Lifting Galois representations

Ravi Ramakrishna Cornell University

October 12, 1999

\underline{Section 1: Introduction} Let p ≥ 5 be a prime and k be a finite extension of F_{p} the field of p elements. Let ρ̄ : G_{Q} = Gal(Q̄/Q) → GL_{2}(k) be a continuous Galois representation. If ρ̄ is odd (i.e. detρ̄(c) = −1 where c is complex conjugation) and absolutely irreducible then Serre has conjectured in [Se1] that ρ̄ “comes from” a modular newform f of prescribed weight k(ρ̄), level N(ρ̄) and character ω(ρ̄). Following work of Eichler-Shimura, Deligne and Deligne-Serre there would then exist a representation ρ_{f} : G_{Q} → GL_{2}(O), where O is the ring of integers of some finite extension of W (k), the ring of Witt vectors of k, with the following properties:

• ρ_{f} is unramified at primes q not dividing N(ρ̄)p.

• For such q let F rob_{q} denote the conjugacy class of Frobenius at q. Then Trace(ρ_{f}(F rob_{q})) = a_{q}, an algebraic integer which is the eigenvalue of the Hecke operator T_{q} acting on the newform f with coefficients in O.

• If π is a uniformizer of O then ρ_{f} mod π is equivalent to ρ̄.

Serre’s Conjecture thus trivially implies, in Mazur’s language, that there is a characteristic zero deformation of ρ̄. Conversely, the existence of such a deformation might be regarded as evidence for Serre’s Conjecture. In [Kh] Khare has shown that if one is willing to allow additional ramification at a finite set of primes not dividing N(ρ̄)p then there is a deformation to mod p^{2}, i.e. a representation ρ_{2} : G_{Q} → GL_{2}(W (k)/p^{2}). His method works for ρ̄ even as well. If ρ̄ is reducible Khare has shown that ρ̄ can be deformed to W (k) if additional ramification is allowed.

The main result of this paper is that with several technical hypotheses on an absolutely irreducible ρ̄ one can deform ρ̄ to W (k). As in Khare’s work more ramification must be allowed and the methods are independent of the parity of ρ̄. We do not know if these deformations are potentially semistable at p in the sense of Fontaine. We also do not know whether, for unramified primes, the trace of Frobenii are algebraic.

After twisting ρ̄ by a character we may (and do) assume the fields fixed by the kernel of ρ̄ and the kernel of its associated projective representation are ramified at the same set of primes. We denote these fields Q(ρ̄) and Q(Ad^{0}ρ̄) throughout this paper. Note that any of twist ρ̄ by a continuous k̄^{∗}-valued character can only increase the set of ramified primes of ρ̄. Let S denote the union of this common set of primes, {p}, and the infinite prime. Let G_{S} = Gal(Q_{S}/Q) where Q_{S} is the maximal extension of Q ramified only at primes in S. Note ρ̄ factors through G_{S} and that with the possible exception of the prime p and the infinite prime all twists of ρ̄ are ramified at all primes of S. For a place v let G_{v} = Gal(Q̄_{v}/Q_{v}). Let Ad^{0}ρ̄ denote the set of 2×2 trace zero matrices with entries in k and G_{S} action through ρ̄ and by conjugation. Let (Ad^{0}ρ̄)^{∗} be its Cartier dual and N and N^{d} be the maximal subgroups of G_{S} that act trivially on Ad^{0}ρ̄ and (Ad^{0}ρ̄)^{∗} respectively. Note Q(Ad^{0}ρ̄) is the fixed field of N. Let D = Q(Ad^{0}ρ̄)∩Q(μ_{p}) and K be the composite Q(Ad^{0}ρ̄)Q(μ_{p}).

Observe Gal(K/D) ≃ Gal(Q(Ad^{0}ρ̄)/D) × Gal(Q(μ_{p})/D). Finally note that for Ad^{0}ρ̄ absolutely irreducible, the minimal field of definition of the representation of G_{S} on the three dimensional k space Ad^{0}ρ̄ may in fact be a proper subfield of k. We call this minimal field k̃. This is discussed in section 6. Note k ≠ k̃ in the odd example of section 8. Note k ≠ k̃ in the odd example of section 8.

[figure: Q; D; Q(μ_{p}); Q(Ad^{0}ρ̄); K]

Theorem 1Let ρ̄ : G_{S} → GL_{2}(k) be an absolutely irreducible Galois representation where the characteristic p of k is greater than or equal to 5. Assume Ad^{0}ρ̄ is an absolutely irreducible k[G_{S}] module, that H^{1}(G_{S}/N, Ad^{0}ρ̄) and H^{1}(G_{S}/N^{d}, (Ad^{0}ρ̄)^{∗}) are trivial, and that H^{2}(G_{v}, Ad^{0}ρ̄) = 0 for all v ∈ S. Finally suppose there is an element

a × b ∈ Gal(Q(Ad^{0}ρ̄)/D) × Gal(Q(μ_{p})/D) ≃ Gal(K/D) ⊆ Gal(K/Q)

such that a corresponds to an element in the (projective) image of ρ̄ whose eigenvalues have ratio t ∈ F_{p}^{∗}, where t ≠ ±1, and b ∈ Gal(Q(μ_{p})/Q) → (Z/p)^{∗} maps to the element t. Let r = dim_{k}H^{2}(G_{S}, Ad^{0}ρ̄) Then there is a set of primes Q = {q_{1}, q_{2}, .., q_{r}} disjoint from S and a representation ρ : G_{S∪Q} → GL_{2}(W (k)) such that ρ ≡ ρ̄ mod p.

While the above theorem is stated quite generally it is not immediately clear whether the hypotheses can actually be satisfied in a case of interest! The following theorem is a corollary of Theorem 1. Recall that Serre’s weight, k(ρ̄), is an invariant of ρ̄ |_{G_{p}=Gal(Q̄_{p}/Q_{p})}, and therefore makes sense for ρ̄ even as well as odd.

Theorem 2 Let p ≥ 7. Assume the image of ρ̄: G_{S} → GL_{2}(k) contains SL_{2}(k). Also assume if l ∈ S and l ≠ p then l ≢ ±1 mod p, that k(ρ̄) ≠ p − 1 or 2p, and that k(ρ̄) ≢ 2 mod p + 1. Let r = dim_{k}H^{2}(G_{S}, Ad^{0}ρ̄). Then there is a set of primes Q = {q_{1}, q_{2}, .., q_{r}} disjoint from S and a representation ρ : G_{S∪Q} → GL_{2}(W (k)) such that ρ ≡ ρ̄ mod p.

The method is to deform ρ̄ one step at a time from mod p^{n} to mod p^{n+1} as in [R1-3]. Such deformation questions have been considered in [Ma1], [BM], [B1-3], [Bö], and [Kh]. As obstructions to deformation problems lie in H^{2}(G_{S}, Ad^{0}ρ̄) the exact sequence

0 → � _{S}^{2}(Ad^{0}ρ̄) → H^{2}(G_{S}, Ad^{0}ρ̄) → ⊕_{v∈S}H^{2}(G_{v}, Ad^{0}ρ̄)

is extremely important. A typical approach for lifting to characteristic zero (for ρ̄ not known to be modular) has been to work with explicit examples where H^{2}(G_{S}, Ad^{0}ρ̄) or at least � _{S}^{2}(Ad^{0}ρ̄) can be shown to be trivial. Here we assume the right hand term in the exact sequence is trivial. Thus � _{S}^{2}(Ad^{0}ρ̄) = H^{2}(G_{S}, Ad^{0}ρ̄) or more loosely ‘all obstructions to deformation questions arise from class group problems’.

By the work of Poitou-Tate the kernel of H^{1}(G_{S}, (Ad^{0}ρ̄)^{∗}) → ⊕_{v∈S}H^{1}(G_{v}, (Ad^{0}ρ̄)^{∗}), which is denoted � _{S}^{1}((Ad^{0}ρ̄)^{∗}), and � _{S}^{2}(Ad^{0}ρ̄) are dual. (The duality uses the fact that S contains the infinite place.) For us � _{S}^{1}((Ad^{0}ρ̄)^{∗}) will play the role of the dual Selmer group in [Wi] and [TW]. Following those papers, we carefully choose a set Q of auxiliary primes that annihilates � _{S∪Q}^{1}((Ad^{0}ρ̄)^{∗}). It will then turn out that global obstructions to deformation problems, allowing ramification in S ∪ Q, need only be studied locally at primes in Q. The method of [R1-3] can then be used to remove these local obstructions. We cannot guarantee that ρ in the theorems will be ramified at the primes in Q. We only know ρ exists and is unramified outside S ∪ Q.

In this paper we study the cohomology of Ad^{0}ρ̄ as opposed to that of Adρ̄, the set of all 2 × 2 matrices over k. For our purposes this means we are fixing the determinant of all deformations of ρ̄ that we consider once and for all. Let χ denote the cyclotomic character, both mod p and in characteristic zero. We then have detρ̄ = ω(ρ̄)χ^{k(ρ̄)−1}. Let ω̃ be the Teichmüller lift of ω. For definiteness we may assume the fixed determinant of all our deformations is ω̃χ^{k(ρ̄)−1}.

The author would like to thank the referee for numerous helpful suggestions. In particular, the referee suggested the formulation and proof of Lemma 7 which leads to the corrected proof we give for Lemma 8.

\underline{Notations}

p: A prime ≥ 5. k : A finite field of characteristic p ≥ 5. k̃: A subfield of k which is the minimal field of definition of Ad^{0}ρ̄ and (Ad^{0}ρ̄)^{∗}. k̄: A separable closure of k. k̂: A finite extension of k. k(ϕ): The one dimensional space k with Galois action by the character ϕ. W (k): The ring of Witt vectors of k. S: A finite set of places (including p and ∞). Q: A finite set of primes disjoint from S. Q_{v}: The completion of Q at the place v. Q_{S}: The maximal separable extension of Q unramified outside the places of S. G_{Q}: Gal(Q̄/Q). G_{S}: Gal(Q_{S}/Q) G_{v}: Gal(Q̄_{v}/Q_{v}). I_{v}: The inertia subgroup of G_{v}. 1: The two by two identity matrix over W (k) or W (k)/p^{m} for suitable m. χ: The cyclotomic character. ρ̄: A continuous representation from G_{S} to GL_{2}(k). k(ρ̄): Serre’s weight for ρ̄. Ad^{0}ρ̄: The trace zero two by two matrices over k with G_{S} action through ρ̄ and by conjugation. (Ad^{0}ρ̄)^{∗}: The Cartier dual of Ad^{0}ρ̄. Ãd^{0}ρ̄: A descent of Ad^{0}ρ̄ to its minimal field of definition k̃. (Ãd^{0}ρ̄)^{∗}: A descent of (Ãd^{0}ρ̄)^{∗} to its minimal field of definition k̃.

\underline{Section 2: Deformation Theory} We give a short introduction to deformation theory. See [Ma1], [BM], [Bo1] and [Bo2] for details and more results.

Let π̄ : H → GL_{d}(k) be an absolutely irreducible continuous representation of a profinite group H. Suppose H^{1}(H, Ad^{0}π̄) is finite dimensional. Let C be the category of Artinian local rings with residue field k where the morphisms are homomorphisms that induce the identity map on k. Let R be in C. We call two lifts γ_{1} and γ_{2} of π̄ to GL_{n}(R) strictly equivalent if γ_{1} = Aγ_{2}A^{−1} for some A congruent to the identity matrix modulo the maximal ideal m_{R} of R. We call a strict equivalence class of lifts of π̄ to R a deformation of π̄ to R.

Mazur studied the deformations of π̄ and proved the following fundamental theorem in [Ma1].

Theorem A There is a complete local Noetherian ring R^{un} with residue field k and a continuous homomorphism π̃ : H → GL_{d}(R^{un}) such that

1) Reduction of π̃ modulo the maximal ideal of R^{un} gives π̄

2) For any ring R in C and any deformation γ of π̄ to GL_{d}(R) there is a unique homorphism ϕ : R^{un} → R in C such that ϕ ∘ π̃ = γ as deformations.

Moreover, if π̄ is not absolutely irreducible the statements hold except the ϕ in part 2 may not be unique. We call R^{un} the universal deformation ring associated to H and π̄ in the absolutely irreducible case. We call R^{un} the versal ring associated to H and π̄ otherwise. In either case we have the following fact. Fact R^{un} is a quotient of W (k)[[T_{1}, T_{2}, ...T_{m}]] where m = dim_{k}H^{1}(H, Ad^{0}π̄).

The elements of H^{1}(H, Ad^{0}π̄) correspond to the deformations of π̄ to k[ϵ] = k[X]/(X^{2}), the dual numbers of k. Given f ∈ H^{1}(H, Ad^{0}π̄) the corresponding deformations of π̄ to the dual numbers is given by π_{f}(σ) = (I + ϵf(σ))π̄(σ).

Let π_{n} be a deformation of π̄ to GL_{d}(W (k)/p^{n}). We ask if π_{n} deforms to GL_{d}(W (k)/p^{n+1}). The obstruction to deforming π_{n} to GL_{d}(W (k)/p^{n+1}) lies in H^{2}(H, Ad^{0}π̄). If this obstruction is trivial π_{n} deforms to some π_{n+1} and pr ∘ π_{n+1} = π_{n} where pr : W (k)/p^{n+1} → W (k)/p^{n} is the canonical projection. When such a lift exists one sees that H^{1}(H, Ad^{0}π̄) acts on the set of deformations of π_{n} to GL_{d}(W (k)/p^{n+1}). For f ∈ H^{1}(H, Ad^{0}π̄) the action is given by (f.π_{n+1})(σ) = (I + p^{n}f(σ))(π_{n+1}(σ)). If π̄ is absolutely irreducible H^{1}(H, Ad^{0}π̄) acts on the deformations of π_{n} to GL_{d}(W (k)/p^{n+1}) as a principal homogeneous space.

Mazur has also shown that modifications could be made so related functors with the ordinary restriction are also representable. Here H is a Galois group and we insist that when restricted to a suitable inertia group I_{p} that we only consider lifts π of π̄ whose restriction to I is of the form

( \begin{matrix}ψ & ∗ \\ 0 & 1 \end{matrix} ). See [Ma1] and [Ma2] for details.

\underline{Section 3: Local at q_{i} Deformation Theory} Let q_{i} ≢ ±1 mod p (and q_{i} ≠ p) be a prime (at which we eventually wish to allow ramification) and let G_{q_{i}} = Gal(Q̄_{q_{i}}/Q_{q_{i}}). Suppose ρ̄: G_{q_{i}} → GL_{2}(k) is unramified at q_{i} and ρ̄(F rob_{q_{i}}) has (necessarily distinct) eigenvalues q_{i}x_{i} and x_{i}. We need to study the deformation theory of this local at q_{i} representation to understand the role the q_{i} play in Theorems 1 and 2. The characteristic polynomial p(z) of ρ̄(F rob_{q_{i}}) is z^{2} − x_{i}(q_{i} + 1)z + q_{i}x_{i}^{2} has coefficients in k. But q_{i} is a rational integer so its reduction mod p lies in F_{p} and is therefore in k. Since x_{i}(q_{i} + 1) ∈ k we see x_{i} ∈ k so p(z) has distinct roots in k. Thus we may assume

ρ̄(F rob_{q_{i}}) = ( \begin{matrix}q_{i}x_{i} & 0 \\ 0 & x_{i} \end{matrix} ) for a suitable choice of basis of the two dimensional k vector space on

which G_{q_{i}} acts.

Recall we denote the cyclotomic character by χ. By k(ϕ) we mean the one dimensional k vector space with Galois action by the character ϕ and by k we denote the one dimensional k vector space with trivial Galois action.

Lemma 1 Let q_{i} be as above. Then H^{2}(G_{q_{i}}, Ad^{0}ρ̄) is one dimensional and H^{1}(G_{q_{i}}, Ad^{0}ρ̄) is two dimensional.

Proof: Note that with G_{q_{i}} action, Ad^{0}ρ̄ ≃ k ⊕ k(χ) ⊕ k(χ^{−1}). Since we are assuming q_{i} ≢ 1 mod p we see H^{0}(G_{q_{i}}, Ad^{0}ρ̄) is one dimensional. Observe (Ad^{0}ρ̄)^{∗} ≃ k(χ) ⊕ k ⊕ k(χ^{2}). Thus for q_{i} ≢ ±1 mod p we see H^{0}(G_{q_{i}}, (Ad^{0}ρ̄)^{∗}) is one dimensional. By local duality H^{0}(G_{q_{i}}, (Ad^{0}ρ̄)^{∗}) is dual to H^{2}(G_{q_{i}}, Ad^{0}ρ̄) which is thus one dimensional. The local Euler-Poincare characteristic gives the result for H^{1}(G_{q_{i}}, Ad^{0}ρ̄).

We want to consider deformations of ρ̄ to W (k)/p^{n}. As q_{i} ≠ p, such deformations factor through the Galois group of the maximal tamely ramified extension Q_{q_{i}}^{t} over Q_{q_{i}}. This group is well understood. See Chaiptre II, §5.6 of [Se2]. Thus we may choose topological generators of the tame quotient of G_{q_{i}}, σ_{q_{i}} and τ_{q_{i}} subject to the relation σ_{q_{i}}τ_{q_{i}}σ_{q_{i}}^{−1} = τ_{q_{i}}^{q_{i}}. Recall τ_{q_{i}} topologically generates inertia and σ_{q_{i}} maps to Frobenius under the surjection Gal(Q_{q_{i}}^{t}/Q_{q_{i}}) → Gal(F̄_{q_{i}}/F_{q_{i}}). Let x̃_{i} ∈ W (k) be the Teichmüller lift of x_{i}.

We call a deformation (or a representation in this equivalence class) of ρ̄ to W (k)/p^{n} (allowing n = ∞ in the case of a deformation to W (k)) of our “desired form” if it is given by σ_{q_{i}} ↦

( \begin{matrix}q_{i}x̃_{i} & 0 \\ 0 & x̃_{i} \end{matrix} ) and τ_{q_{i}} ↦ ( \begin{matrix}1 & ∗ \\ 0 & 1 \end{matrix} ). The images of σ_{q_{i}} and τ_{q_{i}} satisfy the relation σ_{q_{i}}τ_{q_{i}}σ_{q_{i}}^{−1} = τ_{q_{i}}^{q_{i}}.

We want to deform this to W (k)/p^{n+1}.

Lemma 2 A deformation of our “desired form” to W (k)/p^{n} deforms to W (k)/p^{n+1}.

Proof: One need only lift ∗ from mod p^{n} to mod p^{n+1} to get a deformation to mod p^{n+1} of the “desired form”.

We give a basis for H^{1}(G_{q_{i}}, Ad^{0}ρ̄). Recall that ρ̄(σ_{q_{i}}) = ( \begin{matrix}q_{i}x_{i} & 0 \\ 0 & x_{i} \end{matrix} ) and ρ̄(τ_{q_{i}}) = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ).

A nontrivial unramified deformation to k[ϵ] is given by ρ(σ_{q_{i}}) = ( \begin{matrix}q_{i}x_{i} & 0 \\ 0 & x_{i} \end{matrix} ) + ϵ ( \begin{matrix}q_{i}x_{i} & 0 \\ 0 & −x_{i} \end{matrix} )

and ρ(τ_{q_{i}}) = 1 + ϵ ( \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ). Let g ∈ G_{q_{i}} and let n_{g} be the image of g under the composite map

G_{q_{i}} → Ẑ → Z/p. Then r_{q_{i}}(g) = ( \begin{matrix}n_{g} & 0 \\ 0 & −n_{g} \end{matrix} ).

A nontrivial ramified deformation to the dual numbers is given by ρ(σ_{q_{i}}) = ( \begin{matrix}q_{i}x_{i} & 0 \\ 0 & x_{i} \end{matrix} ) +

ϵ ( \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ) and ρ(τ_{q_{i}}) = 1 + ϵ ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ). We see σ_{q_{i}}τ_{q_{i}}σ_{q_{i}}^{−1} = τ_{q_{i}}^{q_{i}} holds. The corresponding

cohomology class is given by s_{q_{i}}(σ_{q_{i}}) = ( \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix} ) and s_{q_{i}}(τ_{q_{i}}) = ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ).

Clearly the images of r_{q_{i}} and s_{q_{i}} in H^{1}(G_{q_{i}}, Ad^{0}ρ̄) are linearly independent and therefore they span the two dimensional space H^{1}(G_{q_{i}}, Ad^{0}ρ̄). Note that both r_{q_{i}} and s_{q_{i}}, or more precisely their corresponding deformations to k[ϵ], cut out Z/p extensions of Q_{q_{i}}(ρ̄), the extension of Q_{q_{i}} fixed by the kernel of ρ̄. Any linear combination ur_{q_{i}} +vs_{q_{i}} with u, v ≠ 0 in k cuts out the unique Z/p×Z/p extension of Q_{q_{i}}(ρ̄).

Also note that for any deformation to mod W (k)/p^{n}, n ≥ 2, of our “desired form”, acting on it by the cohomology class s_{q_{i}} leaves it in the “desired form”. One sees this by noting

(1 + p^{n−1}s_{q_{i}}(σ_{q_{i}})) ( \begin{matrix}q_{i}x̃_{i} & 0 \\ 0 & x̃_{i} \end{matrix} ) = ( \begin{matrix}q_{i}x̃_{i} & 0 \\ 0 & x̃_{i} \end{matrix} )

and

(1 + p^{n−1}s_{q_{i}}(τ_{q_{i}})) ( \begin{matrix}1 & ∗ \\ 0 & 1 \end{matrix} ) = ( \begin{matrix}1 & ∗ + p^{n−1} \\ 0 & 1 \end{matrix} ) = ( \begin{matrix}1 & ˜∗ \\ 0 & 1 \end{matrix} ) .

We call a cohomology class that preserves the “desired form” null. If a cohomology class does not preserve the desired form we call it nonnull. Observe r_{q_{i}} is nonnull.

Proposition 1 Let ρ̄ : G_{q_{i}} → GL_{2}(k) be unramified and given by ρ̄(σ_{q_{i}}) = ( \begin{matrix}q_{i}x_{i} & 0 \\ 0 & x_{i} \end{matrix} ) with

q_{i} ≢ ±1 mod p. Fix f ∈ H^{1}(G_{q_{i}}, Ad^{0}ρ̄) such that f and s_{q_{i}} are linearly independent. Let ρ_{n} be a deformation of ρ̄ to mod p^{n} of the “desired form” and ρ_{n+1} be any deformation of ρ_{n} to mod p^{n+1}. Then there is an ν_{n+1} ∈ k such that (ν_{n+1}f).ρ_{n+1} is of our “desired form.” Thus ρ̄ can be lifted to W (k) one step at a time with adjustments made at each step only by a multiple of f. In particular, we may take for f any nonzero multiple of the unramified cohomology class r_{q_{i}}.

Proof: Note that from the discussion in section 2 we see ρ_{n+1} differs from the “desired form” by the action of some element ν_{n+1}f + μ_{n+1}s_{q_{i}} of the two dimensional space H^{1}(G_{q_{i}}, Ad^{0}ρ̄). That is, (ν_{n+1}f + μ_{n+1}s_{q_{i}}).ρ_{n+1} is of the “desired form”. Since the one dimensional subspace of null cohomology classes preserves this form we see (ν_{n+1}f).ρ_{n+1} is of the “desired form”.

Remark: One possibility is that our characteristic zero representation may have τ_{q_{i}} ↦ 1, that is it might be unramified.

Section 4: Local at l ≠ p Obstructions

Lemma 3 H^{2}(G_{v}, Ad^{0}ρ̄) ≠ 0 if and only if Ad^{0}ρ̄ has a one dimensional quotient on which G_{v} acts by the cyclotomic character χ.

Proof: In this lemma v is any finite prime, including p. By local duality we see H^{2}(G_{v}, Ad^{0}ρ̄) ≠ 0 if and only if H^{0}(G_{v}, (Ad^{0}ρ̄)^{∗}) ≠ 0. This happens exactly when (Ad^{0}ρ̄)^{∗} has a G_{v} stable one dimensional subspace, that is when Ad^{0}ρ̄ has a one dimensional quotient (by a G_{v} stable two dimensional subspace) on which G_{v} acts by χ.

The aim of this section is to prove the proposition below.

Proposition 2 Let p ≥ 5 and l be a prime such that l ≠ p, l ≢ ±1 mod p. Suppose ρ̄: G_{l} → GL_{2}(k) and the action of G_{l} on Ad^{0}ρ̄ is ramified. Then H^{2}(G_{l}, Ad^{0}ρ̄) = 0.

Remark: We choose p ≥ 5 as all primes (other than 3 itself) are congruent to ±1 mod 3. While the conditions of Proposition 2 are sufficient they are not necessary for the conclusions. We have chosen these strong hypotheses because they are easy to state and verify.

Proof: We know H^{2}(G_{l}, Ad^{0}ρ̄) = 0 if and only if Ad^{0}ρ̄ does not have a one dimensional quotient (by a two dimensional G_{l} stable subspace) on which G_{l} acts by χ. As l ≠ p this action is unramified and as l ≢ ±1 mod p we see χ has order at least 3. Let G̃ and Ĩ be the images of G_{l} and its inertia subgroup I_{l} under the composite map G_{l} → GL_{2}(k) → P GL_{2}(k).

We need the following lemma to prove the proposition.

Lemma 4 If [G̃ : Ĩ] ≤ 2 then Ad^{0}ρ̄ has no one dimensional quotient on which G_{l} acts by χ. Thus H^{2}(G_{l}, Ad^{0}ρ̄) = 0 in such cases.

Proof: The order of χ would have to be 1 or 2, i.e. l ≡ ±1 mod p. We are excluding these cases.

We recall the classification in [Di1] of two dimensional local mod p Galois representations. Note that we are reversing the roles of l and p in [Di1] and the classification is often only given up to a twist. Since Ad^{0}ρ̄ as a k[G_{l}] module is insensitive to twists by k̄^{∗}-valued characters this does not affect our computations. We do not need to worry about possible difficulties arising from scalar extensions associated to these twists or about the minimal field of defintion of Ad^{0}ρ̄ as H^{2}(G_{l}, Ad^{0}ρ̄) ⊗_{k} k̄ ≃ H^{2}(G_{l}, Ad^{0}ρ̄ ⊗_{k} k̄) where k̄ is a separable closure of k. Thus we extend scalars and consider Ad^{0}ρ̄ and (the representation space of) ρ̄ as k̄[G_{l}] module.

In his classification Diamond has four cases that he calls P, S, V and H. We recall these below under the hypotheses that l ≢ ±1 mod p and ρ̄ is ramified. We do not give all of Diamond’s equivalent formulations of each case.

• P : ρ̄ is twist equivalent to a representation of the form ( \begin{matrix}ψ & 0 \\ 0 & 1 \end{matrix} ) for some ramified character

ψ.

• S : ρ̄ is twist equivalent to ( \begin{matrix}χ & u \\ 0 & 1 \end{matrix} ) where χ is the cyclotomic character and u is a not

coboundary.

• V : Ĩ is cyclic of order not divisible by p and G̃ is dihedral of twice that order.

• H : a) Ĩ is dihedral of order 2l^{r} for some r ≥ 1 and G̃ is dihedral of order dividing 4l^{r}, or b) l = 2, Ĩ (respectively G̃) is isomorphic to D_{4} (respectively A_{4}), A_{4} (respectively A_{4}), or A_{4} (respectively S_{4}).

Proof of Proposition 2: In case V the hypotheses of Lemma 4 are satisfied so Proposition 2 holds in this case.

In case P we see ρ̄ is a direct sum of the trivial character and a ramified character. One easily sees that as a k̄[G_{l}] module Ad^{0}ρ̄ becomes a direct sum of three one dimensional k̄[G_{l}] modules. One of these is trivial and the other two have ramified G_{l} action. Thus H^{2}(G_{l}, Ad^{0}ρ̄) is nontrivial only if χ is the trivial character, that is l ≡ 1 mod p. We are excluding this case as well.

In case S we have ρ̄ = ( \begin{matrix}χ & u \\ 0 & 1 \end{matrix} ). Then Ad^{0}ρ̄ has a unique one dimensional quotient and G_{l}

acts on this quotient by χ^{−1}. Since l ≢ −1 mod p we know χ^{−1} ≠ χ. Thus H^{2}(G_{l}, Ad^{0}ρ̄) = 0 in this case.

It remains to consider case H. Here the hypotheses of Lemma 2 are satisfied except when l = 2, Ĩ ≃ D_{4} and G̃ ≃ A_{4}. By Lemma 3 we know that for H^{2}(G_{l}, Ad^{0}ρ̄) not to be trivial there must be a one dimensional quotient of Ad^{0}ρ̄ on which G_{l} acts via χ. This quotient must be by a two dimensional G_{l} stable subspace. Since #G̃ = 12 and p ≥ 5 we see that G_{l} acts on Ad^{0}ρ̄ via a quotient of order prime to p. By the theory of representations of finite groups of order prime to the characteristic we see Ad^{0}ρ̄ is a semisimple representation of A_{4}. Since A_{4} acts faithfully on Ad^{0}ρ̄ we easily see Ad^{0}ρ̄ is an irreducible three dimensional representation of A_{4}. Thus there are no two dimensional invariant subspaces so we must have H^{2}(G_{l}, Ad^{0}ρ̄) = 0 in this last case.

Proposition 2 is now proved.

\underline{Section 5: Local at p Obstructions} Recall that Serre attaches in [Se1] a weight k(ρ̄) to a mod p Galois representation that depends only on its restriction to I, the inertia subgroup of G_{p}. (We use I for inertia in this section to be consistent with [Se1]). This definition therefore applies to ρ̄ even as well as odd.

Proposition 3 Let p ≥ 5 and ρ̄: G_{p} → GL_{2}(k) be given. If k(ρ̄) ≠ p − 1 or 2p and k(ρ̄) ≢ 2 mod (p + 1) then H^{2}(G_{p}, Ad^{0}ρ̄) = 0.

Remark: As before the hypotheses are sufficient for the conclusion but not necessary. See for example section 11 of [Ma3]. As in section 4 (of this paper) we need not worry about possible scalar extensions since such extensions do not change the dimension of the cohomology groups. Thus we consider Ad^{0}ρ̄ as a k̄[G_{p}] module.

Proof: Following section 2 of [Se1] we will separate the cases when inertia acts (through ρ̄) via characters of level two and characters of level one. By Lemma 3, to show H^{2}(G_{p}, Ad^{0}ρ̄) = 0, it suffices to show that Ad^{0}ρ̄ has no one dimensional quotient on which G_{p} acts via χ. Here χ is ramified. We will study the k̄[I] module Ad^{0}ρ̄ and show that if the conditions of Proposition 3 are satisfied then Ad^{0}ρ̄ has no one dimensional quotient on which I acts via χ.

If inertia acts through ρ̄ via characters of level two, then by section 2.2 of [Se1] we see ρ̄ |_{G_{p}}

is irreducible and ρ̄ |_{I}= ( \begin{matrix}ϕ & 0 \\ 0 & ϕ^{p} \end{matrix} ) where ϕ is a character of level two. Thus as a k̄[I] module

Ad^{0}ρ̄ decomposes as a direct sum of 3 one dimensional spaces, one with trivial I action, one with I action via ϕ^{p−1} and one with I action via ϕ^{1−p}. To show H^{2}(G_{p}, Ad^{0}ρ̄) = 0 it suffices to show neither ϕ^{p−1} nor ϕ^{1−p} equal χ.

From [Se1] we see that we can write ϕ = ψ^{a+pb} where ψ is a fundamental character of level two and 0 ≤ a, b ≤ p − 1. Thus ϕ^{p−1} = ψ^{a(p−1)+b(p^{2}−p)}. Recall χ = ψ^{p+1} and that ψ has order p^{2} − 1.

Lemma 5 With ϕ as above, neither ϕ^{p−1} nor ϕ^{1−p} equal χ. Therefore if I acts (through ρ̄^{ss}) via characters of level two then H^{2}(G_{p}, Ad^{0}ρ̄) = 0.

Proof: Setting ϕ^{p−1} = χ we see ψ^{a(p−1)+b(p^{2}−p)} = ψ^{p+1} or equivalently a(p − 1) + b(p^{2} − p) ≡ p + 1 mod(p^{2} − 1). This becomes (a − b)(p − 1) ≡ p + 1 mod (p^{2} − 1). Since p − 1 divides p^{2} − 1 this last equation holds mod (p − 1) and becomes 0 ≡ 2 mod(p − 1) so p = 3. We are excluding this case. The computation for ϕ^{1−p} is similar.

We now turn to the case where inertia acts through ρ̄ via characters of level one.

If wild inertia acts trivially, ρ̄ |_{I}= ( \begin{matrix}χ^{a} & 0 \\ 0 & χ^{b} \end{matrix} ) where 0 ≤ a ≤ b ≤ p − 2. Keeping in mind p ≥ 5

we see Ad^{0}ρ̄ has a one dimensional quotient on which I acts by χ only if b − a = 1. In that case, by section 2.3.2 of [Se1] we have k(ρ̄) = 1 + pa + b = 2 + a(p + 1). We are excluding such weights from our consideration. Note if ρ̄ is unramified at p then a = b = 0 and Serre defines k(ρ̄) to be p.

If wild inertia acts nontrivially, ρ̄ |_{I}= ( \begin{matrix}χ^{β} & ∗ \\ 0 & χ^{α} \end{matrix} ) where 0 ≤ α ≤ p − 2 and 1 ≤ β ≤ p − 1 and

ρ̄ |_{I} is a nontrivial extension class. One easily sees that Ad^{0}ρ̄ has a one dimensional quotient on which I acts by χ if and only if χ^{α−β} = χ, that is if and only if α − β ≡ 1 mod(p − 1). (Recall χ has order p − 1). From the bounds on α and β we see −(p − 1) ≤ α − β ≤ p − 3 so α − β = 1 or −(p − 2). In the first case α = β + 1 so following [Se1] k(ρ̄) = 1 + pβ + β + 1 = 2 + β(p + 1). In the second case β = p − 2 or p − 1 so α = 0 or 1 respectively and k(ρ̄) = p − 1 or 2p respectively. (Recall p ≠ 3.) As we have excluded these cases Proposition 3 is now proved.

For the lemma below we treat Ad^{0}ρ̄ and (Ad^{0}ρ̄)^{∗} as k[G_{p}] and k[I] modules. We also will extend scalars to the quadratic extension k̂ of k to diagonalize ρ̄ |_{I} as necessary.

Lemma 6 Ad^{0}ρ̄ ≄ (Ad^{0}ρ̄)^{∗} as F_{p}[G_{p}] modules.

Proof: Recall we are assuming p ≥ 5. In particular we know p ≠ 2. It suffices to show Ad^{0}ρ̄ and (Ad^{0}ρ̄)^{∗} are not isomorphic as F_{p}[I] modules.

Consider first the case where ρ̄ acts through I via fundamental characters of level one. We study Ad^{0}ρ̄^{ss}, the semisimplification of Ad^{0}ρ̄ as a F_{p}[I] module. One easily sees (Ad^{0}ρ̄)^{ss} ≃ k ⊕ k(χ^{m}) ⊕ k(χ^{−m}) as k[I] modules for some integer m. Thus ((Ad^{0}ρ̄)^{∗})^{ss} ≃ k(χ) ⊕ k(χ^{1−m}) ⊕ k(χ^{1+m}). As p ≠ 2 it is easy to see these semisimplifications are not isomorphic as k[I] modules. Since k is finite over F_{p}, counting F_{p} eigenspaces shows (Ad^{0}ρ̄)^{ss} and ((Ad^{0}ρ̄)^{∗})^{ss} are not isomorphic as F_{p}[I] modules. Thus Ad^{0}ρ̄ and (Ad^{0}ρ̄)^{∗} are not isomorphic as F_{p}[G_{p}] modules.

Suppose now I acts via character of level two. We know that (after a possible quadratic

extension of scalars to a field k̂) that ρ̄ |_{I}= ( \begin{matrix}ψ^{a+pb} & 0 \\ 0 & ψ^{b+pa} \end{matrix} ) so

Ad^{0}ρ̄ ≃ k̂ ⊕ k̂(ψ^{(p−1)(b−a)}) ⊕ k̂(ψ^{(p−1)(a−b)})

as k̂[I] modules. We see

(Ad^{0}ρ̄)^{∗} ≃ k̂(χ) ⊕ k̂(χψ^{(p−1)(a−b)}) ⊕ k̂(χψ^{(p−1)(b−a)})

as k̂[I] modules. If the k[G_{p}] modules Ad^{0}ρ̄ and (Ad^{0}ρ̄)^{∗} are isomorphic as F_{p}[G_{p}] modules we would necessarily have that the k̂[G_{p}] modules Ad^{0}ρ̄ and (Ad^{0}ρ̄)^{∗} are isomorphic as F_{p}[G_{p}] modules and thus as F_{p}[I] modules. Without loss of generality we would then have χ = ψ^{(p−1)(a−b)}. Thus Ad^{0}ρ̄ ≃ k̂ ⊕ k̂(χ) ⊕ k̂(χ^{−1}) as k̂[I] modules. Since p ≠ 2 this is not self-dual as a k̂[I] module. Counting F_{p} eigenspaces we see the k̂ modules Ad^{0}ρ̄ and (Ad^{0}ρ̄)^{∗} are not isomorphic as F_{p}[I] modules. Thus Ad^{0}ρ̄ and (Ad^{0}ρ̄)^{∗} are not isomorphic as F_{p}[G_{p}] modules. The lemma is now proved.

\underline{Section 6: Descents} In the introduction we alluded to the fact that the minimal field of definition of the three dimensional representation of G_{S} on Ad^{0}ρ̄ could conceivably be smaller than k. We have the following fact which is Lemma 6.13 of [DS]

Fact Let τ : G → GL_{n}(k) be a semisimple representation. Let k̃ be the subfield of k generated be the coefficients of the characteristic polynomials of elements of G. Then τ is realizable over k̃, i.e. it is isomorphic to a semisimple representation ϕ : G → GL_{n}(k̃) and ϕ ⊗k̃ k ≃ τ. The descent to k̃ is semisimple and unique up to non-canonical isomorphism.

I am grateful to the referee for pointing out the following lemma and its proof.

Lemma 7 Let τ : G → GL_{n}(k̃) be absolutely irreducible with minimal field of definition k̃. Let V = k̃^{n} be the representation space. Suppose W is a nonzero F_{p} subspace of V stable under the action of G. Then W = V.

Proof: First we show V is a semisimple F_{p}[G] module. Let V_{0} be a nonzero irreducible G stable F_{p} subspace of V. If V_{0} = V then V is a simple F_{p}[G] module. Assume V_{0} ≠ V. Since V_{0} is G stable so is its k̃ span. Since τ is irreducible we see the k̃ span of V_{0} is all of V. For α ∈ k̃, α ≠ 0, consider αV_{0}, the F_{p} subspace of V consisting of all multiples of elements of V_{0} by α. Clearly αV_{0} is G stable and an irreducible F_{p}[G] module. Note also that for α, β ∈ k̃, α, β ≠ 0, we have αV_{0} and βV_{0} intersect trivially or they are equal. Since the k̃ span of V_{0} is all of V we see V is contained in the span of simple F_{p}[G] modules and therefore V itself is a semisimple F_{p}[G] module.

To prove the lemma it suffices to show V is irreducible as an F_{p}[G] module. To show this it is enough to show the injection k̃ → End_{F_{p}[G]}(V) is an isomorphism. For if V were a direct sum of more than one irreducible constituent End_{F_{p}[G]}(V) would contain noninvertible elements.

We will show that [k̃ : F_{p}] = dim_{F_{p}}(End_{F_{p}[G]}(V)) by extending scalars and checking the equivalent statement [k̃ : F_{p}] = dimk̃ (End_{k̃[G]}(k̃ ⊗_{F_{p}} V)) = dimk̃ (End_{k̃[G]}((k̃ ⊗_{F_{p}} k̃) ⊗k̃ V)).

Note that k̃ ⊗_{F_{p}} k̃ ≃ ∏_{s} k̃ where the isomorphism is as left k̃ algebras and the product is indexed by the elements s ∈ Gal(k̃/F_{p}) and the sth factor has right k̃ algebra structure via the automorphism s of k̃. Thus End_{k̃[G]}(k̃ ⊗_{F_{p}} V) = End_{k̃[G]}(∏_{s} V_{s}) where V_{s} is the absolutely irreducible k̃[G] module obtained from V by base change by s. Note End_{k̃[G]}V_{s} = k̃.

We claim that for s, t distinct in Gal(k̃/F_{p}) that V_{s} and V_{t} are not isomorphic as k̃[G] modules so Hom_{k̃[G]}(V_{s}, V_{t}) = 0. Since k̃ is the minimal field of definition of τ and st^{−1} is not trivial in Gal(k̃/F_{p}) there must be some g ∈ G so that the characteristic polynomials over k̃ of the action of g on V_{s} and V_{t} are distinct. The irreducible k̃[G] modules V_{s} and V_{t} are thus not isomorphic so End_{k̃[G]}(k̃ ⊗_{F_{p}} V) = ∏_{s} End_{k̃[G]}(V_{s}) = ∏_{s} k̃ which has dimension #Gal(k̃/F_{p}) = [k̃ : F_{p}] over k̃ as desired.

Denote by Ãd^{0}ρ̄ and (Ãd^{0}ρ̄)^{∗} descents of Ad^{0}ρ̄ and (Ad^{0}ρ̄)^{∗} to k̃.

Lemma 8 Suppose ρ̄: H → GL_{2}(k) is absolutely irreducible and that Ãd^{0}ρ̄ is absolutely irreducible as a representation of H. Let f ∈ H^{1}(H, Ãd^{0}ρ̄) be non-zero. Let ρ : H → GL_{2}(k[ϵ]) be the deformation to the dual numbers corresponding to f. Let A and B be the kernels of ρ̄ and ρ respectively. Suppose also that H^{1}(H/A, Ãd^{0}ρ̄) is trivial. In terms of the choice of f, the F_{p}[H] module A/B can be naturally endowed with a structure of a k̃ vector space so that Ãd^{0}ρ̄ = (A/B)⊗k̃ k as k[H] modules. In particular A/B is a simple F_{p}[H] module and has cardinality (#k̃)^{3}.

Proof: Consider the inflation-restriction sequence

0 → H^{1}(H/A, Ãd^{0}ρ̄) → H^{1}(H, Ãd^{0}ρ̄) → H^{1}(A, Ãd^{0}ρ̄)^{H/A}.

Since f ≠ 0 and H^{1}(H/A, Ãd^{0}ρ̄) = 0 we see see that f ∈ H^{1}(H, Ãd^{0}ρ̄) maps to a nonzero element f̃ of H^{1}(A, Ãd^{0}ρ̄)^{H/A} = Hom_{H}(A, Ãd^{0}ρ̄). The last equality follows because A acts trivially on Ãd^{0}ρ̄. So f gives rise to a nonzero H equivariant map f̃ : A → Ad^{0}ρ̄. That the kernel of f̃ is B follows from the explicit description of the correspondance between elements of H^{1}(H, Ãd^{0}ρ̄) and deformations of ρ̄ to the dual numbers described in section 2. Let k̃ be the minimal field of definition of the representation of H on Ãd^{0}ρ̄. We know the image of A/B under f̃ is a nonzero F_{p}[H] submodule of Ãd^{0}ρ̄. By Lemma 7 the image equals Ãd^{0}ρ̄ and we are done.

\underline{Section 7: Global Methods} We recall our set-up for Theorem 1 in preparation for its proof. We have ρ̄: G_{S} → GL_{2}(k) a continuous absolutely irreducible Galois representation where k is a finite field of characteristic ≥ 5. We assume Ad^{0}ρ̄ is absolutely irreducible. This necessarily implies that (Ad^{0}ρ̄)^{∗} is an absolutely irreducible k[G_{S}] module as well. We assume H^{1}(G_{S}/N, Ad^{0}ρ̄) and H^{1}(G_{S}/N^{d}, (Ad^{0}ρ̄)^{∗}) are trivial where N and N^{d} are the maximal subgroups of G_{S} that act trivially on Ad^{0}ρ̄ and (Ad^{0}ρ̄)^{∗} respectively. Let Q(Ad^{0}ρ̄) and Q(Ad^{0}ρ̄^{∗}) be the fixed fields of N and N^{d}. Put D = Q(Ad^{0}ρ̄) ∩ Q(μ_{p}) and K = Q(Ad^{0}ρ̄)Q(μ_{p}). We assume that H^{2}(G_{v}, Ad^{0}ρ̄) = 0 for all v ∈ S. Also recall there exists an element a × b ∈ Gal(K/Q) as described in Theorem 1.

[figure: Q; D; Q(μ_{p}); Q(Ad^{0}ρ̄); K]

We refer the reader to [Ha] and [Mi] for the main theorems of global Galois cohomology. Recall Ãd^{0}ρ̄ and (Ãd^{0}ρ̄)^{∗} are decents of Ad^{0}ρ̄ and (Ad^{0}ρ̄)^{∗} to k̃. Henceforth we study these objects. The truth of our cohomological assumptions for Ad^{0}ρ̄ and (Ad^{0}ρ̄)^{∗} implies the truth of the corresponding statements for Ãd^{0}ρ̄ and (Ãd^{0}ρ̄)^{∗}.

Studying the exact sequence

0 → � _{S}^{2}(Ãd^{0}ρ̄) → H^{2}(G_{S}, Ãd^{0}ρ̄) → ⊕_{v∈S}H^{2}(G_{v}, Ãd^{0}ρ̄)

we see we have an isomorphism � _{S}^{2}(Ãd^{0}ρ̄) → H^{2}(G_{S}, Ãd^{0}ρ̄). Let r be the common k̃ dimension of these cohomology groups. Recall that by global Poitou-Tate duality � _{S}^{2}(Ãd^{0}ρ̄) is dual to � _{S}^{1}((Ãd^{0}ρ̄)^{∗}), the kernel of the map H^{1}(G_{S}, (Ãd^{0}ρ̄)^{∗}) → ⊕_{v∈S}H^{1}(G_{v}, (Ãd^{0}ρ̄)^{∗}). Also using global duality and the global Euler characteristic we find H^{1}(G_{S}, Ãd^{0}ρ̄) is r or r + 2 dimensional as ρ̄ is even or odd. Let {g_{1}, g_{2}, .., g_{r}} be a basis of � _{S}^{1}((Ãd^{0}ρ̄)^{∗}) and {f_{1}, f_{2}, .., f_{r}} be linearly independent in H^{1}(G_{S}, Ãd^{0}ρ̄). Our plan, is for each i, 1 ≤ i ≤ r, to find a prime q_{i} ∉ S such that

• q_{i} ≢ ±1 mod p and ρ̄(F rob_{q_{i}}) has eigenvalues with ratio q_{i}. By the remarks at the beginning of section 3 the distinct eigenvalues of ρ̄(F rob_{q_{i}}) will lie in k.

• g_{i} |_{G_{q_{i}}} ≠ 0.

• f_{i} |_{G_{q_{i}}} ≠ 0 is unramified and therefore nonnull at q_{i} by the results of section 3.

• For i ≠ j we have f_{i} |_{G_{q_{j}}} = 0 and g_{i} |_{G_{q_{j}}} = 0.

Letting Q = {q_{1}, q_{2}, .., q_{r}} we will then have � _{S∪Q}^{1}((Ãd^{0}ρ̄)^{∗}) is trivial and by global duality that

� _{S∪Q}^{2}(Ãd^{0}ρ̄) = 0. The map H^{2}(G_{S∪Q}, Ãd^{0}ρ̄) → ⊕_{v∈S∪Q}H^{2}(G_{v}, Ãd^{0}ρ̄) is therefore injective. Since

(Ãd^{0}ρ̄)^{∗} is assumed absolutely irreducible we have H^{0}(G_{S}, (Ãd^{0}ρ̄)^{∗}) = 0 and the H^{2} restriction map above is surjective by Chapter I, Theorem 4.10 of [Mi]. Since H^{2}(G_{v}, Ãd^{0}ρ̄) is assumed trivial for v ∈ S the map H^{2}(G_{S∪Q}, Ãd^{0}ρ̄) → ⊕_{v∈Q}H^{2}(G_{v}, Ãd^{0}ρ̄) is an isomorphism. Thus obstructions to deformation problems for ρ̄ with ramification in S ∪ Q need only be analyzed at exactly the primes in Q. But for each q_{i} we have, by our choice of the q_{i}, a nonnull cohomology class in H^{1}(G_{S}, Ãd^{0}ρ̄) available, namely f_{i}. We use these as in Proposition 1 to at each stage put the deformation in our “desired form” at q_{i}. Since f_{i} |_{G_{q_{j}}} = 0 for i ≠ j adjusting by f_{i} will not change the deformation at q_{j}.

The idea of seeking the set Q to annihilate � _{S∪Q}^{1}((Ãd^{0}ρ̄)^{∗}) comes from the work of [Wi] and [TW]. However, unlike the situation in [Wi] and [TW] our auxiliary primes cannot be congruent to 1 mod p. Had we chosen them so the f_{i} would have been null for q_{i}. Thus f_{i} could not be used to bring the local deformation problem at q_{i} to an unobstructed form. (The ‘shape’ of the local at q_{i} deformation theory for q_{i} ≡ 1 mod p is different than that given in section 3. See the lemma in the appendix of [TW].) Chebotarev’s theorem will provide us with the q_{i}.

[figure: Q; D; Q(μ_{p}); Q(Ad^{0}ρ̄); K; M_{i}; L_{i}; L̃_{i}]

Recall N and N^{d} are normal subgroups of G_{S} that fix Q(Ad^{0}ρ̄) and Q((Ad^{0}ρ̄)^{∗}) respectively.

Lemma 9 1) Let f̃_{i} be the image of f_{i} in H^{1}(N, Ãd^{0}ρ̄)^{G_{S}/N} = Hom_{G_{S}}(N, Ãd^{0}ρ̄). The kernel of f̃_{i} fixes a field L̃_{i} ⊆ Q_{S} Galois over Q with Gal(L̃_{i}/Q(Ad^{0}ρ̄)) isomorphic to the k̃[G_{S}] module Ãd^{0}ρ̄. The exact sequence

1 → Gal(L̃_{i}/Q(Ad^{0}ρ̄)) → Gal(L̃_{i}/Q) → Gal(Q(Ad^{0}ρ̄)/Q) → 1

splits.

2) Let g̃_{i} be the image of g_{i} in H^{1}(N^{d}, (Ãd^{0}ρ̄)^{∗})^{G_{S}/N^{d}} = Hom_{G_{S}}(N, (Ãd^{0}ρ̄)^{∗}). The kernel of g̃_{i} fixes a field M_{i} ⊆ Q_{S} Galois over Q with Gal(M_{i}/K) isomorphic to the k̃[G_{S}] module (Ãd^{0}ρ̄)^{∗}. The sequence

1 → Gal(M_{i}/K) → Gal(M_{i}/Q) → Gal(K/Q) → 1

splits.

Proof: Recall H^{1}(G_{S}/N, Ãd^{0}ρ̄) is assumed trivial, N acts trivially on Ãd^{0}ρ̄, and Ãd^{0}ρ̄ is an absolutely irreducible k[G_{S}] module. Consider the inflation-restriction sequence

0 → H^{1}(G_{S}/N, (Ãd^{0}ρ̄)^{N}) → H^{1}(G_{S}, Ãd^{0}ρ̄) → H^{1}(N, Ãd^{0}ρ̄)^{G_{S}/N} → H^{2}(G_{S}/N, (Ãd^{0}ρ̄)^{N}).

Then f_{i} ∈ H^{1}(G_{S}, Ãd^{0}ρ̄) maps to non-zero element f̃_{i} of H^{1}(N, Ãd^{0}ρ̄)^{G_{S}/N} = Hom_{G_{S}}(N, Ãd^{0}ρ̄) as desired. The image of f̃_{i} is a nonzero F_{p}[G_{S}] submodule of Ãd^{0}ρ̄. The map f̃_{i} : N → Ãd^{0}ρ̄ is surjective by Lemma 8. Let L̃_{i} be the fixed field of the kernel of f̃_{i}. Observe Gal(L̃_{i}/Q(Ad^{0}ρ̄)) inherits a k̃ structure from Ãd^{0}ρ̄. The sequence splits because f̃_{i} maps to zero in H^{2}(G_{S}/N, Ãd^{0}ρ̄) by exactness.

For the second part we note N^{d} ⊇ Gal(Q_{S}/K) and H^{1}(G_{S}/N^{d}, (Ãd^{0}ρ̄)^{∗}) is assumed trivial. Replace Ad^{0}ρ̄ and N by (Ad^{0}ρ̄)^{∗} and N^{d} in the exact sequence above. If the fixed field of N^{d} is not K let M_{i} be the composite of K and the field fixed by the kernel of g̃_{i}.

Let L_{i} be the composite L̃_{i}K. Note [L_{i} : K] = [L̃_{i} : Q(Ad^{0}ρ̄)] = (#k̃)^{3} as [K : Q(Ad^{0}ρ̄)] = [Q(μ_{p}) : D] is prime to p. Let L̂_{j} be the composite of all the L_{i} except L_{j}. Define M̂_{j} similarly. Let L be the composite of all the L_{i} and M the composite of all the M_{i}. Let F = LM.

[figure: Q; D; Q(μ_{p}); Q(Ad^{0}ρ̄); K; M̂_{i}; L̂_{i}; L_{i}; M_{i}; L; M; F]

Lemma 10 1) L_{i} ∩ L̂_{i} = K.

2) M_{i} ∩ M̂_{i} = K

3) L ∩ M = K.

Proof: Recall Gal(L̃_{i}/Q(Ad^{0}ρ̄)) ≃ Gal(L_{i}/K) ≃ Ãd^{0}ρ̄ as k̃[G_{S}] modules by Lemma 9. Since Ãd^{0}ρ̄ is a simple F_{p}[G_{S}] module and the L_{i} and L̂_{i} are Galois over Q we see L_{i} ∩ L̂_{i} = L_{i} or K. If the intersection is L_{i} we would have L_{i} ⊆ L̂_{i}. Suppose this happens. Then, as a k̃[G_{S}] module, Gal(L/K) has at most r − 1 copies of Ãd^{0}ρ̄ in its Jordan-Hölder sequence. Consider the inflation-restriction sequence

0 → H^{1}(Gal(K/Q), (Ãd^{0}ρ̄)^{Gal(L/Q)}) → H^{1}(Gal(L/Q), Ãd^{0}ρ̄) → H^{1}(Gal(L/K), Ãd^{0}ρ̄)^{Gal(K/Q)}.

The first term is H^{1}(G_{S}/N, (Ãd^{0}ρ̄)^{N}) which is assumed trivial. As Gal(L/K) acts trvially on Ãd^{0}ρ̄ the last term becomes Hom_{G_{S}}(Gal(L/K), Ãd^{0}ρ̄) and is at most r − 1 dimensional. Thus H^{1}(Gal(L/Q), Ãd^{0}ρ̄) is at most r−1 dimensional. This contradicts the independence of {f_{1}, f_{2}, .., f_{r}}, so the L_{i} ∩ L̂_{i} = K. The second part is handled similarly

For the third part, observe the simple terms in the composition series for the k̃[G_{s}] module Gal(L/K) are all Ãd^{0}ρ̄. Those of Gal(M/K) are all (Ãd^{0}ρ̄)^{∗}. By Lemma 6 these are, after scalar extension to k, nonisomorphic F_{p}[G_{p}] modules so they are nonisomorphic k̃[G_{S}] modules. Thus L ∩ M = K.

Remark: Gal(L/K) and Gal(M/K) inherit k̃ structures from the k̃ modules Gal(L_{i}/K) and Gal(M_{i}/K).

Lemma 11 1) Gal(L/K) ≃ (Ãd^{0}ρ̄)^{r} as k̃[G_{S}] modules and the exact sequence

1 → Gal(L/K) → Gal(L/Q) → Gal(K/Q) → 1

splits.

2) Gal(M/K) ≃ ((Ãd^{0}ρ̄)^{∗})^{r} as k̃[G_{S}] modules and the exact sequence

1 → Gal(M/K) → Gal(M/Q) → Gal(K/Q) → 1

splits.

Proof: The L_{i} are linearly disjoint over K by Lemma 10. Since each Gal(L_{i}/K) is isomorphic to the k̃[G_{S}] module Ãd^{0}ρ̄ we see Gal(L/K) ≃ (Ãd^{0}ρ̄)^{r}. The splitting follows from Lemma 9. This proves part 1. The proof of part 2 is similar.

Lemma 12 Gal(F/K) ≃ Gal(L/K) × Gal(M/K).

Proof: Immediate from part 3 of Lemma 10.

Lemma 13 Gal(F/Q) ≃ Gal(F/K)× Gal(K/Q).

Proof: We have that the exact sequences

1 → Gal(L/K) → Gal(L/Q) → Gal(K/Q) → 1

and

1 → Gal(M/K) → Gal(M/Q) → Gal(K/Q) → 1

both split by Lemma 11. Since Gal(F/K) ≃ Gal(L/K)×Gal(M/K) we see (using part 3 of Lemma 10) that the exact sequence

1 → Gal(F/K) → Gal(F/Q) → Gal(K/Q) → 1

splits.

In Theorem 1 we assume an element a×b ∈ Gal(Q(Ad^{0}ρ̄)/D)×Gal(Q(μ_{p})/D) ≃ Gal(K/D) ⊆ Gal(K/Q) exists where a is in the projective image of ρ̄, the eigenvalues of a have ratio t and b corresponds to the class of t ∈ (Z/p)^{∗} where t ≢ ±1 mod p. Note that we insist that t lie in the multiplicative group of the prime field F_{p}. Let c = a × b. Let d be the (multiplicative) order of t mod p. Note d is prime to p and c has order d in Gal(K/Q).

Lemma 14 1) Consider the action of c on Gal(L_{i}/K) ≃ Ãd^{0}ρ̄. There are nontrivial elements of Gal(L_{i}/K) on which c acts trivially.

2) There are nontrivial elements of Gal(M_{i}/K) on which c acts trivially.

Proof: 1) One easily sees that b acts trivially on Ad^{0}ρ̄. (Not Ãd^{0}ρ̄!). Using that the projective element a has distinct eigenvalues with ratio t, we see that Ad^{0}ρ̄ decomposes under the action of c into one dimensional k vector spaces. Thus the action of c is via multiplication by t, 1 and 1/t respectively. As t ∈ F_{p} we see Ad^{0}ρ̄ decomposes as an F_{p} vector space into c eigenspaces with c acting by t, 1 and 1/t.

Recall we have an injection Ãd^{0}ρ̄ → Ad^{0}ρ̄ by extending scalars and this injection is G_{S} equivariant. Thus the F_{p} vector space Ãd^{0}ρ̄ decomposes into eigenspaces with eigenvalues t, 1, and 1/t under the action of c. Suppose there are no eigenspaces with eigenvalue 1. Then under the scalar extension map Ãd^{0}ρ̄ → Ad^{0}ρ̄ the image of Ãd^{0}ρ̄ lies in the two dimensional k vector space spanned by the one dimensional k eigenspaces on which c acts by t and 1/t. Since the image of Ãd^{0}ρ̄ in Ad^{0}ρ̄ is G_{S} stable its k span is also G_{S} stable. But the k span is a two dimensional k subspace of Ad^{0}ρ̄. This contradicts the absolute irreducibility of Ad^{0}ρ̄.

2) The proof is identical, except the action of c on (Ad^{0}ρ̄)^{∗} decomposes (Ad^{0}ρ̄)^{∗} into one dimensional k vector spaces with eigenvalues 1, t and t^{2}.

Lemma 15 Let α_{i} ∈ Gal(L/K) ≃ (Ãd^{0}ρ̄)^{r} be the element all of whose entries are 0 except the the ith entry which is a nonzero element on which c acts trivially. Put

β_{i} = α_{i}× c ∈ Gal(L/K)× Gal(K/D) ⊆ Gal(L/K)× Gal(K/Q) ≃ Gal(L/Q).

Then β_{i} has order pd in Gal(L/Q). Let u_{i} be a prime of Q unramified in L with Frobenius in the conjugacy class of β_{i} in Gal(L/Q). Then for j ≠ i the primes above u_{i} in K split completely from K to L_{j} but these primes do NOT split completely from K to L_{i}.

Proof: The order of β_{i} in Gal(L/Q) is the least common multiple of the orders of α_{i} and c, namely pd. The remaining statements follow from projecting β_{i} ∈ Gal(L/Q) to Gal(L_{j}/Q) and Gal(L_{i}/Q) and observing the order of the projections are d and pd respectively.

Corollary 1 f_{i} |_{G_{u_{i}}} ≠ 0 and is unramified. Therefore by the results of section 3 it is nonnull at u_{i}. For j ≠ i we have f_{j} |_{G_{u_{i}}} = 0.

Proof: By the choice of c prior to Lemma 14 we see u_{i} is a prime as in section 3 and all the results there apply. By the choice of u_{i} we see G_{u_{i}} acts on Ad^{0}ρ̄ through a quotient of order d. If we complete the extension L_{i}/Q at a prime above u_{i} the degree of the local extension is pd. This follows from Lemma 15 and the fact that Ãd^{0}ρ̄ is a p-group. Observe (1 + ϵf_{i})ρ̄ |_{G_{u_{i}}} is the restriction to G_{u_{i}} of the deformation to the dual numbers k[ϵ] corresponding to f_{i} ∈ H^{1}(G_{S}, Ad^{0}ρ̄). The projective representation associated to this representation has image of order pd. Since the order of the image of the projective representation associated to ρ̄ |_{G_{u_{i}}} is d we see f_{i} |_{G_{u_{i}}} ≠ 0 and it is clearly unramified at u_{i}. That f_{i} |_{G_{u_{j}}} = 0 follows from the fact that the local degree at u_{i} of L_{j}/Q is d.

Lemma 16 Let γ_{i} ∈ Gal(M/K) ≃ ((Ãd^{0}ρ̄)^{∗})^{r} be the element with all entries 0 except the ith which is a nonzero element on which c acts trivially. Put

δ_{i} = γ_{i}× c ∈ Gal(M/K)× Gal(K/D) ⊆ Gal(M/K)× Gal(K/Q) ≃ Gal(M/Q).

Then δ_{i} has order pd in Gal(M/Q). Let v_{i} be a prime with Frobenius in the conjugacy class of δ_{i} in Gal(M/Q). Then for j ≠ i the primes above v_{i} in K split completely from K to M_{j}, but they do NOT split completely from K to M_{i}.

Proof: The proof is the same as Lemma 15.

Corollary 2 g_{i} |_{G_{v_{i}}} ≠ 0 and for j ≠ i we have g_{j} |_{G_{v_{i}}} = 0.

Proof: The proof is as in Corollary 1.

Proposition 4 Let

η_{i} ∈ Gal(F/Q) ≃ Gal(F/K)× Gal(K/Q) ≃ (Gal(L/K) × Gal(M/K))× Gal(K/Q)

be the element (α_{i} × γ_{i})× c. Let q_{i} be a prime of Q unramified in F with Frobenius in the conjugacy class of η_{i} in Gal(F/Q). Let Q = {q_{1}, q_{2}, .., q_{r}}. Then � _{S∪Q}^{1}((Ãd^{0}ρ̄)^{∗}) = 0.

Proof: Note that (α_{i} × γ_{i})× c ∈ Gal(F/Q) projects to β_{i}, δ_{i}, and c in Gal(L/Q), Gal(M/Q) and Gal(K/Q) respectively. Thus if q_{i} is a prime with Frobenius in the conjugacy class of (α_{i} ×γ_{i})× c ∈ Gal(F/Q) we see q_{i} is a prime as in section 3 by Corollary 1. By Corollaries 1 and 2 we see that f_{i} |_{G_{q_{i}}} ≠ 0, g_{i} |_{G_{q_{i}}} ≠ 0 and for i ≠ j that f_{i} |_{G_{q_{j}}} = 0 and g_{i} |_{G_{q_{j}}} = 0.

Consider the natural injection � _{S∪Q}^{1}((Ãd^{0}ρ̄)^{∗}) → � _{S}^{1}((Ãd^{0}ρ̄)^{∗}). Let h ∈ � _{S}^{1}((Ãd^{0}ρ̄)^{∗}). Then

h = ∑_{i=1}^{r} ν_{i}g_{i} where ν_{i} ∈ k̃ and we see h |_{G_{q_{i}}} = ν_{i}g_{i}. If h ∈ � _{S∪Q}^{1}((Ãd^{0}ρ̄)^{∗}) then h |_{G_{q_{i}}} is trivial so we must have ν_{i} = 0 for all 1 ≤ i ≤ r, that is h = 0.

Proposition 5 The map H^{2}(G_{S∪Q}, Ãd^{0}ρ̄) → ⊕_{v∈Q}H^{2}(G_{v}, Ãd^{0}ρ̄) is an isomorphism.

Proof: For all v ∈ S we assume H^{2}(G_{v}, Ãd^{0}ρ̄) = 0. Thus the right side of the above map is actually ⊕_{v∈S∪Q}H^{2}(G_{v}, Ãd^{0}ρ̄). Injectivity follows from the fact that � _{S∪Q}^{2}(Ãd^{0}ρ̄) is dual to

� _{S∪Q}^{1}((Ãd^{0}ρ̄)^{∗}) which is trivial. Surjectivity follows from Chapter I, Theorem 4.10 of [Mi].

Theorem 1Let ρ̄ : G_{S} → GL_{2}(k) be an absolutely irreducible Galois representation where the characteristic p of k is greater than or equal to 5. Assume Ad^{0}ρ̄ is an absolutely irreducible k[G_{S}] module, that H^{1}(G_{S}/N, Ad^{0}ρ̄) and H^{1}(G_{S}/N^{d}, (Ad^{0}ρ̄)^{∗}) are trivial, and that H^{2}(G_{v}, Ad^{0}ρ̄) = 0 for all v ∈ S. Finally suppose there is an element

a × b ∈ Gal(Q(Ad^{0}ρ̄)/D) × Gal(Q(μ_{p})/D) ≃ Gal(K/D) ⊆ Gal(K/Q)

such that a corresponds to an element in the (projective) image of ρ̄ whose eigenvalues have ratio t ∈ F_{p}^{∗}, where t ≠ ±1, and b ∈ Gal(Q(μ_{p})/Q) → (Z/p)^{∗} maps to the element t. Let r = dim_{k}H^{2}(G_{S}, Ad^{0}ρ̄) Then there is a set of primes Q = {q_{1}, q_{2}, .., q_{r}} disjoint from S and a representation ρ : G_{S∪Q} → GL_{2}(W (k)) such that ρ ≡ ρ̄ mod p.

Proof: Let Q be as in Proposition 4. We apply induction to deform from W (k)/p^{n} to W (k)/p^{n+1} one step at a time. We note that for all the q_{i}, ρ̄ |_{G_{q_{i}}} is of the “desired form” of section 3. This is the base case in our induction. Suppose we have a deformation ρ_{n} : G_{S∪Q} → GL_{2}(W (k)/p^{n}) with ρ_{n} |_{G_{q_{i}}} of the desired form for 1 ≤ i ≤ r. Since the local at q_{i} deformation problems (to mod p^{n+1}) are then unobstructed we can by Proposition 5 deform ρ_{n} to mod p^{n+1}, that is there is a ρ̃: G_{S∪Q} → GL_{2}(W (k)/p^{n+1}) with ρ̃ ≡ ρ_{n} mod p^{n}. The local at q_{i} representations ρ̃ |_{G_{q_{i}}} may not be of the “desired form”. However we know f_{i} is nonnull at q_{i} by Corollary 1. Using Proposition 1, we can alter ρ̃ by an appropriate multiple ν_{i} ∈ k of f_{i} so that (ν_{i}f_{i}).ρ̃ |_{G_{q_{i}}} is of the “desired form” at q_{i}. Let ρ_{n+1} = (∑ ν_{i}f_{i}).ρ̃. As f_{j} |_{G_{q_{i}}} is trivial for j ≠ i we have by Corollary 1 that ρ_{n+1} |_{G_{q_{i}}} = (ν_{i}f_{i}).ρ̃ |_{G_{q_{i}}} and the deformation problem is unobstructed at q_{i}. Then ρ_{n+1} = (∑ ν_{i}f_{i}) .ρ̃ is of the “desired form” at G_{q_{i}} for 1 ≤ i ≤ r. The induction is now complete and the theorem follows.

We now prove Theorem 2. Recall Q(ρ̄) is the field fixed by the kernel of ρ̄ and that we assume the image of ρ̄, denoted Imρ̄, contains SL_{2}(k).

Lemma 17 Let ρ̄ : G_{S} → GL_{2}(k) be such that Imρ̄ ⊇ SL_{2}(k) where the characteristic p of k is greater than or equal to 5. Then ρ̄ and Ad^{0}ρ̄ are absolutely irreducible and the minimal field of definition of Ad^{0}ρ̄ is k.

Proof: If ρ̄ is not absolutely irreducible then it is conjugate (over k̄) to a representation over k̄ that is upper triangular. This implies Imρ̄ is solvable. As we assume p ≥ 5 we know Imρ̄ is not solvable.

To show absolute irreducibility of Ad^{0}ρ̄ it suffices to show the action of SL_{2}(k) on the two by two matrices with trace 0 over k is absolutely irreducible. It suffices to prove the special case that the action of SL_{2}(F_{p}) on the two by two matrices with trace 0 over F_{p} is absolutely irreducible.

Since p is odd we may cite [CR], example 17.17. The absolutely irreducible representations of SL_{2}(F_{p}) are given by actions of SL_{2}(F_{p}) on the d + 1 dimensional space M_{d} of homogeneous

polynomials in two variables (say X and Y) of degree d over F̄_{p}. Let g = ( \begin{matrix}r & s \\ t & u \end{matrix} ) ∈ SL_{2}(F_{p}).

Then g acts on M_{d} by g.X = rX +sY and g.Y = tX +uY. When d is odd note that −1 ∈ SL_{2}(F_{p}) acts nontrivially on M_{d}. Since −1 acts trivially on Ad^{0}ρ̄ we see the Jordan-Hölder sequence for Ad^{0}ρ̄ can contain only M_{d} with d even. The only possibilities are the one dimensionsal space M_{0} and the three dimensional space M_{2}. If M_{0} occurs in the Jordan-Hölder sequence it occurs with multiplicity three and as in the first paragraph of this proof one sees P SL_{2}(F_{p}) is solvable for p ≥ 5, a contradiction. The absolutely irreducible representation M_{2} is Ad^{0}ρ̄ and we are done.

Let q = #k. As for the minimal field of definition of Ad^{0}ρ̄, consider the the elements of Ãd^{0}ρ̄

stabilized by the group of matrices U of order q of the form ( \begin{matrix}1 & ∗ \\ 0 & 1 \end{matrix} ) ∈ SL_{2}(k). Since we have a

p group acting on a p group there are nonzero elements of Ãd^{0}ρ̄ fixed by the action of U. Clearly the k̃ span of these elements in Ãd^{0}ρ̄ is pointwise fixed under the action of U. If this k̃ span were greater than one dimensional then on extending scalars to k we see U would pointwise fix a two (or greater) dimensional k subspace of Ad^{0}ρ̄. It is easy to see this is not the case. Let l_{1} be the one dimensional k̃ subspace of Ãd^{0}ρ̄ pointwise fixed by U. We ask what other elements of SL_{2}(k) pointwise fix l_{1}. These elements of SL_{2}(k) pointwise fix the k span of l_{1} in Ad^{0}ρ̄ and we easily see that ±U is the stabilizer of l_{1}. Let x_{0} ∈ l_{1} be nonzero. The orbit of x_{0} under the action of SL_{2}(k) has cardinality #SL_{2}(k)/(2#U) = (q^{2} − 1)/2. So Ãd^{0}ρ̄ contains at least distinct (q^{2} − 1)/2 elements. If the minimal field of definition k̃ is strictly smaller than k then #k̃ ≤ q^{1/2}. Thus Ãd^{0}ρ̄ contains at most q^{3/2} distinct elements. Since q ≥ 5 we see q^{3/2} < (q^{2} − 1)/2, a contradiction. The minimal field of definition of Ad^{0}ρ̄ is k.

Remark: Since, with the hypotheses of Theorem 2, we have k̃ = k we will use Ad^{0}ρ̄ and (Ad^{0}ρ̄)^{∗} for the rest of this section.

Lemma 18 Let p ≥ 5 and suppose the image of ρ̄ : G_{S} → GL_{2}(k) contains SL_{2}(k). Recall D = Q(Ad^{0}ρ̄)∩Q(μ_{p}). Then [D : Q] = 1 or 2. As [Q(μ_{p}) : Q] = p−1 we have that Gal(Q(μ_{p})/D) is not the trivial group.

Proof: As D ⊆ Q(μ_{p}), we see D/Q is abelian so Gal(Q(ρ̄)/D) contains the commutator subgroup of Gal(Q(ρ̄)/Q) = Imρ̄. As Imρ̄ ⊇ SL_{2}(k) and #k ≥ 4 we see this commutator subgroup is just SL_{2}(k). As D ⊆ Q(Ad^{0}ρ̄), the fixed field of the projective representation associated to ρ̄, we have that Gal(Q(ρ̄)/D) contains the scalar matrices in Imρ̄ which we denote Z. Thus Gal(Q(ρ̄)/D) contains Z · SL_{2}(k) as a normal subgroup.

Let A be in Imρ̄ with determinant a ∈ k. Then A^{2} ∈ Imρ̄ has determinant a^{2} so A^{2} =

( \begin{matrix}a & 0 \\ 0 & a \end{matrix} ) · X where X ∈ SL_{2}(k), that is A^{2} ∈ Z · SL_{2}(k). Since Imρ̄/SL_{2}(k) ⊆ k^{∗} is cyclic we

have [Imρ̄: Z · SL_{2}(k)] = 1 or 2. We conclude [D : Q] = 1 or 2.

Lemma 19 If #k ≥ 7 and Imρ̄ ⊇ SL_{2}(k) then H^{1}(G_{S}/N, Ad^{0}ρ̄) is trivial.

Proof: Let B′ be the Borel subgroup of upper triangular matrices in Imρ̄. Since [Imρ̄: B′] is easily seen to be prime to p it suffices to show H^{1}(B′, M^{0}) = 0 where M^{0} is the set of two by two trace zero matrices over k. But as a B′ module M^{0} has a 3 step filtration. By computing the cohomology of the various one dimensional subquotients the result follows. See Lemma 1.2 of [Fl]. Note that during this argument the fact that B′ contains diagonal elements the ratio of whose eigenvalues is not ±1 is used. This is why we require #k ≥ 7. Alternatively, if k = F_{5} and Imρ̄ = GL_{2}(F_{5}) the same methods can be used to show H^{1}(GL_{2}(F_{5}), Ad^{0}ρ̄) = 0.

Lemma 20 [NN^{d} : N] is prime to p.

Proof: Note that NN^{d}/N^{d} ≃ N/(N ∩ N^{d}). Recall that N = Gal(Q_{S}/Q(Ad^{0}ρ̄)) and that N^{d} ⊇ Gal(Q_{S}/K). So N ∩ N^{d} ⊇ Gal(Q_{S}/K). Thus N/(N ∩ N^{d}) is isomorphic to a subquotient of Gal(K/Q(Ad^{0}ρ̄)) which is in turn isomorphic to Gal(Q(μ_{p})/D) which has order prime to p.

Lemma 21 If p ≥ 5 and Imρ̄ ⊇ SL_{2}(k) then H^{1}(G_{S}/N^{d}, (Ad^{0}ρ̄)^{∗}) = 0.

Proof: We apply the inflation-restriction sequence to G_{S}/N^{d} and its normal subgroup NN^{d}/N^{d}. The quotient is G_{S}/NN^{d} and since N^{d} fixes (Ad^{0}ρ̄)^{∗} we see (Ad^{0}ρ̄)^{∗}^{NN^{d}/N^{d}} = (Ad^{0}ρ̄)^{∗}^{N}. We get the exact sequence

0 → H^{1}(G_{S}/NN^{d}, (Ad^{0}ρ̄)^{∗}^{N}) → H^{1}(G_{S}/N^{d}, (Ad^{0}ρ̄)^{∗}) → H^{1}(NN^{d}/N, Ad^{0}ρ̄)^{G_{s}/NN^{d}}

where the last term is trivial as NN^{d}/N has order prime to p by Lemma 20. As N acts trivially on Ad^{0}ρ̄ we see the action of N on (Ad^{0}ρ̄)^{∗} is χ |_{N}, which is nontrivial by Lemma 18. so ((Ad^{0}ρ̄)^{∗})^{N} = 0. Thus the left term in the sequence is trivial so H^{1}(G_{S}/N^{d}, (Ad^{0}ρ̄)^{∗}) = 0.

Theorem 2 Let p ≥ 7. Assume the image of ρ̄: G_{S} → GL_{2}(k) contains SL_{2}(k). Also assume if l ∈ S and l ≠ p then l ≢ ±1 mod p, that k(ρ̄) ≠ p − 1 or 2p, and that k(ρ̄) ≢ 2 mod p + 1. Let r = dim_{k}H^{2}(G_{S}, Ad^{0}ρ̄). Then there is a set of primes Q = {q_{1}, q_{2}, .., q_{r}} disjoint from S and a representation ρ : G_{S∪Q} → GL_{2}(W (k)) such that ρ ≡ ρ̄ mod p.

Proof: Recall N and N^{d} are the kernels of the actions of G_{S} on Ad^{0}ρ̄ and (Ad^{0}ρ̄)^{∗} respectively. By Lemmas 19 and 21 we have H^{1}(G_{S}/N, Ad^{0}ρ̄) and H^{1}(G_{S}/N^{d}, (Ad^{0}ρ̄)^{∗}) are trivial. The hypotheses on the ramified primes and the weight k(ρ̄) imply, by the results of sections 4 and 5, that H^{2}(G_{v}, Ad^{0}ρ̄) = 0 for all v ∈ S. Since p is odd we easily see H^{2}(G_{∞}, Ad^{0}ρ̄) = 0. By Lemma 17 we see ρ̄ is absolutely irreducible and Ad^{0}ρ̄ is an absolutely irreducible k[G_{S}] module.

By Theorem 1 it remains to find a × b ∈ Gal(Q(Ad^{0}ρ̄)/D) × Gal(Q(μ_{p})/D) ≃ Gal(K/D) ⊆ Gal(K/Q) such that a corresponds to an element in the (projective) image of ρ̄ whose eigenvalues have ratio t ∈ F_{p}^{∗}, t ≠ ±1, and b ∈ Gal(Q(μ_{p})/Q) → (Z/p)^{∗} maps to the element t. as in Theorem 1.

Since p ≥ 7 there is an x ∈ F_{p}^{∗} such that x^{2} ≠ ±1 in F_{p}^{∗}. As Gal(Q(μ_{p})/Q) ≃ F_{p}^{∗} we consider the element x^{2} in Gal(Q(μ_{p})/Q). Since x^{2} is a 2nd power and [D : Q] = 1 or 2 by Lemma 18 we see x^{2} ∈ Gal(Q(μ_{p})/D).

Recall that by the proof of Lemma 18 that Gal(Q(ρ̄)/D) ⊇ SL_{2}(k). We have that ( \begin{matrix}x & 0 \\ 0 & x^{−1} \end{matrix} ) ∈

Gal(Q(ρ̄)/D). We consider the projection of this element, x̃, in Gal(Q(Ad^{0}ρ̄)/D), i.e. its image in the projective representation associated to ρ̄. Finally, we take a × b to be x̃ × x^{2} ∈ Gal(Q(Ad^{0}ρ̄)/D) × Gal(Q(μ_{p})/D) ≃ Gal(K/D) ⊆ Gal(K/Q). Theorem 1 now applies.

\underline{Section 8: Examples} We give an even and odd example that illustrate the theorems.

Consider the polynomial j(x) = x^{7} − 22x^{6} + 141x^{5} − 204x^{4} − 428x^{3} + 768x^{2} + 320x − 512 of [ZM]. The splitting field of j(x) is a totally real field with Galois group over Q isomorphic to P SL_{2}(F_{7}). We abuse notation for a moment and call this splitting field Q(Ad^{0}ρ̄). Zeh-Marschke has shown that there is a quadratic extension (which we call Q(ρ̄)) of Q(Ad^{0}ρ̄) Galois over Q with Gal(Q(ρ̄)/Q) ≃ SL_{2}(F_{7}). This extension gives our even ρ̄. The discriminant of j(x) is 2^{50}19^{4}367^{2}. As mentioned in the introduction prior to the statement of Theorem 1, we may assume these are the only ramified primes in Q(ρ̄)/Q. These primes are not congruent to ±1 mod 7 and ρ̄ is unramified at p = 7 so by Serre’s definition of weight we see k(ρ̄) = p = 7. The hypotheses of Theorem 2 are satisfied. Thus there is a finite set of primes Q and a deformation ρ of ρ̄ with ρ : G_{S∪Q} → SL_{2}(Z_{7}). That this representation is surjective follows from [Se3], Chapter IV, Lemma 3.

In section 5 of [Se1] Serre has given an absolutely irreducible odd representation ρ̄ : G_{Q} → GL_{2}(F_{49}). The image of Galois in P GL_{2}(F_{49}) is P SL_{2}(F_{7}) and is the splitting field of the polynomial h(x) = x^{7} − 7x + 3. (The polynomial is due to Trinks). Note k = F_{49} and k̃ = F_{7}. The splitting field of h(x) is complex and is our Q(Ad^{0}ρ̄). As h(x) has discriminant 3^{8}7^{8} we see Q(Ad^{0}ρ̄)/Q is ramified only at 3,7 and ∞, and we may assume Q(ρ̄)/Q is ramified only at 3,7 and ∞. Since 3 is not congruent to ±1 mod 7 we have H^{2}(G_{3}, Ad^{0}ρ̄) = 0. Serre has shown that a twist of ρ̄ is weight 3 so by Proposition 3 we have H^{2}(G_{7}, Ad^{0}ρ̄) = 0.

We see that H^{1}(G_{S}/N, Ãd^{0}ρ̄) = H^{1}(P SL_{2}(F_{7}), Ad^{0}ρ̄). This last cohomology group is trivial (essentially) by Lemma 19 so H^{1}(G_{S}/N, Ãd^{0}ρ̄) = 0. Similarly, H^{1}(G_{S}/N^{d}, (Ãd^{0}ρ̄)^{∗}) = 0 essentially by Lemma 21.

As Q(Ad^{0}ρ̄)/Q is an extension with Galois group the simple group P SL_{2}(F_{7}) we see Q(Ad^{0}ρ̄)∩ Q(μ_{p}) = Q so in this example D = Q.

[figure: Q; Q(μ_{p}); Q(Ad^{0}ρ̄); K; Q(ρ̄)]

Serre has shown Imρ̄ ⊃ SL_{2}(F_{7}) so ( \begin{matrix}2 & 0 \\ 0 & 1/2 \end{matrix} ) ∈ Imρ̄. Let a be the projection of this element

in Gal(Q(Ad^{0}ρ̄)/Q). Consider a × b = ( \begin{matrix}2 & 0 \\ 0 & 1/2 \end{matrix} ) × 4 ∈ Gal(Q(Ad^{0}ρ̄)/Q) × Gal(Q(μ_{7})/Q) ≃

Gal(K/Q). This element satisfies the last hypothesis of Theorem 1 and we see ρ̄ deforms to W (F_{49}) after allowing ramification at an additional finite set of primes.

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