ^{2} = τ^{n} = 1 , στσ^{−1} = τ^{−1}⟩

for some n ≥ 2. We want to recall a few elementary facts, for which the reader is referred to [13], about this situation. The field M contains a quadratic extension K/ℚ_{p} corresponding to the cyclic subgroup ⟨τ⟩ of D_{n}. (There is exactly 1 such quadratic extension in M (i.e. such that M/K is cyclic) if n ≥ 3, and if n = 2 we let K denote any of the 3 quadratic extensions in M.) The Galois group of M/K is then cyclic of order n, so that the restriction of \overline{ρ} to Gal (\overline{ℚ}_{p}/K) is given by a character χ of Gal (\overline{ℚ}_{p}/K). Conversely, if K/ℚ_{p} is a given quadratic extension and χ is a non-trivial character of Gal (\overline{ℚ}_{p}/K), then the field M cut out by χ is Galois over ℚ_{p} with dihedral Galois group if and only if χ∘ver_{K/ℚ_{p}} vanishes, where ver_{K/ℚ_{p}} denotes the transfer. If this condition is fulfilled, χ then gives rise to a unique projective representation \overline{ρ} : Gal (\overline{ℚ}_{p}/ℚ_{p}) → P GL_{2}(ℂ) of dihedral type. Any lifting ρ of \overline{ρ} has (up to equivalence) the form:

ρ = Ind_{K/ℚ_{p}}(ψ),

where Ind_{K/ℚ_{p}} means induction from Gal (\overline{ℚ}_{p}/K) to Gal (\overline{ℚ}_{p}/ℚ_{p}), and where ψ is a character of Gal (\overline{ℚ}_{p}/K) with:

ψ(σgσ^{−1}) = χ(g)ψ(g), g ∈ Gal (\overline{ℚ}_{p}/K),

where σ denotes any element of Gal (\overline{ℚ}_{p}/ℚ_{p}) − Gal(\overline{ℚ}_{p}/K). The Artin conductor of ρ is:

A(ρ) = D(K/ℚ_{p})N_{K/ℚ_{p}}(℘_{K}^{c_{K}(ψ)}),

where D(K/ℚ_{p}) is the discrimant of K/ℚ_{p} and N_{K/ℚ_{p}} : K → ℚ_{p} the norm, and its determinant is:

det (ρ) = ε · (ψ ∘ ver_{K/ℚ_{p}}),

where ε is the quadratic character corresponding to K/ℚ_{p}. Furthermore, if φ is a character of Gal (\overline{ℚ}_{p}/ℚ_{p}), then:

ρ ⊗ φ = Ind_{K/ℚ_{p}}(ψ) ⊗ φ = Ind_{K/ℚ_{p}}(ψ · res (φ)),

where res is the restriction to Gal (\overline{ℚ}_{p}/K).

Viewing χ and ψ as characters of K^{x} and φ as a character of ℚ_{p}^{x}, we now see (by classfield theory) that the problem of section 1 amounts to the following:

Given a quadratic extension K/ℚ_{p} and a character χ of K^{x} which vanishes on ℚ_{p}^{x}, determine for a character ψ of K^{x} such that:

ψ(\frac{σx}{x}) = χ(x) for all x ∈ K^{x},

where σ denotes the generator of Gal (K/ℚ_{p}), the following:

(1) the restriction of ψ to the group of units of ℤ_{p},

and

(2) the conductor of ψ · (φ ∘ N_{K/ℚ_{p}}), where φ runs through the characters of ℚ_{p}^{x}.

This is done by theorem 1 below together with the remarks following it. We have chosen to state and prove a slightly more general result, because this costs little extra effort and because we want to make it clear what sort of problems one would have to solve, if one wanted to consider the situation for an arbitrary p-adic groundfield.

Proposition 1: Suppose that l is a prime number, that K/ℚ_{p} is a finite extension and that K^{x} contains the l′th roots of unity. Let L/K be a Galois extension with Galois group G ≅ ℤ/ℤl, and let σ be a generator of G. Denote by σ − 1 the endomorphism x ↦ x^{−1}σx of L^{x}.

(1) Let i ∈ ℕ. An element x ∈ K^{x} belongs to (L^{x})^{σ−1}U_{L}^{i} if and and only if

x^{p} ∈ N_{L/K}(U_{L}^{i}).

(2) Suppose that L/K is unramified. Then: (U_{L}^{i})^{σ−1} ≤ U_{L}^{i} for all i ∈ ℕ,

and the homomorphism:

U_{L}^{i}/U_{K}^{i}U_{L}^{i+1} → U_{L}^{i}/U_{L}^{i+1}

induced by σ − 1 is injective.

(3) Suppose that L/K is ramified with ramification groups:
G = G_{0} = … = G_{t} ≠ G_{t+1} = 0

(where t is a non-negative integer). If i ∈ ℕ with l|i, we have: (U_{L}^{i})^{σ−1} ≤ U_{L}^{i+t+1}. If i ∈ ℕ with l ∤ i, then: (U_{L}^{i})^{σ−1} ≤ U_{L}^{i+t}, and the homomorphism:

U_{L}^{i}/U_{L}^{i+1} → U_{L}^{i+t}/U_{L}^{i+t+1}

induced by σ − 1 is an isomorphism.

Proof:

(1) This is a trivial consequence of Hilbert’s theorem 90.

(2) Clearly, σ − 1 maps U_{L}^{i} into itself for all i ∈ ℕ. Choose π = π_{K} as a prime element of L. If u ∈ U_{L}^{i} − U_{L}^{i+1} and u^{−1}σu ∈ U_{L}^{i+1}, then choose a ∈ E_{L} such that u is represented by 1 + aπ^{i} modulo U_{L}^{i+1}. One then finds:

\frac{σ(1 + aπ^{i})}{1 + aπ^{i}} ≡ 1 + (σa − a)π^{i} mod ℘_{L}^{2i},

whence a is not a unit. One deduces a ∈ K^{x} and so u ∈ U_{K}^{i}U_{L}^{i+1}.

(3) Suppose that L/K is wildly ramified, i.e. t > 0, i.e. l = p. Let π be a prime element for L. Then:

σπ = π + uπ^{t+1} ,

where u is a unit, since σ ∈ G_{t} − G_{t+1}. If i ∈ ℕ and b ∈ O_{L}, then as σb ≡ b mod ℘_{L}^{t+1} one easily finds:

\frac{σ(1 + bπ^{i})}{1 + bπ^{i}} ≡ 1 + iu(σb)π^{i+t} mod ℘_{L}^{i+t+1} .

From this the assertions follow immediately. The case t = 0 is similar, but simpler. Q.E.D.

We want to consider the situation of proposition 1 in the case that K = ℚ_{p}(μ_{p}) and l = p , i.e. L/K is a Galois extension with Galois group G ≅ ℤ/ℤp . Let σ be a generator of G . Recall that the group of 1-units of K has a basis, as a ℤ_{p}-module, of the form:

ζ, η_{2}, …, η_{p} ,

where ζ is a primitive p′th root of unity and η_{i} has level exactly i (i.e. η_{i} ∈ U_{K}^{i}−U_{K}^{i+1}) for i = 2, …, p (cf. [7] pp. 246-247). Here, and in what follows, we suppose that a choice of the elements η_{2}, …, η_{p} has been fixed. Put:

U′_{K} = ⟨η_{2}, …, η_{p}⟩ .

Let χ be a character of L^{x} which vanishes on K^{x} . Let c = 1 if χ is unramified and c = c_{L}(χ) otherwise.

Suppose first that L/K is ramified with t = p−1, where t is defined as in proposition 1 , and that χ is wildly ramified, i.e. c > 1 . Let the integer a be such that c ≡ a(p) and 1 ≤ a ≤ p . Using [12], V, one finds:

U_{K}^{\frac{1}{p} (c−a)+p} = N_{L/K}(U_{L}^{c+p−1}) ,

so that if u ∈ U′_{K} with u^{p} ∈ U_{K}^{\frac{1}{p} (c−a)+p} , then there is x ∈ L^{x} such that:

u ≡ \frac{σx}{x} mod U_{L}^{c+p−1} .

If x, y ∈ L^{x} and:

\frac{σx}{x} ≡ \frac{σy}{y} mod U_{L}^{c+p−1} ,

put z = x/y . Then z^{−1}σz ∈ U_{L}^{c+p−1} , and since c ≥ 1 , we see that z ∈ K^{x}U_{L}^{1} . If z ∈ K^{x} , then χ(x) = χ(y) . Otherwise, choose i ∈ ℕ largest possible such that z ∈ K^{x}U_{L}^{i} . Then p ∤ i , since U_{L}^{j} ≤ K^{x}U_{L}^{j+1} , if p | j . So, proposition 1 gives that z^{−1}σz ∉ U_{L}^{i+p} ; as z^{−1}σz ∈ U_{L}^{c+p−1} , we have i ≥ c = c_{L}(χ) , hence χ(z) = 1 . Since η_{2}, …, η_{p} form a basis of U′_{K} , we infer the existence of a character ψ_{2} on U′_{K} satisfying the following requirements: For i = 2, …, p let s_{i} ≥ 1 be smallest possible such that:

η_{i}^{p^{s_{i}}} ∈ U_{K}^{\frac{1}{p} (c−a)+p} ,

and let x_{i} ∈ L^{x} be such that:

η_{i}^{p^{s_{i}−1}} ≡ \frac{σx_{i}}{x_{i}} mod U_{L}^{c+p−1} .

The requirements are then:

ψ_{2}(η^{p_{i}^{s_{i}−1}}) = χ(x_{i}) , i = 2, …, p .

We let ψ_{2} denote any such character. One easily sees, that ψ_{2} has the following property: If u ∈ U′_{K} and x ∈ L^{x} are such that:

u ≡ \frac{σx}{x} mod U_{L}^{c+p−1} ,

then:

ψ_{2}(u) = χ(x) .

Suppose then that L/K is unramified. By a similar, but simpler argument, one now infers the existence of a character ψ_{2} on U′_{K} satisfying: Let for i = 2, … p the integer s_{i} ≥ 1 be smallest possible such that:

η_{i}^{p^{s_{i}}} ∈ U_{L}^{c} .

Then there are x_{i} ∈ L^{x} such that:

η_{i}^{p^{s_{i}−1}} ≡ \frac{σx_{i}}{x_{i}} mod U_{L}^{c} , i = 2, …, p ,

and the requirements are:

ψ_{2}(η_{i}^{p^{s_{i}−1}}) = χ(x_{i}) , i = 2, …, p .

Denote by ψ_{2} any such character. One finds that ψ_{2} has the property: If u ∈ U′_{K} and x ∈ L^{x} are such that:

u ≡ \frac{σx}{x} mod U_{L}^{c} ,

then:

ψ_{2}(u) = χ(x) .

Theorem 1: Suppose that l and p are prime numbers, and that K is a finite extension of ℚ_{p} containing the l′th roots of unity. Let L/K be a Galois extension with Galois group G ≅ ℤ/ℤl and let σ be a generator of G. If L/K is unramified, put t = 0. If L/K is ramified, we denote by t ≥ 0 the break in the ramification filtration of G:

G = G_{0} = … = G_{t} ≠ G_{t+1} = 0 .

If l = p, we make the assumption that K = Q_{p}(μ_{p}), and furthermore that t ≥ p − 1, if L/K is ramified.

Let χ be a non-trivial character on L^{x} which vanishes on K^{x}. Let α ∈ K^{x} be such that L = K(α^{1/l}), and let the primitive l′th root of unity be such that:

σα^{1/l} = ζα^{1/l}.

Let ψ_{1} be a character of μ_{l^{∞}}(K) satisfying the following requirements:

ψ_{1} = 1 , if L ≠ K(\sqrt{−1}) and χ(α^{1/l}) = 1 ,
ψ_{1}(ζ) = χ(α^{1/l}) , if L ≠ K(\sqrt{−1}) and χ(α^{1/l}) ≠ 1 ,
ψ_{1} = 1 , if L = K(\sqrt{−1}) and χ(1 + \sqrt{−1}) = 1 ,
ψ_{1}(−1) = χ(1 + \sqrt{−1})^{2} , if L = K(\sqrt{−1}) and χ(1 + \sqrt{−1}) ≠ 1 .

(Note that if also L = K(β^{1/l}), then χ(α^{1/l}) ≠ 1 ⟺ χ(β^{1/l}) ≠ 1 .)

Define:

c = \{ \begin{matrix}1, & if χ is unramified \\ c_{L}(χ), & if χ is ramified \end{matrix} .

If l ≠ p , let U′_{K} be U_{K}^{1} , and put U′_{K} = ⟨η_{2}, …, η_{p}⟩ if l = p . Let ψ_{2} be the trivial character on U′_{K} , if either l ≠ p or if l = p and L/K is ramified with either t ≥ p or (t = p − 1 and c = 1 ). Otherwise, i.e. if l = p and L/K is either unramified of ramified with (t = p − 1 and c > 1) , let ψ_{2} be a character on U′_{K} of the type described immediately after proposition 1 .

Finally, denote by U_{0} the group of roots of unity in K^{x} of order prime to lp .

Then there exists a character ψ on L^{x} such that:

(i) ψ(\frac{σx}{x})  =  χ(x) for all x ∈ L^{x} ,
(ii) ψ(⟨π_{L}⟩U_{0})  =  1 ,
(iii) ψ|μ_{l^{∞}}(K)  =  ψ_{1} ,
(iv) ψ|U′_{K}  =  ψ_{2} ,

and

(v) c_{L}(ψ)  =  c + t .

Furthermore, if φ is a character of K^{x} and N_{L/K} denotes the norm map L^{x} → K^{x} , then:

c_{L}(ψ · (φ ∘ N_{L/K})) = max {c + t , c_{L}(φ ∘ N_{L/K})} ,

and for the number c_{L}(φ ∘ N_{L/K}):

c_{L}(φ ∘ N_{L/K}) = c_{K}(φ) , if L/K is unramified ,

and if L/K is ramified:

\begin{aligned}c_{L}(φ ∘ N_{L/K}) & = lc_{K}(φ) + (1 − l)t + 1 , if c_{K}(φ) ≥ t + 2 , \\ c_{L}(φ ∘ N_{L/K}) & ≤ t + 1 , if c_{K}(φ) ≤ t + 1 . \end{aligned}

Proof: Using the fact that if ξ ∈ (L^{x})^{σ−1} ∩ μ_{l^{∞}}(L) , then N_{L/K}(ξ) = 1 and:

σξ = ζ^{a}ξ with a ≢ 0 (l) , if ξ ∉ K^{x} ,

one easily finds:

(∗)  (L^{x})^{σ−1} ∩ μ_{l^{∞}}(L) = \{ \begin{matrix}⟨\sqrt{−1}⟩ , & if L = K(\sqrt{−1}) \\ ⟨ζ⟩ & otherwise \end{matrix} .

Define the character ψ_{0} on (L^{x})^{σ−1} by:

ψ_{0}(\frac{σx}{x}) = χ(x) for x ∈ L^{x} ;

this is well-defined since χ vanishes on K^{x} . It now follows from (*) and the definition of ψ_{1} that there is a character on (L^{x})^{σ−1}μ_{l^{∞}}(L) whose restriction to (L^{x})^{σ−1} and μ_{l^{∞}}(K) respectively is ψ_{0} and ψ_{1} respectively.

If y ∈ (L^{x})^{σ−1}μ_{l^{∞}}(L) ∩ U′_{K} , then:

y^{l} = N_{L/K}(y) ∈ μ_{l^{∞}}(K) ∩ U′_{K} = {1} .

So, if l ≠ p we have y = 1 , since y is a 1-unit. If l = p , it also follows that y = 1 , since y ∈ U′_{K} and U′_{K} is torsion free for l = p .

We deduce the existence of a character on (L^{x})^{σ−1}μ_{l^{∞}}(L)U′_{K} whose restriction to (L^{x})^{σ−1} , μ_{l^{∞}}(K) and U′_{K} respectively is ψ_{0} , ψ_{1} and ψ_{2} respectively. We fix one such character and denote it by abuse of notation by ψ_{0} .

Denote by i_{0} the smallest non-negative integer such that:

U_{L}^{i_{0}} ∩ (L^{x})^{σ−1} ≤ ker (ψ_{0}) .

We claim that:

(∗∗)  i_{0} = c + t .

Note that i_{0} ≥ 1 , because (L^{x})^{σ−1} ≤ U_{L}^{0} and because ψ_{0} cannot be trivial on (L^{x})^{σ−1} since χ is non-trivial.

Let us show (**) in the case where L/K is unramified and χ is wildly ramified. Then c + t = c_{L}(χ) + t . We note that c ≢ 1 (l) . This follows once we note that if i ∈ ℕ is divisible by l , then:

U_{L}^{i} ≤ K^{x}U_{L}^{i+1} .

There is an x ∈ U_{L}^{c−1} with χ(x) ≠ 1 . Now proposition 1 gives that x^{−1}σx ∈ U_{L}^{c−1+t} − U_{L}^{c+t} , since c − 1 is not divisible by l . Hence i_{0} ≥ c + t . On the other hand, suppose that x ∈ L^{x} is such that x ∉ K^{x} and x^{−1}σx ∈ U_{L}^{c+t} . Let i be largest possible such that l ∤ i and such that there is a y ∈ U_{L}^{i} with x ≡ y mod K^{x} . Then y ∉ U_{L}^{i+1} ; for if l ∤ i+1 , this is clear, and otherwise there is a y_{1} ∈ U_{L}^{i+2} with y ≡ y_{1} mod K^{x} and l ∤ i + 2 . As l ∤ i , proposition 1 gives that x^{−1}σx = y^{−1}σy ∉ U_{L}^{i+t+1} . So: i ≥ c , whence χ(x) = χ(y) = 1 . We conclude that i_{0} ≤ c + t .

The proof of (**) in the other cases, i.e. L/K unramified or L/K ramified but χ unramified, is similar but simpler.

Concerning the norm map N_{L/K} : L^{x} → K^{x} we note the following: If L/K is unramified, then:

N_{L/K}(U_{L}^{i}) = U_{K}^{i} for all i ≥ 0 ,

and if L/K is ramified, we have:

N_{L/K}(U_{L}^{lx+(1−l)t+1}) = … = N_{L/K}(U_{L}^{lx+(1−l)t+l}) = U_{K}^{x+1} for x ≥ t ,

and

N_{L/K}(U^{x+1}) ≤ U_{K}^{x+1} for 0 ≤ x ≤ t ,

cf. [12], V. From this, the remarks in the statement of the theorem about the number c_{L}(φ ∘ N_{L/K}) for a character φ of K^{x} immediately follow.

We now claim that:

(∗ ∗ ∗)  ⟨π_{L}⟩U_{0}U_{L}^{i_{0}} ∩ (L^{x})^{σ−1}μ_{l^{∞}}(L)U′_{K} ≤ ker (ψ_{0}) .

The rest of the theorem follows from (***). For if (***) holds, then we know from harmonic analysis that there is a character ψ on the locally compact group L^{x} whose restriction to the compact group (L^{x})^{σ−1}μ_{l^{∞}}(L)U′_{K} is ψ_{0} and which vanishes on the closed subgroup ⟨π_{L}⟩U_{0}U_{L}^{i_{0}} . If ψ is any such character, then ψ satisfies (i), (ii), (iii), and (iv) in the statement of the theorem and c_{L}(ψ) is at the most i_{0} = c + t . Furthermore, by definition of i_{0} there is an x ∈ U_{L}^{i_{0}−1} ∩ (L^{x})^{σ−1} with ψ(x) ≠ 1 . Hence c_{L}(ψ) is exactly i_{0} . If φ is any character on K^{x} , then φ ∘ N_{L/K} vanishes on (L^{x})^{σ−1} and in particular (φ ∘ N_{N/L})(x) = 1 . It follows that:

c_{L}(ψ · (φ ∘ N_{L/K})) = max {c_{L}(ψ) , c_{L}(φ ∘ N_{L/K})} .

Concerning the proof of (***):

Suppose that y ∈ ⟨π_{L}⟩U_{0}U_{L}^{i_{0}} ∩ (L^{x})^{σ−1}μ_{l^{∞}}(L)U′_{K} . As y ∈ (L^{x})^{σ−1}μ_{l^{∞}}(L)U′_{K} , y is a unit, so that we can write:

y = u_{0}u = \frac{σx}{x} · ξu_{1} ,

with u_{0} ∈ U_{0} , u ∈ U_{L}^{i_{0}} , x ∈ L^{x} , ξ ∈ μ_{l^{∞}}(L) and u_{1} ∈ U′_{K} . Then N_{L/K}(y) = N_{L/K}(ξ)u_{1}^{l} , so there is an s ∈ ℕ such that:

N_{L/K}(y)^{l^{s}} is a 1−unit.

On the other hand, N_{L/K}(y) = u_{0}^{l}N_{L/K}(u) , hence:

u_{0}^{l^{s+1}} is a 1−unit.

Since u_{0} is a root of unity of order prime to lp , we deduce:

u_{0} = 1 .

One now splits the discussion up into 4 cases: (l ≠ p or l = p) and (L/K unramified or L/K ramified). Let us for example consider the case l = p and L/K ramified. We have K = ℚ_{p}(μ_{p}) which has ramification index e = p − 1 over ℚ_{p} . Since the 1-units η_{2}, …, η_{p} have level > \frac{e}{p−1} , it follows that if λ ∈ U′_{K} has level exactly i , then λ^{p} has level exactly i + e . Choose a such that 1 ≤ a ≤ p and c ≡ a (p) , and put w = \frac{1}{p}(c − a) + t . Now,

N_{L/K}(ξ)u_{1}^{p} = N_{L/K}(u) ∈ N_{L/K}(U_{L}^{c+t}) = U_{K}^{w+1} ,

and so N_{L/K}(ξ) and u_{1}^{p} both belong to U_{K}^{w+1} . As N_{L/K}(ξ) is a power of ζ , we must have N_{L/K}(ξ) = 1 , so ξ ∈ (L^{x})^{σ−1} . Consequently, there is x_{0} ∈ L^{x} such that:

y = u = \frac{σx_{0}}{x_{0}} · u_{1} .

Suppose first that t ≥ p or (t = p − 1 and c = 1). In both cases we have:

t ≥ p − \frac{p − a}{p − 1} ,

and this gives:

u_{1} ∈ U_{K}^{w+1−e} ≤ U_{L}^{p(w+1−e)} ≤ U_{L}^{c+t} ,

hence x_{0}^{−1}σx_{0} ∈ U_{L}^{c+t} . We then deduce that x_{0} ∈ K^{x}U_{L}^{c} . For, since c ≥ 1 , we must have x_{0} ∈ K^{x}U_{L}^{1} , so if x_{0} ∉ K^{x} we choose i largest possible such that x_{0} ∈ K^{x}U_{L}^{i} ; then p ∤ i , and proposition 1 gives \frac{σx_{0}}{x_{0}} ∉ U_{L}^{i+t+1} , hence i ≥ c . We then get:

ψ_{0}(y) = χ(x_{0})ψ_{0}(u_{1}) = ψ_{2}(u_{1}) = 1 ,

since ψ_{2} is trivial. Suppose then that t = p − 1 and c > 1 . Now,

u_{1}^{−1} ≡ \frac{σx_{0}}{x_{0}} mod U_{L}^{c+t} ,

so from the properties of ψ_{2} we obtain:

ψ_{0}(y) = χ(x_{0})ψ_{2}(u_{1}) = χ(x_{0})χ(x_{0}^{−1}) = 1 .

Q.E.D.

Remarks:

(1) As is clear from the proof of theorem 1, there is, in the setting of theorem 1 for an arbitrary groundfield K (containing the l’th roots of unity), always a character ψ on L^{x} such that ψ(x^{−1}σx) = χ(x) for all x ∈ L^{x} and such that c_{L}(χ) = c + t in the notation of the theorem. A special case of this result was given in [3] but without a discussion of the possible behaviour of the restriction of ψ to K^{x}. It is also clear from the above proof, that the explicit construction of a possible choice of this restriction depends in the general case on a detailed knowledge of the structure of the 1-unit group of K and in particular on the structure of the quotients U_{K}^{1}/U_{K}^{i} for i ∈ ℕ . It is also clear that a possible ψ for the case p ≠ 2 , K = ℚ_{p}(μ_{p}) , and L/K wildly ramified with t < p − 1 may be constructed, but we have avoided that since it is unnecessary for our purposes.

(2) The value c + t is the smallest possible value of c_{L}(ψ) if ψ is a character on L^{x} with ψ(x^{−1}σx) = χ(x) for all x ∈ L^{x} ; this follows immediately from the definition of i_{0}(= c + t) in the proof of the theorem.

(3) The value of c_{L}(ψ · (φ ∘ N_{L/K})) , where ψ is as in the theorem and φ is a character on K^{x} , is computed immediately alone from the knowledge of c_{K}(φ) . This is clear if L/K is unramified or if L/K is ramified and c_{K}(φ) ≥ t + 2 ; if L/K is ramified and c_{K}(φ) ≤ t + 1 , we get:

c_{L}(ψ · (φ ∘ N_{L/K})) = c_{L}(ψ) = c + t ,

since then c_{L}(φ ∘ N_{L/K}) ≤ 1 + t and c ≥ 1 .

(4) From the remarks preceding proposition 1 it is clear that the problem of section 1 for representations of dihedral type is solved by theorem 1 once we explicate the character ψ_{2} in the cases where it is not a priori trivial. There are 3 such cases: K = ℚ_{2} and L = ℚ_{2}(\sqrt{α}) , where α is −3, −1 or 3 , and χ is a character on L^{x} vanishing on ℚ_{2}^{x} such that χ is wildly ramified if α is −1 or 3 . In all 3 cases U′_{K} is the group generated by 5, and a few simple computations now reveal that we can choose for ψ_{2} any character on ⟨5⟩ satisfying the following:

Let c be defined as in theorem 1.

\underline{If α = −3:} Put ψ_{2} = 1 if c ≤ 2 . If c ≥ 3 , we require:

ψ_{2}(5^{2^{c−3}}) = χ(1 + η · 2^{c−1}) ,

where η is a primitive 3’rd root of unity.

\underline{If α = −1 or α = 3:} Here L/K is wildly ramified and since c > 1 , the number c = c_{L}(χ) must be even. Put ψ_{2} = 1 if c = 2 . If c ≥ 4 , we require:

ψ_{2}(5^{2^{\frac{1}{2} c−2}}) = χ(1 + π^{c−1}) ,

where π is any prime element of L .

3. The “primitive” case

Let p be a prime number and K/ℚ_{p} a finite extension. Let us consider a projective representation:

\overline{ρ} : Gal (\overline{ℚ_{p}}/K) → P GL_{2}(ℂ) ,

such that Im (\overline{ρ}) is isomorphic to A_{4} or S_{4} . We want to recall a few facts concerning this situation; we refer to [3] or [18].

First, we must necessarily have p = 2, cf. [3], pp. 18-20. Let M denote the fixed field of ker(\overline{ρ}) , and put G = Gal(M/K) so that \overline{ρ} is given by an embedding of G in P GL_{2}(ℂ) . The group G contains a unique normal subgroup V isomorphic to the Klein 4-group, and we have G/V either cyclic of order 3 or isomorphic to S_{3} . Let L denote the fixed field of V . Then M/L is totally, wildly ramified, and L/K is at the most tamely ramified. If G ≅ S_{4} , the quadratic extension K_{0}/K contained in L must then be unramified, and since L/K is not abelian, L/K_{0} is tamely ramified of degree 3 . Let e denote the ramification index of L/K , so that e is 1 or 3. Since V has no proper subgroup which is normal in G , we see that the ramification groups for M/L are all either V or 0 ; define t ≥ 1 such that:

V = V_{0} = … = V_{t} ≠ V_{t+1} = 0

is the sequence of ramification groups for M/L .

For every lifting ρ of \overline{ρ} the restriction of ρ to Gal (\overline{ℚ}_{2}/M) has the form:

ρ(g) = ( \begin{matrix}χ(g) & 0 \\ 0 & χ(g) \end{matrix} ) ,

where χ is a character of Gal (\overline{ℚ}_{2}/M) ; we refer to χ as the \underline{central character} of the lifting ρ . The Artin conductor ℘_{K}^{a(ρ)} is related to the conductor of χ by:

(1)  a (ρ) = \frac{1}{2e}(c_{M}(χ) + 3t + 4e − 1) .

The representation \overline{ρ} has a lifting ρ with central character χ such that:

(2)  c_{M}(χ) = 3t + 1 , hence a (ρ) = \frac{3}{e} t + 2 ,

and ℘_{K}^{3t/e+2} is the minimal value of the Artin conductor of a lifting of \overline{ρ} . Furthermore, if ρ is a lifting with this minimal Artin conductor and χ is its central character, then there is an u ∈ U_{M}^{3t} with:

(3)  N_{M/K}(u) = 1 and χ(u) ≠ 1 .

Here, (1) and (2) are the principal statements of [3], chap. 2, and [18], section 3. The existence of u ∈ U_{M}^{3t} with (3) follows from the proof of minimality of 3t + 1 in (2), cf. [18], section 3.

Now we want to study the norm map N_{M/K} : M → K . Let W be a subgroup of V of order 2 and let L_{0} be the fixed field of W . It is easy to see that the ramification groups for M/L_{0} and L_{0}/L are the following:

W = W_{0} = … = W_{t} ≠ W_{t+1} = 0 ,

and

(V/W) = (V/W)_{0} = … = (V/W)_{t} ≠ (V/W)_{t+1} = 0 .

We conclude that:

U_{L_{0}}^{x+1} = N_{M/L_{0}}(U_{M}^{2x−t+1}) = N_{M/L_{0}}(U_{M}^{2x−t+2}) for x ≥ t ,

and

U_{L_{0}}^{x} ≥ N_{M/L_{0}}(U_{M}^{x}) for 1 ≤ x ≤ t ,

and similarly for L_{0}/L , cf [12], chap. 5. Hence:

U_{L}^{x+1} = N_{M/L}(U_{M}^{i}) for 4x − 3t + 1 ≤ i ≤ 4x − 3t + 4 , if x ≥ t ,
U_{L}^{x} ≥ N_{M/L}(U_{M}^{x}) for 1 ≤ x ≤ t .

Using again [12], chap. 5, we furthermore obtain:

U_{K}^{x+1} = N_{L/K}(U_{L}^{ex+a}) for x ≥ 0 , 1 ≤ a ≤ e .

Combined with the above, we find for 1 ≤ a ≤ e:

(4)  U_{K}^{x+1} = N_{M/K}(U_{M}^{i}) for 4ex − 3t + 4a − 3 ≤ i ≤ 4ex − 3t + 4e ,

if ex ≥ t − a + 1 , x ≥ 0 , and

(5)  U_{K}^{x+1} ≥ N_{M/K}(U_{M}^{ex+a}) if 0 ≤ x ≤ \frac{t − a}{e} .

We conclude that if φ is a character on K^{x} with conductor c_{K}(φ) = c , then:

(6)  c_{M}(φ ∘ N_{M/K}) ≤ ec − e + 1 , if c ≤ \frac{t − 1}{e} + 1 ,

and

(7)  c_{M}(φ ∘ N_{M/K}) = 4ec − 3t − 4e + 1 , if c ≥ \frac{t + 1}{e} + 1 ,

since in the latter case: c − 2 ≥ \frac{1}{e}(t − e + 1) and c − 1 ≥ \frac{t}{e} so that:

U_{K}^{c} = N_{M/K}(U_{M}^{4ec−3t−4e+1}) ,

and

U_{K}^{c−1} = N_{M/K}(U_{M}^{4ec−3t−4e}) .

Using the above it is now easy to prove the following proposition which is a reformulation of a result due to E.-W. Zink (see [18]).

Proposition 2: Let \overline{ρ} : Gal (\overline{ℚ_{2}}/K) → P GL_{2}(ℂ) be a representation with Im (\overline{ρ}) isomorphic to A_{4} or S_{4} . If ρ is any lifting of \overline{ρ} with minimal Artin conductor, then for any character φ of K^{x} we have, retaining the above notation, for the exponent a (ρ ⊗ φ) of the Artin conductor of ρ ⊗ φ :

\begin{aligned}a (ρ ⊗ φ) & = \frac{3}{e} t + 2 for c_{K}(φ) ≤ \frac{3t}{2e} + 1 , \\ a (ρ ⊗ φ) & = 2c_{K}(φ) for c_{K}(φ) ≥ \frac{3t}{2e} + 1 . \end{aligned}

Proof: Let ρ be any lifting of \overline{ρ} with minimal Artin conductor ℘_{K}^{3t/e+2} and let χ be the central character of ρ . Let φ be a character of K^{x} and put c = c_{K}(φ) . Now, ρ ⊗ φ is also a lifting of \overline{ρ} and its central character is:

χ · (φ ∘ N_{M/K}) .

According to (2) and (3) above, we have c_{M}(χ) = 3t + 1 and there is an u ∈ U_{M}^{3t} with N_{M/K}(u) = 1 and χ(u) ≠ 1 . So, χ · (φ ∘ N_{M/K}) will not vanish on u , and from this we conclude that:

c_{M}(χ · (φ ∘ N_{M/K})) = max {3t + 1 , c_{M}(φ ∘ N_{M/K})} .

Suppose that c ≥ \frac{3t}{2e} + 1 . We claim that c ≥ \frac{t+1}{e} + 1 . This is clear if t ≥ 2 . If t = e = 1 , then c ≥ 3 = 1 + \frac{2}{e} , and if t = 1, e = 3 , then \frac{2}{e} + 1 < 2 ≤ c . From (7) we conclude that:

c_{M}(φ ∘ N_{M/K}) = 4ec − 3t − 4e + 1 ≥ 3t + 1 ,

hence c_{M}(χ(φ ∘ N_{M/K})) = 4ec − 3t − 4e + 1 , and:

\begin{aligned}a (ρ ⊗ φ) & = \frac{1}{2e}(c_{M}(χ(φ ∘ N_{M/K})) + 3t + 4e − 1) \\ & = 2c . \end{aligned}

Suppose then that c ≤ \frac{3t}{2e} + 1 . If c > \frac{t−1}{e} + 1 , then e(c − 1) ≥ t , so that according to (4):

U_{K}^{c} = N_{M/K}(U_{M}^{4ec−3t−4e+1}) ,

whence:

c_{M}(φ ∘ N_{M/K}) ≤ 4ec − 3t − 4e + 1 ≤ 3t + 1 .

If c ≤ \frac{t−1}{e} + 1 , then (6) gives:

c_{M}(φ ∘ N_{M/K}) ≤ ec − e + 1 ≤ t < 3t + 1 .

So, c_{M}(χ · (φ ∘ N_{M/K})) = c_{M}(χ) = 3t + 1 in any case, and a (ρ ⊗ φ) = \frac{3}{e}t + 2 . Q.E.D.

We shall now restrict the discussion to the groundfield K = ℚ_{2} . We know, see [17], that M/ℚ_{2} is a finite extension with Galois group isomorphic to A_{4} or S_{4} if and only if M is one of the following 4 fields:

M_{1} = ℚ_{2}(ζ_{7} , \sqrt{1 + 2ζ_{7}} , \sqrt{1 + 2ζ_{7}^{2}} , \sqrt{1 + 2ζ_{7}^{4}}) ,

where ζ_{7} is a primitive 7’th root of unity; put:

L = ℚ_{2}(ζ_{3}, π) ,

where ζ_{3} is a primitive 3’rd root of unity and π^{3} = 2 , and let α be the automorphism of L with απ = ζ_{3}π ; put:

M_{i} = L(\sqrt{x_{i}} , \sqrt{αx_{i}} , \sqrt{α^{2}x_{i}}) for i = 2, 3, 4,

where x_{2} = 3(1 + π)(1 + π^{2}) , x_{3} = 3(1 + π) and x_{4} = 1 + π^{2} . We have:

Gal (M_{1}/ℚ_{2}) ≅ A_{4} and Gal (M_{i}/ℚ_{2}) ≅ S_{4} for i = 2, 3, 4 .

In the above notation we have the values e = 1, 3, 3, 3 and t = 1, 5, 5, 1 respectively for the extensions M_{i}/ℚ_{2} , i = 1, 2, 3, 4 respectively.

The following theorem solves the problem of section 1 for 2-dimensional, projective Galois representations over ℚ_{2} of type A_{4} or S_{4} .

Theorem 2: Let \overline{ρ} : Gal (\overline{ℚ_{2}}/ℚ_{2}) → P GL_{2}(ℂ) be a representation such that Gal (M/ℚ_{2}) is isomorphic to A_{4} or S_{4} , where M is the fixed field of ker(\overline{ρ}) . Then \overline{ρ} has a lifting ρ such that its determinant character ε = det(ρ) , viewed as a character of ℚ_{2}^{x} , and the Artin conductors 2^{a(ρ⊗φ)} of the twist ρ ⊗ φ, where φ is any character of ℚ_{2}^{x} with conductor c = c_{ℚ_{2}}(φ) , satisfy the following.

(I)  If M = M_{1} : ε(−1) = −1 , ε(5) = 1 , and :

a (ρ ⊗ φ) = \{ \begin{matrix}5 & for c ≤ 2 \\ 2c & for c ≥ 3 . \end{matrix}

(II)  If M = M_{2} : ε(−1) = −1 , ε(5) = 1 , and :

a (ρ ⊗ φ) = \{ \begin{matrix}7 & for c ≤ 3 \\ 2c & for c ≥ 4 . \end{matrix}

(III)  If M = M_{3} : ε(−1) = ε(5) = 1 , and :

a (ρ ⊗ φ) = \{ \begin{matrix}7 & for c ≤ 3 \\ 2c & for c ≥ 4 . \end{matrix}

(IV)  If M = M_{4} : ε(−1) = ε(5) = 1 , and

a (ρ ⊗ φ) = \{ \begin{matrix}3 & for c ≤ 1 \\ 2c & for c ≥ 2 . \end{matrix}

Proof: Let ρ be a lifting of \overline{ρ} with minimal conductor and let χ be its central character. Hence c_{M}(χ) = 3t + 1 , where t = 1, 5, 5, 1 respectively if M = M_{i} , i = 1, 2, 3, 4 respectively. The restriction of det (ρ) to Gal (\overline{ℚ}_{2}/M) is given by:

det (ρ)(g) = ( \begin{matrix}χ(g) & 0 \\ 0 & χ(g) \end{matrix} ) , g ∈ Gal (\overline{ℚ}_{2}/M) ,

hence, if ε = det (ρ) is viewed as a character of ℚ_{2}^{x} and χ as a character of M^{x} , we have:

(∗)  ε ∘ N_{M/ℚ_{2}} = χ^{2} .

We must determine the restriction of ε to the group of (1-)units of ℚ_{2} . Now, the image of the norm map M^{x} → ℚ_{2}^{x} coincides with the image of the norm M_{0}^{x} → ℚ_{3}^{x} , where M_{0}/ℚ_{2} is the maximal abelian extension contained in M , and since M_{0} is in any case unramified, we have N_{M/ℚ_{2}}(U_{M}) = U_{ℚ_{2}} . Hence it suffices to study the behaviour of χ^{2} on U_{M} . Now, if χ^{2} is trivial on U_{M} , then ε is unramified, hence ε = ψ^{2} for some unramified character ψ on ℚ_{2}^{x} . Then ρ ⊗ ψ^{−1} still has minimal conductor and the square of its central character is 1 . By replacing ρ by ρ ⊗ ψ^{−1} if necessary, we may assume that if χ^{2} is non-trivial, it is non-trivial on U_{M} .

Now, the minimal order among the orders of central characters of liftings of \overline{ρ} is 4, 4, 2, 2 respectively for the cases M = M_{i} , i = 1, 2, 3, 4 respectively, cf. [18], section 2, or [2], where it is shown how to compute this order using a criterion of Serre. Let ρ_{1} be a lifting of \overline{ρ} whose central character χ_{1} has this minimal order. There is a character ψ of ℚ_{2}^{x} such that:

ρ = ρ_{1} ⊗ ψ ,

if ψ is viewed as a character of Gal (\overline{ℚ}_{2}, /ℚ_{2}) , and this means:

(∗∗)  χ = χ_{1} · (ψ ∘ N_{M/ℚ_{2}}) .

One now proceeds with an individual discussion in each of the 4 cases M = M_{i}, i = 1, 2, 3, 4. Let us for example consider the case:

\underline{M = M_{3}:} Suppose that χ^{2} is non-trivial. Then χ^{2} is non-trivial on U_{M} , and since (**) gives:

χ^{2} = ψ^{2} ∘ N_{M/ℚ_{2}} ,

because χ_{1}^{2} = 1 , we deduce that ψ^{2} is non-trivial on U_{ℚ_{2}} . Since ψ^{2}(−1) = 1 , we then see that ψ^{2} has conductor at least 3 . Then (7) above gives:

c_{M}(χ^{2}) = c_{M}(χ^{2} ∘ N_{M/ℚ_{2}}) = 12c_{ℚ_{2}}(ψ^{2}) − 26 ≥ 10 ,

which is impossible: If u ∈ U_{M}^{8}, then u^{2} ∈ U_{M}^{16}, and since c_{M}(χ) = 3t + 1 = 16, we have c_{M}(χ^{2}) ≤ 8. Hence χ^{2} = 1 , and ε(−1) = ε(5) = 1 .

The statements about the Artin conductor of the twists ρ ⊗ φ follow immediately from proposition 2 because ρ is a lifting with minimal Artin conductor.

Q.E.D.

4. Examples

4.0. We shall now illustrate the preceding 3 sections with some examples which will be analysed further in VI.

Denote for squarefree n ∈ ℤ by χ_{n} the character of the quadratic field ℚ(\sqrt{n})/ℚ .

We will enumerate irreducible Galois representations:

(+)  ρ : Gal (\overline{ℚ}/ℚ) → GL_{2}(ℂ)

with Artin conductor N and determinant character ε in the following cases:

N = p , ε = χ_{−p} with p ∈ {487, 751, 887, 919, 2083},
N = 4p , ε = χ_{−p} with p ∈ {487, 751, 887, 919},

and

N = 2^{5}p, ε = χ_{−p} with p ∈ {73, 193}.

Note that the numbers 73, 193, 487, 751, 887, 919 and 2083 are prime numbers.

We shall fix the following notation. The symbol ρ denotes a representation (+). If ρ is such a representation, we denote by \overline{ρ} the associated projective representation, by K/ℚ the extension cut out by \overline{ρ} , and for a prime number l by G_{l} the (isomorphism class of the) image under \overline{ρ} of some decomposition group over l. Also, N and ε denote the Artin conductor and determinant character of ρ respectively.

In the following we shall occasionally make use of the following simple principle: Suppose that G is the absolute Galois group of a number field L (finite extension of ℚ), that V is a finite-dimensional complex vectorspace and that:

R : G → Aut (V)

is a continuous representation. If M/L is the Galois extension cut out by R and M_{0}/L is a cyclic subextension, then the class field theoretic conductor of M_{0}/L divides the Artin conductor of R . This follows from the corresponding local statement:

Lemma: Suppose that p is a prime number, L/ℚ_{p} a finite extension, M/L a finite Galois extension with Galois group G , N a normal subgroup of G with G/N cyclic, M_{0} the fixed field of N in M , and that:

R : G → Aut (V)

is a finite-dimensional, faithful complex representation. Then the conductor of M_{0}/L divides the Artin conductor of R .

Proof: Let:

G ≥ G_{0} ≥ G_{1} ≥ … ≥ G_{i} ≥ …

be the chain of ramification groups in G and let g_{i} denote the order of G_{i} . Define the numbers t, s_{j} and d_{j} such that:

codim V^{G_{i}} = d_{0} ,  i = 0, …, s_{1}  ,
⋮
codim V^{G_{i}} = d_{t} ,  i = s_{t} + 1, …, s_{t+1} ,
codim V^{G_{i}} = d_{t+1} = 0 , i ≥ s_{t+1} + 1 ,

where d_{0} > d_{1} > … > d_{t} > d_{t+1} = 0 , and where V^{G_{i}} means the space of fixed points of G_{i} in V . Letting φ_{M/L} denote the Herbrand function of M/L , the Artin conductor of R is by definition ℘_{K}^{a(R)} where we have with s_{0} = −1 :

a(R) = ∑_{j=0}^{t} ∑_{i=s_{j}+1}^{s_{j+1}} d_{j} · \frac{g_{i}}{g_{0}}

= d_{0} ∑_{i=0}^{s_{1}} \frac{g_{i}}{g_{0}} + ∑_{j=1}^{t} d_{j} · (∑_{i=0}^{s_{j+1}} \frac{g_{i}}{g_{0}} − ∑_{i=0}^{s_{j}} \frac{g_{i}}{g_{0}} )

= d_{0} + d_{0}φ_{M/L}(s_{1}) + ∑_{j=1}^{t} d_{j}(φ_{M/L}(s_{j+1}) − φ_{M/L}(s_{j}))

= d_{0} + ∑_{j=1}^{t+1}(d_{j−1} − d_{j})φ_{M/L}(s_{j}) .

Let c be the largest integer with (G/N)_{c} ≠ 1 . Then the conductor of M_{0}/L is ℘_{L}^{γ} with:

γ = 1 + φ_{M_{0}/L}(c) .

If δ ∈ ℝ with δ > s_{t+1} then:

V^{G_{δ}} = V ,

and so G_{δ} = 1 , since R is faithful. Thus:

1 = (G_{δ}N)/N = (G/N)_{φ_{M/M_{0}} (δ)} ,

and so c < φ_{M/M_{0}}(δ) ; hence c ≤ φ_{M/M_{0}}(s_{t+1}) . Then:

φ_{M_{0}/L}(c) ≤ φ_{M_{0}/L}(φ_{M/M_{0}}(s_{t+1})) = φ_{M/L}(s_{t+1}) ,

i.e.:

γ ≤ 1 + φ_{M/L}(s_{t+1}) ≤ d_{0} + d_{t}φ_{M/L}(s_{t+1}) ≤ a(R) .

Q.E.D.

4.1. Let us now consider representations ρ of A_{5}-type with N = 2^{α}p , where α = 0, 2 or 5 , p is an odd prime number, and ε = χ_{−p} .

First, since p is odd the group G_{p} must be dihedral or cyclic. However, if G_{p} were dihedral, then \overline{ρ} would not have a lifting for which the p-part of its conductor were p ; this is an immediate consequence of theorem 1 of section 2. Hence G_{p} is cyclic, and since ε is quadratic, G_{p} is cyclic of order 2. As p is odd, we then find that the p-part of the discriminant of a “root field” of K , i.e. a subfield of K of degree 5 over ℚ , is p^{2} .

If α = 0 , we should then look for A_{5}-fields K for which the discriminant of as root field is p^{2} . Note that we must have p ≡ 3 (4) . Table 1, which will be discussed in II , reveals that for p ≤ 2083 there is exactly 1 such field K , namely the splitting field of the polynomial:

x^{5} + 8x^{3} + 7x^{2} + 172x + 53 ;

here, p = 2083 . The splitting field of this polynomial over ℚ_{2083} is:

ℚ_{2083}(\sqrt{−2083}) .

Using the theorem of Tate mentioned in section 1, we find that each of the 2 projective representations (corresponding to the 2 different embeddings A_{5} ↪ P GL_{2}(ℂ)) associated with this field has a lifting with conductor 2083 and determinant character χ_{−2083} . Any such lifting may be twisted, without changing conductor or determinant, with a quadratic character with conductor dividing 2083 , i.e. with 1 or χ_{−2083} . We conclude then that there are exactly 4 (inequivalent) representations ρ of A_{5}-type with N = 2083 and ε = χ_{−2083} . (The analysis of representations with prime conductor is by the way well-known, cf. [13]).

Let us then proceed with the case α = 2 and p ≡ 3 (4) . A priori the group G_{2} is cyclic of order 2 , dihedral of order 4 or 6 , or isomorphic to A_{4} . Since ε = χ_{−p} has conductor p , the cyclic case is excluded. Since α = 2 , one finds that the dihedral group of order 4 and A_{4} are also excluded. So, G_{2} must be dihedral of order 6 , i.e. it corresponds to the tamely ramified extension of order 3 of the unramified quadratic extension of ℚ_{2} . This gives the contribution 2^{2} to the discriminant of a root field of K . Accordingly, we must in this case look for A_{5}-fields K for which the discriminant of a root field is 2^{2} · p^{2} , where p is a prime number ≡ 3 (4) .Table 1 shows that for p ≤ 1041 there are exactly 4 such fields, namely: for p = 487, 751, 887, 919 the splitting field of the polynomial:

\begin{matrix}x^{5} & − & 7x^{3} & − & 17x^{2} & + & 18x & + & 73 & , \\ x^{5} & − & 8x^{3} & + & 10x^{2} & + & 160x & + & 128 & , \\ x^{5} & + & 10x^{3} & + & 10x^{2} & + & 44x & + & 56 & , \\ x^{5} & − & 8x^{3} & + & 28x^{2} & − & 40x & + & 48 & , \end{matrix}

respectively. In each case one finds that the splitting field of the polynomial over ℚ_{2} is indeed the tamely ramified extension of degree 3 of ℚ_{2}(\sqrt{5}) . Over ℚ_{p} the splitting field is ℚ_{p}(\sqrt{p}) in the cases p = 751, 887 or 919, and it is ℚ_{p}(\sqrt{−p}) for p = 487 . Using the theorem of Tate and theorem 1 above, we find in each case that any of the 2 projective representations associated with K lifts to a representation with N = 4p and ε = χ_{−p} , and that any such lifting can be twisted, without changing conductor or determinant, with a quadratic character with conductor dividing p . We conclude that for p = 487, 751, 887 or 919 there are exactly 4 representations of A_{5}-type with N = 4p and ε = χ_{−p} .

We then turn to the case α = 5 , ε = χ_{−p} . Again, the group G_{2} is either cyclic or order 2 , dihedral of order 4 or 6 , or isomorphic to A_{4} , but here none of these possibilities can be a priori excluded. We only find, using theorem 1 and the condition α = 5 , some restrictions if G_{2} has order 4 . Working through the various cases, one finds that the 2-part of the discriminant of a root field of K is in any case bounded by 2^{6} . Consequently, we want to look for A_{5}-fields for which the discriminant of a root field is 2^{β} · p^{2} , where p is a prime number and β ≤ 6 . Table 1 shows that for p = 73 or p = 193 there is exactly 1 such field, namely:

for p = 73 the splitting field of : x^{5} + 2x^{3} − 4x^{2} − 2x + 4 ,
and for p = 193 the splitting field of : x^{5} + 10x^{3} − 26x^{2} + 11x + 30 .

In both cases, β = 6 , and the splitting field over ℚ_{2} is the unique A_{4}-extension of ℚ_{2} ; over ℚ_{p} the splitting field is ℚ_{p}(\sqrt{−p}) . Using theorem 2 of section 3, we find in both cases (p = 73 or p = 193) that any of the 2 projective representations associated with the splitting field of the polynomial has a lifting with conductor 2^{5} · p whose determinant character is quadratic with conductor 2^{2} ·p ; this determinant character is thus χ_{−p} . It follows furthermore from theorem 2 that any such lifting can be twisted, without changing the conductor or determinant, with a quadratic character with conductor dividing 2^{2} · p . There are 4 such characters, namely 1, χ_{−1}, χ_{p} and χ_{−p} , and so we conclude that there are for p = 73 or p = 193 precisely 8 representations ρ of A_{5}-type with N = 2^{5} · p and ε = χ_{−p} .

4.2. We shall now find the representations of A_{4}- or S_{4}-type where (N, ε) is one of the possibilities mentioned in 4.0. The A_{4}-case may be dealt with quickly, since it is impossible for a simple reason: Suppose that ρ is an A_{4}-type representation. The projective kernel field K then contains a cyclic Galois extension L/ℚ of degree 3 . There exists an odd prime number l which ramifies in L . The group G_{l} is not isomorphic to A_{4} since l is odd, and as a subgroup of A_{4} it must then be cyclic of order 3 . It follows that the determinant character of ρ has order divisible by 3 . Since ε^{2} = 1 in any of our cases, the A_{4}-case will not occur.

Let us then turn to representations of S_{4}-type. If ρ is a representation of S_{4}-type, then the projective kernel field K contains a unique Galois extension L/ℚ with Galois group isomorphic to S_{3} . Let M/ℚ denote the quadratic extension in L . Since we require ρ to have conductor of the form 2^{α}p (where p is one of the prime numbers 73, 193, 487, 751, 887, 919, 2083), we see that L/ℚ , and thus M/ℚ is unramified outside {2, p} ; for p = 2083 , we have that L/ℚ is unramified outside {p} , since we require ρ to have conductor p in this case. In the other cases, 2 is either ramified or decomposed in M/ℚ which means that L/M must be unramified over 2 . Since we require the determinant character of ρ to be χ_{−p} , we find as in the preceding subsection that L/M is also unramified over p . Hence L/M is an unramified Galois extension of degree 3 and so the class number of M must be divisible by 3 . The class numbers h of the quadratic fields ℚ(\sqrt{a}) which a priori (recall the lemma of section 4.0.) are candidates for M are given by the following table:

a h  a h  a h  a  h
-1 1  -919 19  751 1  - 2 · 73 16
-2 1  -2083 7  887 1  2 · 73 2
-73 4  2 1  919 1  -2 · 193 20
-193 4  73 1  2083 1  2 · 193 2
-487 7  193 1
-751 15  487 1
-887 29

So, we only have to consider the case:

M = ℚ(\sqrt{−751}) ,

and here there does in fact exist a unique S_{3}-extension of ℚ containing M , namely the splitting field L of the polynomial:

x^{3} − 11x − 15 .

Now we have to ask whether L can be embedded in an S_{4}-extension of ℚ unramified outside {2, 751} . This is in fact possible as follows from the table in [6]: The splitting field of the polynomial:

x^{4} + 6x^{3} + 13x^{2} + 11x + 1

is an S_{4}-extension K/ℚ ramified only over 751 ; the discriminant of a quartic subfield of K is −751 , and K is unique with these properties. As is well-known (cf. [13]) or easily seen, this means that there are exactly 2 representations ρ with N = 751 and ε = χ_{−751} . Now, using the method of [2] one may verify that this field K is in fact the only possibility even if we allow K/L to be ramified over 2 : We consider the elliptic curve:

E : y^{2} = x^{3} − 11x − 15

over ℚ . Its conductor is N_{E} = 2 · 751 and its discriminant is Δ_{E} = −2^{4} · 751 . Now, from [2] we know that S_{4}-extensions K/ℚ containing the above S_{3}-extension L/ℚ are given by elements:

φ ∈ H^{1}(Gal (\overline{ℚ}/ℚ), E(\overline{ℚ})_{2})\\{0\} .

One may further show, cf. [5], that if we want to produce representations ρ with conductor of the form 2^{α} · 751 , then it suffices to consider non-trivial elements φ in the Selmer group S_{{2}}(E, ℚ)_{2} . This group can be determined by an algorithm of Birch and Swinnerton-Dyer, and one finds that it has order 2 , generated by δ(P) , where P = (\frac{17}{4}, \frac{31}{8}) is a rational point on E , and δ is the natural injection:

E(ℚ)/2E(ℚ) → H^{1}(Gal (\overline{ℚ}/ℚ), E(\overline{ℚ})_{2}) .

The corresponding S_{4}-extension of ℚ is then obtained by adjoining to ℚ the xcoordinates of points Q ∈ E(\overline{ℚ}) with 2Q = P . One finds then that this S_{4}-extension is the splitting field of the polynomial:

x^{4} − 17x^{3} + 22x^{2} + 307x + 376 ,

which of course is the same field as is obtained from the table in [6].

4.3. Finally, we shall briefly discuss representations ρ of dihedral type in the cases listed in 4.0. These are most easily obtained by global means, and let us recall how (cf. [13]): Any representation ρ of dihedral type has the form:

ρ = Ind (ψ) ,

where ψ is a character on Gal (\overline{ℚ}/M) , where M/ℚ is some quadratic extension, with σ · ψ ≠ ψ for a (any) element σ ∈ Gal (\overline{ℚ}/ℚ) which does not belong to Gal (\overline{ℚ}/M) , and where Ind means induction from Gal (\overline{ℚ}/M) to Gal (\overline{ℚ}/ℚ) . If ψ is viewed as a character on the idele classes of M , then the determinant character of ρ , viewed as a character on the idele classes of ℚ , is:

det (ρ) = ω · (ψ|C_{ℚ}) ,

where ω is the quadratic character of M/ℚ and ψ|C_{ℚ} means restriction of ψ to the idele classes of ℚ . The Artin conductor of ρ is:

|D(M/ℚ)| · N_{M/ℚ}(c_{M}(ψ)) ,

where D(M/ℚ) is the discriminant of M/ℚ and c_{M}(ψ) is the (class field theoretic) conductor of ψ . If ρ_{i} = Ind (ψ_{i}) , i = 1, 2 , are 2 such representations where ψ_{i} are characters on the idele classes of the \underline{same} quadratic field M/ℚ , then ρ_{1} and ρ_{2} are equivalent if and only if:

either ψ_{1} = ψ_{2} or ψ_{1} = σ · ψ_{2} (σ as above) .

The case of dihedral type representations with prime conductor p is well-known, cf. [13], and we shall not reproduce the simple arguments, but merely state the facts:

We must have p ≡ 3 (4) and with the above notation it suffices to consider the field M = ℚ(\sqrt{−p}) . The non-trivial automorphism σ of this extension M/ℚ , considered as acting on characters of the class group of M , fixes only the trivial character. The class number h of M is thus odd, and we obtain precisely \frac{1}{2}(h − 1) inequivalent representations of dihedral type with conductor p. The determinant character of any of these representations is χ_{−p} . Consequently, we have for p = 487, 751, 887, 919, 2083 exactly 3, 7, 14, 9, 3 representations of dihedral type with conductor p and determinant character χ_{−p} , respectively.

Let us then consider representations of dihedral type with conductor 4p and determinant χ_{−p} , where p is an odd prime number ≡ 3 (4) . A priori, we have 2 possibilities for the field M : M = ℚ(\sqrt{p}) or M = ℚ(\sqrt{−p}) . However, if M = ℚ(\sqrt{p}) , then the character ψ would have to be unramified with an \underline{odd} restriction to the idele classes of ℚ , which is easily seen to be impossible (for example by computing the appropriate ray class number). If M = ℚ(\sqrt{−p}) with p ≡ 7 (8) , then 2 decomposes:

(2) = ℘_{1}℘_{2} in M ;

since the ray class numbers of M corresponding to the cycles (2), ℘_{1}^{2} and ℘_{2}^{2} all coincide with the class number of M , this case is excluded. (By the way, if p ≡ 3 (8) , 2 is inert in M , and one finds that the ray class number corresponding to the cycle (2) is 3 times the class number h of M , and that we have exactly h representations of dihedral type with conductor 4p and determinant χ_{−p}). Hence, if p = 487, 751, 887, or 919 , there are no representations of dihedral type with conductor 4p and determinant χ_{−p} (and there are precisely 7 such representations if p = 2087).

Suppose now that p is a prime number ≡ 1 (8) , and let us ask for dihedral-type representations with N = 2^{5} · p and ε = χ_{−p} . Keeping the above notation, the field M must be one of the following:

ℚ(\sqrt{±2}) , ℚ(\sqrt{−1}) , ℚ(\sqrt{±2 · p}) , ℚ(\sqrt{±p}) .

If M is one of the fields ℚ(\sqrt{±2}) , ℚ(\sqrt{±2 · p}) , then 2 ramifies:

(2) = ℘^{2} in M ,

and the p-part of the conductor of the character ψ should be ℘^{2} . But then the conductor of ψ|C_{ℚ} would certainly not be divisible by 2^{3} , which would be required by these cases. If M = ℚ(\sqrt{−p}) , then 2 is again ramified in M : (2) = ℘^{2} , but now the character ψ should have conductor ℘^{3} and trivial restriction to C_{ℚ} ; this is easily seen to be impossible (consider the restriction of ψ to M_{℘}^{x}). Similarly, the case M = ℚ(\sqrt{−1}) is excluded. If then finally M = ℚ(\sqrt{p}) , then 2 decomposes in M :

(2) = ℘_{1}℘_{2} ,

and ψ|C_{ℚ} should be χ_{−p} . The norm of the conductor of ψ should be 2^{5} and so there are various possibilities for the conductor of ψ . Suppose for example that ψ has conductor ℘_{1}^{3}℘_{2}^{2} and consider the restrictions:

ψ_{i} = ψ|M_{℘_{i}}^{x} , i = 1, 2 .

Note that M_{℘_{i}} = ℚ_{2} . Then ψ_{1} cannot be trivial on all elements of the type 1 + b · 2^{2} + …; for otherwise there would be an element:

u = 1 + 2 + b · 2^{2} + …

with ψ_{1}(u) = −1 . Since ψ_{2} has conductor ℘_{2}^{2} we would also have ψ_{2}(u) = −1 . We then see that the restriction ψ|C_{ℚ} would be trivial on ℚ_{2}^{x} which is impossible. So, there is an element y = 1 + b · 2^{2} + … with ψ_{1}(y) ≠ 1 . Since ψ_{2} has conductor ℘_{2}^{2} we conclude that ψ|C_{ℚ} has conductor divisible by 2^{3} , which is impossible. The other possibilities for the conductor of ψ are similarly excluded.

We conclude that if p is a prime number ≡ 1 (8) , then there are no dihedral type representations with conductor 2^{5} · p and determinant character χ_{−p} . This holds then in particular for p = 73 and for p = 193 .

Let us then finally, for use in VI, note the following fact: If p = 73 or p = 193 , then there is at least 1 dihedral-type representation with conductor 2^{2} ·p and determinant character χ_{−p} . This follows, as one may easily verify in both of these cases, because the ideal class group of ℚ(\sqrt{−p}) is cyclic of order 4 and contains an ideal class which is not invariant under the action of the non-trivial automorphism of ℚ(\sqrt{−p})/ℚ .

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