Ann. Math. Québec (2014) 38:61–71 DOI 10.1007/s40316-014-0012-4

Linking numbers and the tame Fontaine–Mazur conjecture John Labute To Helmut Koch on his 80th birthday.

Received: 15 May 2013/Accepted: 6 August 2014 / Published online: 28 June 2014 © Fondation Carl-Herz and Springer International Publishing Switzerland 2014

Abstract Let p be an odd prime, let S be a finite set of primes q ≡ 1 mod p but q  ≡ 1 mod p 2 and let G S be the Galois group of the maximal p-extension of Q unramified outside of S. If ρ is a continuous homomorphism of G S into GL2 (Z p ), then under certain conditions on the linking numbers of S we show that ρ = 1 if ρ = 1. We also show that ρ = 1 if ρ can be put in triangular form mod p 3 . Résumé Soit p un premier impair et soit S un ensemble fini de premiers q ≡ 1 mod p avec q  ≡ 1 mod p 2 ; soit G S le groupe de Galois de la p-extension maximale de Q non ramifiée en dehors de S. Si ρ est un homomorphisme de G S dans GL2 (Z p ), alors sous certaines conditions sur les nombres d’enlacements de S, nous montrons que ρ = 1 si ρ = 1. Nous montrons aussi que ρ = 1 si ρ peut s’écrire sous forme triangulaire mod p 3 . Keywords

Fontaine–Mazur Conjecture · Pro- p-groups · Fab groups · Lie algebras

Mathematics Subject Classification

11R34 · 20E15 · 12G10 · 20F05 · 20F14 · 20F40

1 Statement of results Let p be a rational prime. Let K be a number field, let S be a finite set of primes of K with residual characteristics  = p and let  S,K be the Galois group of the maximal (algebraic) extension of K unramified outside of S. The Tame Fontaine–Mazur Conjecture (cf. [1], Conj. 5a) states that every continuous homomorphism ρ :  S,K → GLn (Z p )

Research supported in part by an NSERC Discovery Grant. J. Labute (B) Department of Mathematics and Statistics, McGill University, Burnside Hall, 805 Sherbrooke Street West, Montreal, QC H3A 0B9, Canada e-mail: [email protected]

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has a finite image. If ρ is the reduction of ρ mod p, then ρ is trivial if and only if the image of ρ is contained in the standard subgroup GL(1) n (Z p ) = {X ∈ GLn (Z p ) | X ≡ 1 mod p} which is a pro- p-group. Hence, if ρ = 1, the homomorphism ρ factors through G S,K , the (1) maximal pro- p-quotient of  S,K . Since GLn (Z p ) is torsion free, this shows that when ρ = 1 the Tame Fontaine–Mazur Conjecture is equivalent to the following conjecture. (1)

Conjecture 1.1 If ρ : G S,K → GLn (Z p ) is a continuous homomorphism then ρ = 1. Conversely, the truth of Conjecture 1.1 for any number field K implies the Fontaine–Mazur Conjecture. In this paper we will prove Conjecture 1.1 when K = Q for certain sets S. We now let K = Q and G S = G S,Q . To prove Conjecture 1.1 we can assume that the primes in S are congruent to 1 mod p, since these are the only primes different from p that can ramify in a p-extension of Q. We will also assume that the primes in S are not congruent to 1 mod p 2 , which is equivalent to G S /[G S , G S ] being elementary. In this case, we will show that Conjecture 1.1 follows from a Lie theoretic analogue of it when p is odd. We therefore assume that p  = 2 for the rest of the paper. To formulate this analogue, let S = {q1 , . . . , qd } and let l S be the finitely presented Lie algebra over F p generated by ξ1 , . . . , ξd with relators σ1 , . . . , σd , where  σi = ci ξi + i j [ξi , ξ j ] j=i −

with ci = (qi −1)/ p mod p and the linking number i j of (qi , q j ) defined by qi ≡ g j i j mod q j with g j a primitive root mod q j . We call l S the linking algebra of S. Up to isomorphism, it is independent of the choice of primitive roots. Theorem 1.2 There exists a mapping  : Homcont (G S , GL(1) n (Z p )) −→ Hom(l S , gln (F p )) such that ρ = 1 ⇔ (ρ) = 0. Corollary 1.3 If the cup-product H 1 (G S , F p ) × H 1 (G S , F p ) −→ H 2 (G S , F p ) is trivial then Conjecture 1.1 is true for G S . Definition 1.4 (Property F M(n)) A lie algebra g over a field F is said to have Property F M(n) if every n-dimensional representation of g is trivial. Theorem 1.5 If l S has Property F M(k), then Conjecture 1.1 is true for n = k. If |S| ≤ 2 then l S has Property F M(n) for all n, since l S = 0 in this case. However l S may not have Property F M(2) if |S| ≥ 3; for example, if p = 3, S = {7, 31, 229} or

if p = 5 and S = {11, 31, 1021}. However, the number of such sets S is relatively small; for example, if p = 7 and the primes in S are at most 10,000, the set S fails to have Property F M(2) approximately 0.2 % of the time. The following theorem gives necessary and sufficient conditions for Property F M(n) to hold when |S| = 3. Theorem 1.6 Let m i j = −i j /ci . If |S| = 3 and n < p, then Property F M(n) holds if and only if one of the following conditions holds:

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(a) m i j = 0 for some i, j; (b) m i j  = 0 for all i, j and m ik = m jk for some i, j, k with i  = j; (c) m i j  = 0 for all i, j and (m ik − m jk )(m ki m i j − m k j m ji )  = 0 for some i, j, k. These conditions are independent of the choice of primitive roots. Theorem 1.7 If |S| = 3 and n < p, then l S fails to have Property F M(n) if and only if i j  = 0 for all i, j and 13 /c1 = −23 /c2 , 21 /c2 = −31 /c3 , 12 /c1 = −32 /c3 . Theorem 1.8 Let ρ : G S → GL2 (Z p ) be a continuous homomorphism. Then ρ = 1 if ρ can be brought to triangular form mod p 3 . The pro- p-groups G S are very mysterious. They are all fab groups, i.e., subgroups of finite index have finite abelianizations, and for |S| ≥ 4 they are not p-adic analytic. So far no one has given a purely algebraic construction of such a pro- p-group. We call a pro- p-group G a Fontaine–Mazur group if every continuous homomorphism of G into GLn (Z p ) is finite. Again, no purely algebraic construction of such a group exists. In this direction we have the following result. Theorem 1.9 Let G be the pro- p-group with generators x1 , . . . , x2m and relations pc

pc

x1 1 [x1 , x2 ] = 1, x2 2 [x2 , x3 ] = 1, . . . , pc

pc

2m−1 x2m−1 [x2m−1,2m ] = 1, x2m2m [x2m , x1 ] = 1

with ci  ≡ 0 mod p > 2 and m ≥ 2. Then every continuous homomorphism of G into (1) GLn (Z p ) is trivial if n < p.

2 Mild pro- p-groups Let G be a pro- p-group. The descending central series of G is the sequence of subgroups G n defined for n ≥ 1 by p

G 1 = G, G n+1 = G n [G, G n ], p

where G n [G, G n ] is the closed subgroup of G generated by p-th powers of elements of G n and commutators of the form [h, k] = h −1 k −1 hk with h ∈ G and k ∈ G n . The graded abelian group gr(G) = ⊕n≥1 gr n (G) = ⊕n≥1 G n /G n+1 is a graded vector space over F p where gr n (G) is denoted additively. We let ιn : G n → gr n (G) = G n /G n+1 be the quotient map. Since p  = 2, the graded vector space gr(G) has the structure of a graded Lie algebra over F p [π] where π ιn (x) = ιn+1 (x p ), [ιn (x), ιm (y)] = ιn+m ([x, y]). Let G = F/R where F is the free pro- p-group on x1 , . . . , xd and R = (r1 , . . . , rm ) is the closed normal subgroup of F generated by r1 , . . . , rm with ri ∈ F2 . If  pa j  xi [xi , x j ]ai jk mod F3 rk ≡ i≥1

i< j

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and we let ξi = ι1 (x1 ), ρk = ι2 (rk ) in L = gr(F), then L is the free Lie algebra over F p [π] on ξ1 , . . . , ξd and   ai πξi + ai jk [ξi , ξ j ]. ρk = i≥1

i< j

Let r be the ideal of L generated by ρ1 , . . . ρm , let g = L/r and let U be the enveloping algebra of g. Then r/[r, r] is a U -module via the adjoint representation. The sequence ρ1 , . . . , ρm is said to be strongly free if (a) g is a torsion-free F p [π]-module (b) and r/[r, r] is a free U -module on the images of ρ1 , . . . , ρm , in which case we say that the presentation is strongly free. Theorem 2.1 ([3], Theorem 1.1) If G = F/R is strongly free, then r is the kernel of the canonical surjection gr(F) → gr(G), so that gr(G) = L/r. A finitely presented pro- p-group G is said to be mild if it has a strongly free presentation. Let A = Z p [[G]] be the completed algebra of G and let I = Ker(A → F p ) be the augmentation ideal of Z p [[G]]. Then gr(A) = ⊕n≥1 I n /I n+1 is a graded algebra over F p [π] where π can be identified with the image of p in I /I 2 . The canonical injection of G into A sends G n into 1 + I n and gives rise to a canonical Lie algebra homomorphism of gr(G) into gr(A) which is injective if and only if G n = G ∩ (1 + I n ). Theorem 2.2 ([3], Theorem 1.1) If G is mild, the canonical map gr(G) → gr(A) is injective and gr(A) is the enveloping algebra of gr(G). Moreover, R/[R, R] is a free A-module, which implies that cd(G) ≤ 2. We now give a criterion for the mildness of G = G S when p  = 2 and p ∈ / S. The group G S has a presentation F(x1 , . . . , xd )/(r1 , . . . , rd ) where xi is a lifting of a generator of an inertia group at qi and  pc [xi , x j ]i j mod F3 , ri = x i i j=i

which is due to Helmut Koch ([2], Example 11.11). Using the transpose of the inverse of the transgression isomorphism tg : H 1 (R, F p ) F = (R/R p [R, F])∗ −→ H 2 (G, F p ), the relator ri defines a linear form φi on H 2 (G, F p )) such that, if χ1 , . . . , χd is the basis of H 1 (F, F p ) = (F/F p [F, F])∗ with χi (x j ) = δi j , we have φi (χi ∪ χ j ) = −i j if i < j; cf. [2], Theorem 7.23. The set S is said to be a circular set of primes if there is an ordering q1 , . . . , qd of the set S such that

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(a) i,i+1  = 0 for 1 ≤ i < d and d1  = 0, (b) i j = 0 if i, j are odd, (c) 12 23 · · · d−1,d d1  = 1m m,m−1 · · · 32 21 . Theorem 2.3 If S is a circular set of primes, then G S is mild. Theorem 2.4 The set S can be extended to a set S ∪ q, where q ≡ 1 mod p with q  ≡ 1 mod p 2 , in such a way that the pairs (q, qi ), (qi , q) with non-zero linking numbers can be arbitrarily prescribed. Corollary 2.5 The set S can always extended to a set S with G S mild. See Theorem 1.1 of [3] for the proof of Theorem 2.3 and see Proposition 6.1 of [3] for the proof of Theorem 2.4. The proof of Proposition 6.1 in [3] yields the sharper form stated here. Theorem 2.6 There exists a finite set S containing S and consisting of primes q ≡ 1 mod p, with q  ≡ 1 mod p 2 , such that G S is mild and, if n < p, the Lie algebra l S has Property(FM(n)) if l S does.

3 Proof of Theorem 1.2 (1) Let G be a pro- p-group with G/[G, G] ∼ = (Z/ p Z)d and let ρ : G → GLn (Z p ) be a continuous homomorphism. Let k GL(k) n (Z p ) = {X ∈ GLn (Z p ) | X ≡ 1 mod p }. ( j)

(i)

Lemma 3.1 Let X = 1 + pi A ∈ GLn (Z p ) and let Y = 1 + p j B ∈ GLn (Z p ). Then [X, Y ] = 1 + pi+ j [A, B] mod pi+ j+1 and X p = 1 + pi+1 A mod pi+2 , where [A, B] = AB − B A. (2)

Lemma 3.2 If ρ(G)  = 1 then ρ(G)  ⊆ GLn (Z p ). (k)

Proof Let H = ρ(G) and let k ≥ 1 be largest with H ⊆ GLn (Z p ). Let h 1 , . . . h d be a (2k) generating set for H and let h i = I + p k Ni . Then [h i , h j ] ∈ GLn (Z p ), which implies that (2k) [H, H ] ⊆ GLn (Z p ). By assumption, there exists i such that Ni  ≡ 0 mod p. However, p

h i = (1 + p k Ni ) p ≡ 1 + p k+1 Ni mod p k+2 . p

Since Ni  ≡ 0 modulo p we have h i ∈ [H, H ] only if k + 1 ≥ 2k, which implies that k = 1.   Let G = G S . Then G S has the presentation F(x1 , . . . , xd )/(r1 , . . . , rd ) where  pc ri = x i i [xi , x j ]i j mod F3 . j=i

Let ρ(xi ) = 1 + p Ai . Then modulo p 3 we have ⎛ 1 = ρ(ri ) = 1 + p 2 ⎝ci Ai +



⎞ i j [Ai , A j ]⎠ .

j=i

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Hence, if Ai is the image of Ai in gln (F p ), we have  i j [Ai , A j ] = 0. ci Ai + j=i

Thus (ρ)(ξi ) = Ai defines a Lie algebra homomorphism (ρ) : l S −→ sln (F p ). If ρ = 1 then Ai = 0 for all i, which implies (ρ) = 0. Conversely, if ρ  = 1 then by Lemma 3.2 we have Ai  = 0 for some i, which implies (ρ)  = 0.

4 Proof of Theorem 1.8 Without loss of generality, we can assume that G S is mild. Let H = ρ(G S ) and assume that H = ρ(G S )  = 1. Note that H is a subgroup of SL2(Z p ) since G S /[G S , G S ] is finite. After a b a change of basis, we can assume that the matrices ∈ H satisfy p 3 |c and that H is c d generated by the image of  1 1 C= · 0 1 (1)

Let h 1 , . . . , h d be a generating set for H with h 1 , . . . , h d−1 ∈ SLn (Z p ) and h d ≡ C mod p. We have   a bi p ad 1 + p e h i − 1 = p Ai = p i · for i < d and h d − 1 = p cd ci di p f We also have d > 1 since otherwise H is infinite cyclic which is impossible since H/[H, H ] is finite.  a b Lemma 4.1 Let X, Y ∈ GL2 (Z p ) with X = 1 + p A = 1 + p and with Y ≡ C mod c d p. Then  −c a − d − c [X, Y ] ≡ 1 + p mod p 2 . 0 c Proof Let N = Y − 1. Then, working mod p 2 , we have [X, Y ] ≡ (1 + p A)−1 (1 + N )−1 (1 + p A)(1 + N ) ≡ (1 − p A)(1 − N + N 2 − N 3 )(1 + p A)(1 + N ) ≡ (1 − p A − N + p AN + N 2 − N 3 )(1 + p A + N + p AN ) ≡ 1 + p[A, N ] − pN AN ≡ 1 + p[A, N ] − pN [A, N ]  −c a − d − c ≡1+ p · 0 c  

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Lemma 4.2 We have

p hd

1 mod p 2 . 0

0 ≡1+ p 0



1+ pe , so that mod p 2 we have p f  0 ad + f N2 ≡ p , N 3 ≡ 0. 0 0

Proof We have h d = 1 + N with N = pN ≡ p

 0 0

1 , 0

67

p ad p cd

p

Hence we have h d = (1 + N ) p ≡ 1 + pN mod p 2 .

 

Let M = Z p e1 + Z p e2 and let B be the image of A = Z p [[G S ]] in End(M). Let J = ( p, h 1 − 1, . . . , h d − 1) be the augmentation ideal of B. Then J M = Z p e 1 + Z p p e2 and by induction we have J k M = Z p p k−1 e1 + Z p p k e2 for k ≥ 1. It follows that gr(M) =



J k M/J k+1 M

k≥0

is a free F p [π]-module with basis e1 ∈ gr 1 (M), e2 ∈ gr 0 (M). Using the fact that (h i − 1)e1 = p ai e1 + p ci e2 with p 2 |ci we see that gr(h i −1)e1 = ai πe1 . Since the elements gr(h i −1), (i ≤ d) generate  gr(B) = J k /J k+1 , k≥0

the submodule W = F p [π]e1 is invariant under gr(B) and we obtain a homomorphism φ1 : gr(B) → End(W ) = gl1 (F p [π]) with φ1 (gr(h i − 1)) = πai . We want to show that ai is non-trivial mod p for some i < d. (1)

Lemma 4.3 If X = 1 + p A ∈ SLn (Z p ), then tr(A) ≡ 0 mod p. (1)

Proof If X = 1 + pN ∈ SLn (Z p ), we have 1 = det(1 + pN ) ≡ 1 + p tr(N ) mod p 2 , which implies that tr(N ) ≡ 0 mod p.

 

Lemma 4.4 If 1 ≤ i < d and πai = φ1 (gr(h i − 1)) = 0, then (2)

[h i , h d ] ∈ SL2 (Z p ). Proof Since ai ≡ 0 mod p, Lemma 4.3 implies that di ≡ 0 mod p. The result then follows from Lemma 4.1.  

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Lemma 4.5 If πai = φ1 (gr(h i − 1)) = 0 for 1 ≤ i < d, then (2)

[H, H ] ⊆ SL2 (Z p ). Proof The pro- p-group [H, H ] is generated, as a normal subgroup of H , by the elements of the form [h i , h j ] and [h i , h d ] with i, j < d. Since the elements of the form [h i , h j ] with i, j < d are congruent to 1 mod p 2 by the proof of Lemma 3.2, the result follows from Lemma 4.4.   p

So if gr(h i − 1) acts trivially on W for 1 ≤ i < d then h d is not in [H, H ] by Lemmas 4.5 and 4.2, contradicting the fact that H/[H, H ] is elementary. So the homomorphism φ1 : gr(B) → gl1 (F p [π]) is non-trivial. Composing φ1 with the canonical surjection gr(Z p [[G S ]]) → gr(B), we obtain a non-trivial homomorphism φ : gr(Z p [[G S ]]) → gl1 (F p [π]). Composing the canonical map α : gr(G S ) → gr(Z p [[G S ]]) with φ, we get a Lie algebra homomorphism gr (ρ) : gr(G S ) → gl1 (F p [π]). Since G S is mild, α is injective and gr(Z p [[G S ]]) is the enveloping algebra of gr(G S ), which implies that gr (ρ)  = 0 since gr(G S ) generates gr(Z p [[G S ]]). Now G S has the presentation F(x1 , . . . , xd )/(r1 , . . . , rd ) where  pc [xi , x j ]i j mod F3 . ri = x i i j=i

Since G S is mild, we have gr(G S ) = ξ1 , . . . , ξn | ρ1 , . . . ρd , where  i j [ξi , ξ j ]. ρi = ci πξi + j=i

In this case, if gr (ρ)(ξi ) = πu i , then gr (ρ)(ρi ) = π 2 ci u i = 0 and so u i = 0 for all i, which contradicts the fact that gr (ρ)  = 0.

5 Proof of Theorem 1.6 Here |S| = 3 and the relations for l S can be written in the form ⎧ ⎨ ξ1 = m 12 [ξ1 , ξ2 ] + m 13 [ξ1 , ξ3 ], ξ2 = m 21 [ξ2 , ξ1 ] + m 23 [ξ2 , ξ3 ], ⎩ ξ3 = m 31 [ξ3 , ξ1 ] + m 32 [ξ3 , ξ2 ],

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where m i j = −i j /ci . Let r Ai = r (ξi ). Then ⎧ ⎨ A1 A2 ⎩ A3

69

: l S → gln (F p ) be a Lie algebra homomorphism and let = m 12 [A1 , A2 ] + m 13 [A1 , A3 ], = m 21 [A2 , A1 ] + m 23 [A2 , A3 ], = m 31 [A3 , A1 ] + m 32 [A3 , A2 ].

Since r = 0 if A1 , A2 , A3 are linearly dependent, we may assume that A1 , A2 , A3 are linearly independent. Note that each of the above relations can be written in the form Ai = [Ai , Bi ] for some Bi ∈ Gln (F p ). Then, by the following Lemma which was pointed out to us by Nigel Boston, each matrix Ai is nilpotent if n < p. Lemma 5.1 Let A, B be n × n matrices over F p with A = [A, B]. Then A is nilpotent if n < p. Proof Replacing F p be a finite extension Fq , we may assume that A is upper triangular. Then the trace of Aq−1 is k · 1 with 0 ≤ k < p. But the trace of An is zero for any n ≥ 1 since A = [A, B] implies tr(An ) = tr(AB An−1 − B An ) = 0.  

It follows that k = 0 and hence that the characteristic polynomial of A is X n . Remark The proof of the above Lemma is due to Julien Blondeau.

If condition (a) holds we can, without loss of generality, assume that m 12 = 0. Then A1 = [A1 , B1 ] with B1 = m 13 A3 nilpotent, which implies ad(B1 ) nilpotent. Hence A1 = 0 and we are reduced to the case |S| = 2. If condition (b) holds we can, without loss of generality, assume that m 13 = m 23 . Taking a linear combination of the first two equations, we obtain  m 23 a A1 + b A2 = (am 12 − bm 21 )[A1 , A2 ] + a A1 + b A2 , m 13 A3 . m 13 Choose non-zero a, b ∈ F p so that am 12 − bm 21 = 0. Then a A1 + b A2 = [a A1 + b A2 , m 13 A3 ], which implies a A1 + b A2 = 0 since ad(A3 ) is nilpotent. We can then write the equations in the form A2 = c[A2 , A3 ],

A2 = d[A2 , A3 ],

A3 = e[A2 , A3 ],

from which we readily get A1 = A2 = A3 = 0. If condition (c) holds we may, without loss of generality, assume that m 23  = m 13 and m 32 m 21  = m 31 m 12 . For non-zero a, b ∈ F p we consider the equation a A1 + b A2 + A3 = (am 12 − bm 21 )[A1 , A2 ] + (am 13 − m 31 )[A1 , A3 ] + (bm 23 − m 32 )[A2 , A3 ]. Let b = m 12 a/m 21 and choose λ such that am 13 − m 31 = λa. Then bm 23 − m 32 = λb ⇐⇒ am 12 m 23 /m 21 − m 32 = λam 12 /m 21 ⇐⇒ am 12 m 23 − m 32 m 21 = m 12 (am 13 − m 31 ) ⇐⇒ am 12 (m 23 − m 13 ) = m 32 m 21 − m 12 m 31 m 32 m 21 − m 31 m 12 ⇐⇒ a = · m 12 (m 23 − m 13 )

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With this choice of a we have a A1 + b A2 + A3 = [λa A1 + λb A2 , A3 ] = [a A1 + b A2 + A3 , λA3 ], which implies a A1 + b A2 + A3 = 0 since ad(A3 ) is nilpotent. If conditions (a), (b), (c) fail, then





m 31 m 32 m 12 m 13 m 21 m 23

=

=

m 21 m 12 m 32 m 23 m 31 m 13 = 0, which implies



m 31 = k1 m 21 , m 32 = k1 m 12 , m 12 = k2 m 32 , m 13 = k2 m 23 , m 21 = k3 m 31 , m 23 = k3 m 13 ,

for some k1 , k2 , k3 ∈ F∗p . This implies that ki k j = 1 for all i  = j and hence that ki2 = 1 for all i. Since, by hypothesis, ki  = 1, we must have ki = −1 for all i. Then the relators for l S are of the form ⎧ ⎨ ξ1 = a[ξ1 , ξ2 ] + b[ξ1 , ξ3 ], ξ2 = c[ξ2 , ξ1 ] − b[ξ2 , ξ3 ], ⎩ ξ3 = −c[ξ3 , ξ1 ] − a[ξ3 , ξ2 ], with a, b, c ∈ F∗p . After the transformations ξ1  → c−1 ξ1 , ξ2  → a −1 ξ2 , ξ3  → b−1 ξ3 , the relations become

⎧ ⎨ ξ1 = [ξ1 , ξ2 ] + [ξ1 , ξ3 ], ξ2 = [ξ2 , ξ1 ] − [ξ2 , ξ3 ], ⎩ ξ3 = −[ξ3 , ξ1 ] − [ξ3 , ξ2 ].

But these relations are satisfied if we replace ξi by Ai ∈ gl2 (F p ) with    1 0 1 1 0 0 1 1 , A2 = − , A3 = − A1 = − 2 0 0 2 1 0 2 −1

1 , −1

which yields an isomorphism of l S with sl2 (F p ). Thus the only case where Property F M(n) would fail would be when i j  = 0 for all i, j and 13 /c1 = −23 /c2 , 21 /c2 = −31 /c3 , 12 /c1 = −32 /c3 . −i j

Note that, since qi ≡ g j (q1c2 q2c1 )c3

mod q j , this is equivalent to

≡ 1 mod q3 , (q2c3 q3c2 )c1 ≡ 1 mod q1 , (q1c3 q3c1 )c2 ≡ 1 mod q2 .

6 Proof of Theorem 2.6

} such that By Theorem 2.4, we can find a set of primes S = {q1 , . . . , q2d ⎧

if i is odd , ⎨ q2i = qi and i,i+1  = 0

= 0 if i < 2d is even, i,i+1 ⎩ 2d,1  = 0 with all other i, j = 0 if i or j is odd.

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If f is a homomorphism of l S into Gln (F p ), let Ai = f (ξi ). Then, if i is odd, ai Ai + [Ai , Ai+1 ] = 0 for some non-zero ai and if i is even, Ai = [Ai , Bi ] for some matrix Bi . By Lemma 5.1, this implies that Ai is nilpotent if i is even and hence that ad(Ai ) is nilpotent if i is even. But this implies that Ai = 0 if i is odd. That G S is mild follows from the fact that S is a circular set of primes.

7 Proof of Theorem 1.9 (1)

Let ρ be a continuous homomorphism of G into GLn (Z p ). If ρ(xi ) = 1+ p Ai then, modulo p 3 , we have ρ(ri ) = 1 + p 2 (c1 Ai + [Ai , Ai+1 ]) = 0 when i < 2m and ρ(r2m ) = 1 + p 2 (c2m A2m + [A2m , A1 ]) = 0. Hence, if Ai is the image of Ai in Gln (F p ), we have c1 A1 + [A1 , A2 ] = 0, c2 A2 + [A2 , A3 ] = 0, . . . , c2m A2m + [A2m , A1 ] = 0. By Lemma 5.1 we see that ad(Ai ) is nilpotent for all i and hence Ai = 0 for all i. But this implies ρ = 1 since ρ  = 1 implies Ai  = 0 for some i by Lemma 3.2.

References 1. Fontaine, J.-M., Mazur, B.: Geometric Galois representations, elliptic curves, modular forms and Fermat’s last theorem. (Hong Kong 1993), 41–48, Ser. Number Theory, I, Internat, Press, Cambridge, MA (1995) 2. Koch, H.: Galois Theory of p-Extensions. Springer, Berlin (2002) 3. Labute, J.: Mild pro- p-groups and Galois groups of p-extensions of Q. J. Reine Angew. Math. 596, 115–130 (2006)

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