Galois representations

1 Introduction (Vladimir) 1.1 Galois representations Galois representations really mean representations of Galois groups.

Denition 1.1. An Artin representation, ρ, over a eld K is a nite dimensional complex representation of Gal(K/K) which factors through a nite quotient (by an open subgroup). I.e., there exists nite Galois extension F/K, such that ρ comes from a representation of Gal(F/K)

K

G F

G K

Gal(K/K) → Gal(F/K) → GLn(C) Note. e.g., I the trivial representation is the same Artin representation for all F/K √ Example. Let 3 , , 3 2 −1 . The character table F = Q(ζ3, 5) K = Q G = Gal(F/Q) = S3 = s, t|s = t = id, tst = s is id (12) (123) I 1 1 1 So:  1 −1 1 ρ 2 0 −1

• I(s) = I(t) = 1 • (s) = 1,(t) = −1 √ ! − 1 − 3 1 0  • ρ(s) = √2 2 , ρ(t) = 3 1 0 −1 2 − 2

Example. Dirichlet characters: Z/NZ → C multiplicative.

Q(ζN )

∗ G=(Z/NZ)

Q

Hence Dirichlet characters can be seen as representation 1 χ : G → C = GL1(C)

1 Denition 1.2. A mod Galois representation is the same thing with matrices in . l GLn(Fl) Example. Let E/Q be an elliptic curve. We know E(Q)[l] =∼ Z/lZ × Z/lZ. Set F = Q(E[l]), the smallest eld generated by the x-coordinates and y-coordinates of the points of order l. We end up with a Galois group

F

G

Q acts on and preserves addition, i.e., . Therefore we get . G E[l] √g(P + Q)√ = g(P ) +√g(Q) ρ :√G → GL2(Fl) E.g.: Let y2 = x3−5, then E[2] = 0, ( 3 5, 0), (ζ 3 5, 0), (ζ2 3 5, 0) . So take F = (ζ , 3 5), then G = Gal(F/ ) 3 3 √ Q 3 √ Q 3 3 permutes E[2] (we see that G = S3). Let us write down the matrix, so let P = ( 5, 0) and Q = (ζ3 5, 0). 0 1 • g ∈ S be a 3-cycle, ρ(g) = ∈ GL ( ) 3 1 1 2 F2

1 1 • g ∈ S , be a transposition, ρ(g) = ∈ GL ( ) 3 0 1 2 F2

1.2 l-adic representations sep Denition 1.3. A continuous l-adic representation over K is a continuous homomorphism Gal(K /K) → GLd(F) for some nite . F/Ql

Remark. An l-adic representation is continuous if and only if for all n there exists a nite Galois extension Fn/K sep n n such that Gal(K /Fn) → id mod l . I.e., ρ mod l factors through a nite extension Fn/K.   sep 1 + lOF lOF So Gal(K /F1) map to . lOF 1 + lOF

Example. Let E/Q be an elliptic curve: basis for . • P1,Q1 E(Q)[l] basis for 2 , with , • P2,Q2 E(Q)[l ] lP2 = P1 lQ2 = Q1 . • . basis for n , with , . • Pn,Qn E(Q)[l ] lPn = Pn−1 lQn = Qn−1 For dene by , , and g ∈ Gal(Q/Q) 0 ≤ an, bn, cn, dn < l gP1 = a1P1 + bQ1 gQ1 = c1P1 + d1Q1 gPn = n−1
n−1 n−1 n−1 a1 + ··· + anl Pn + (b1 + ··· + bnl )Qn and gQn = (c1 + ··· + cnl )Pn + (d1 + ··· + dnl )Qn. Then

 n−1 n−1  a1 + ··· + l an + . . . c1 + ··· + l cn + ... ρ(g) = n−1 n−1 ∈ GL2(Zl) b1 + ··· + l bn + . . . d1 + ··· + l dn + ...

Note that ρ(g) mod ln tells you what g does to E[ln]. This does give a 2d continuous l-adic representations.

2 Galois Representations: vocabulary (Matthew S) 2.1 Galois Theory of Innite Algebraic Extensions

Notation. G(F/K) := Gal(F/K), GK = G(K/K) the absolute Galois group For this section we assume K is a perfect eld (so every extensions is separable) and F is a normal algebraic extension of K.

2 Example. Let be a prime, and , let be dened as p. is xed by . Naively p K = Fp F = Fp φp φp(x) = x Fp hφpi we would think ∼ , but this is not true at all. To see this, take such that an where GFp = hφpi = Z φ ∈ GFp φ|Fpn = φp is a sequence such that where . This shows . {an} an ≡ am mod m m|n GFp > hφpi Denition 2.1. Let F/K be a Galois extension. For each nite subextension K0 consider G(K0/K). When we have two of them, such that K0 ⊆ K00 consider

G(K00/K) → G(K0/K).

This denes an inverse system of groups. G(F/K) = lim G(K0/K). ←−K0/K B = {left/right cosets of finite index subgroups}

Fact. G(F/K) is Hausdor, compact and totally disconnected. Theorem 2.2. Let F/K be a Galois extension. The map K0 → G(F/K0) is a bijective inclusion reversing corres- pondence between K0 and closed subgroups of G(F/K), H → F H .

Example. Back to the example, G( n / ) = /n , so G = lim /n = . Fp Fp Z Z Fp ←−n Z Z Zb

2.2 Galois groups of Q. Fix , : Q → Qp Q → Qp

Qp Fp

ur Qp Fp

Qp Fp Note ur ∼ . G(Qp /Qp) = GFp ∼ µr = GQp / / G(Qp /Qp) / GFp C C

The kernel of such a map is Ip. Ip admits a large normal p-subgroup, Wp,the wild inertia group. Ip/Wp tame inertia Let , for a Galois extension of if : Θ:GQp  G(K/Qp) Qp

• Θ(Ip) = 0 we say that K is unramied

• Θ(Wp) = 0 then we say that K is tamely ramied

• Θ(Wp) 6= 0 then we say that K is widely ramied Example. Cyclotomic extensions: ∗ ∗ ∗ ∼ , n we have an isomorphism . Let , G(Q(ζm)/Q) = (Z/mZ) Kl = ∪n∈Z>0 Q(ζl ) G(Kl/Q) → Zl l : GQ → Zl dened as for: l(σ). is ramied at and at , For , recall , then , p. σ ∈ GQ,σ(ζ) = ζ Kl ∞ l p 6= l φp (φp) = p φp(ζ) = ζ Conjecture. Any nite group is a discrete quotient of GQ

3 2.3 Restricting the ramication Let be a set of primes including . Let be the maximal extension of unramied outside . Let S {∞} QS Q S . GQ,S = G(QS/Q) Theorem 2.3 (Hermito-Minkowski). Let nite, a nite set of primes, . Then there exists nitely K/Q S d ∈ Z>0 many degree d extensions F/K unramied outside F . In particular is nite. Homcont(GK,S, Z/pZ) Theorem 2.4 (p-niteness condition). Let p be a prime, K a number eld, S a nite set of primes (non- archimedean). Let G ⊂ GK,S which is open then there exists only nitely many continuous homomorphism from G to Z/pZ. Theorem 2.5. If is a nite extension then is topologically nite generated. K Qp GK Conjecture. If , the map is an inclusion • p ∈ S GQp → GQ,S If , the map has kernel exactly . So . • p∈ / S GQp → GQ,S Ip GQp /Ip ,→ GQ,S Suppose now that we have not xed our embedding.

Theorem 2.6 (Chebotarov). Let K/Q be a Galois extension unramied outside a nite set of primes S. Let T ⊇ S be a nite set of primes. For each there exists a well-dened , the union of these classes is p∈ / T [φp] ⊂ G(K/Q) dense in G(K/Q) 2.4 Galois Representations

Denition 2.7. A Galois representation over a topological ring A unramied outside S (a set of primes) is a continuous homomorphism, . ρ : GQ,S → GLn(A) Let M be a free rank n A-module, we can equip it with a G action: g · a = ρ(g) · a. More formally: Suppose we have a free A-module M such that: • G (a pronite group) acts continuously • M = lim M H where H runs over open normal subgroups of G, −→H then we can make M into a A[[G]]-module: A[[G]] = lim A[G/H] where H is as before. We say , a representation of , is : ←−H ρ GQ

• unramied at p if it is trivial on Ip.

• tamely ramied at p if it is trivial on Wp • otherwise it is widely ramied. Proposition 2.8. Let S be any set of primes: 1. An Artin representation, , is determined by trace on such that is unramied ρ : GQ → GLn(C) (ρ(φp)) p∈ / S ρ at p. 2. A semisimple representation, , is determined by the values of trace i where mod l ρ : GQ → GLn(k) (∧ (ρ(φp))) i = 1, . . . , n on primes p∈ / S at which ρ is unramied. If l > n it is sucient to use trace(ρ(φp)) at the same primes. 3. A semisimple -adic representations, , is determined by trace on at which is l ρ : GQ → GLn(A) (ρ(φp)) p∈ / S ρ unramied.

4 2.5 Conductors of representation The inertia group is ltered by u , closed and for Ip Ip C GQ,p u ∈ [−1, ∞] If then u v • u ≤ v Ip ⊃ Ip If , then u and ∞ • u ≤ 0 Ip = Ip Ip = {1} u • Wp = ∪u>0Ip u v • Ip = ∩v

3 Invariants of Artin and l-adic Representations (Céline) Notation.

• πK be a xed uniformiser of K

•OK the ring of integers of K

• νK the normalized valuation on K

• IF/K the inertia group

• FrobF/K for a Frobenius element −1 also called the Geometric Frobenius • ΦF/K = FrobF/K 3.1 Artin Representation

3.1.1 Local polynomials and l-functions Denition 3.1. The local polynomial of an Artin Representation ρ over a local eld K is   P (ρ, T ) = det 1 − ΦF/K T |ρIF/K

IF/K where ρ factors through F/K and ρ is the subspace of IF/K -invariant vectors.

IF/K Remark. P (ρ, T ) is essentially the characteristic polynomial of ΦF/K on ρ Example. Consider √ 3 F = Q5(ζ3, 5)

√ 3 Q5(ζ3) Q5( 5)

K = Q5 ∼ ∼ ∼ We have Gal(F/K) = S3, IF/K = C3 = Gal(F/K(S3)) and FrobF/K = t (an element of order 2). Then

5 For we have (Since C3 ) • P ( ,T ) = det(1 − ΦF/K T | IF/K ) = det(1 − T ) = 1 − T = I I I I I

For (the sign representation) (since IF/K , •  P (, T ) = det(1 − ΦF/J T |IF/K ) = det(1 − (−1)T |) = 1 + T  =  so (t) = −1)

For the -dimensional representation: (since C3 , we have no • ρ 2 P (ρ, T ) = det(1 − ΦF/K T |ρIF/K ) = 1 ρ = 0 invariant subspace)

Denition 3.2. The Artin L-function of an Artin representation over a number eld K is

Y 1 L(ρ, s) = −s PP (ρ, Nm(P) ) P⊂OK where PP (ρ, T ) is the local polynomial of ρ restricted to Gal(KP /KP ) . The Euler product converges to an analytic function if re(s) > 1 Example.

Let be a number eld, then for all , so Q 1 the • K ρ = I PP (I,T ) = 1 − T P L(I, s) = P 1−Nm(P)−s = ζK (s) Dedekind ζ-function of K , the order character of ∼ . Need and • K = Q ρ 2 Gal(Q(S3)/Q) = C2 IF/K p(t)

 , then is totally ramied, hence and IF/K . So p = 3 Q3(S3)/Q3 IF/K = C2 ρ = 0 P3(ρ, T ) = 1  then , and p ≡ 1 mod 3 Qp(S3) = Qp If/K = {e} Pp(ρ, T ) = 1 − T  then is unramied so and . So . p ≡ 2 mod 3 Qp(S3)/Qp IF/K = {e} ρ(t) = −1 Pp(ρ, T ) = 1 + T Putting it together we get

Y 1 L(ρ, s) = 1 − p
p−s p6=3 3 ∞ X n = n−s 3 n=1

the L function of the non-trivial Dirichlet character Z/3Z → C∗ Fact. The Artin L-function of 1-dimensional Artin representation over Q correspond to Dirichlet L-functions of primitive characters.

Basic Properties

1. For ρ1 and ρ2 Artin representations over a local eld K, P (ρ1 ⊕ ρ2,T ) = P (ρ1,T )P (ρ2,T )

f 2. When F/K is a nite extension, ρ an Artin represetnations over F then PF (ρ, T ) = PK (Indρ,T ) where f is the residue degree of F/K.

3. When K is a number eld, L(ρ1 ⊕ ρ2, s) = L(ρ1, s)L(ρ2, s). If F/K is nite, ρ Artin representation over F , then L(ρ, s) = L(Indρ, s) (where the rst one is an Artin L-function over F and the second over K)

Conjecture (Artin). Let ρ 6= I be irreducible Artin representation over a number eld, then its L-function is analytic

6 3.1.2 Conductor

Denition 3.3. The conductor exponent of an Artin representation over a local eld K is nρ = nρ,tame + nρ,wilde, I P∞ 1 Ik where nρ,tame = dim ρ − dim ρ F/K and nρ,wild = dim ρ/ρ where ρfactors through F/K and I = k=1 [I:Ik] F/K  k+1 I = I0, Ik = σ ∈ GL(F/K)|σ(α)α mod π ∀α ∈ OF are the higher ramication group (with lower numbering) So wild inertia and tame inertia I1 = SylρI = I/I1 = We say ρ is unramied (respectively tame) if nρ = 0 (respectively nρ,wilde = 0) if and only if I acts trivial on ρ (respectively I1)

nρ Denition 3.4. The conductor of ρ is the ideal Nρ = (π ) Theorem 3.5 (Artin). np ∈ Z Remark. . Hence nρ1⊕ρ2 = nρ1 + nρ2 Nρ1⊕ρ2 = Nρ1 Nρ2 Theorem 3.6 (Swan's character). Let ρ be an Artin representation over a local eld K which factors through Gal(F/K). Then nρ,wild = hTraceρ, bi where ( 1 − νF (g(πF ) − πF ) for g ∈ IF/K \{e} b(g) = P − h6=e b(h) for g ∈ e Theorem 3.7 (Conductor-Discriminant formula).

F

H

G L

K

Let F/K be Galois, ρ be a representation of H = Gal(F/L). Then n G = (dim ρ) · νK (∆L/K ) + PL/K nρ IndH ρ equivalently dim ρ NIndρ = ∆L/K NmL/K (Nρ) Example. √ 3 F = Q5(ζ3, 5) √ 3 Q5(ζ3) Q5( 5)

K = Q5

Then IF/K = C3, I1 = {1}: as and • nI = 0 nρ,tame = 1 − 1 nρ,wild = 0

• n = 0

• nρ = 2 = 2 − 0 By the Conductor-discriminant formula: 2 (up to units) ∆L/K = NIndS3 = MρNI = 5 C2 I 4 (up to units) ∆F/K = N S3 = Nρ⊕ρ⊕⊕I = 5 Ind{1}I

7 Denition 3.8. The conductor of an Artin representation over a number eld K

Y nP (ρ) Nρ = P P where nP (ρ) is the conductor exponent of ρ restricted to Gal(KP /KP ). Example of Application: Suppose is Galois, . Let and be intermediate with and . Then F/Q Gal(F/Q) = D10 K L [K : Q] = 2 [L : Q] = 5 2 2 SF (s)SQ(s) = SL(s) SK (s)

3.1.3 Functional equations Theorem 3.9. The Artin -function of satises the functional equation 1/2−s where L ρ Λ(ρ, s) = ωA Λ(ρ,b 1 − s) • Y d+(ρ) d−(ρ) Y Λ(s) = L(ρ, s) ΓR(s) ΓR(s + 1) ΓC(s) νreal νcomplex

is the dimension of the eigenspace of the image of complex conjugation at , ∗, • d±(ρ) ± ν ω ∈ C • |ω| = 1 global root number q dimρ • A = Nm(Nρ) |ΛK |

−s/2 • ΓR(s) = π Γ(s/2) −s • ΓC(s) = (2π) Γ(s) ( (s)! s ∈ N • Γ(s) = ∞ s−1 −x 0 x e dx ´ 3.2 l-adic Representations 3.2.1 Local Polynomials Denition 3.10. Let be nite, where with , be a continuous -adic K/Qp ρ : Gal(K/K) → GLd(F) F/Ql l 6= p l representation. The local polynomial of ρ is

P (ρ, T ) = det(1 − Φ T | I ) K/K ρ K/K

3.2.2 Conductor

Denition 3.11. The conductor exponent is n = n + n where n = dim I , n = ρ ρ,tame ρ,wilde ρ,tame ρ/ρ K/K ρ,wild P 1 dim ρ/ρIF/K,k where F/K is a nite extension chosen so that the action of wild inertia factors k≥1 [IF/K ,IF/K,k]   1 + lOF lOF through. We can take F = F1, then the image of Gal(K/F ) lies in and im(I1) = id since it lOF 1 + lOF is a (pro) p-group send into a (pro) l-group.

nρ Denition 3.12. The conductor of ρ is Nρ = (πK ) .

8 4 Decomposition Theorems (Pedro)

Notation.

• Let p and l be distinct primes. • K a p-adic eld •F an l-adic eld

• IL the (absolute) inertia group of a eld L w the (absolute) wild inertia group of a eld • IL L

• ΦL a geometric Frobenius element

4.1 Finite Image of Inertia

Theorem 4.1. Let τ : GK → GLd(F) be an l-adic Galois representation such that τ(IK ) is nite and ΦK acts semisimple, for any choice of ΦK . Then we can write τ = ⊕i(ρi ⊗ χi) (after possible a nite extension of F) where ρi is an l-adic Galois representation which factors through a nite quotient and χi is a one dimensional unramied Galois representation. To show this thing, we use the following:

Proposition 4.2. Let k be a eld of characteristic c ≥ 0, V a nite dimensional vector space, G a group and ρ : G → GL(V ) a representation of G. Assume that there exists a nite index subgroup H ≤ G such that ρ|H is semisimple and c - [G : H]. Then ρ is semisimple. Proof. Choose a subrepresentation W of ρ and let W 0 be k[H]-module such that V = W ⊕ W 0 (As k[H] modules). Consider π 0 / W / V / V/W / 0 ]

f For , take 1 P −1 . u ∈ V/W h(u) = [G:H] g∈G/H gf(g u) Proof of Theorem 4.1. By the previous proposition, we can assume that τ is irreducible. We can take a totally ramied extension L/K such that τ(IL) = 1

hΦLhiΦL0 i

L0 Lnr e,H L Knr e f K

Let 0 be the Galois closure of . Note that nr is generated by and . We have f , L L Gal(L /K) H ΦL ΦL0 = ΦL so doesn't commute with . Pick , we then have −1 −1 , but −1 −1 so ΦL0 ΦL σ ∈ H σ ΦL0 σΦL0 ∈ H σ ΦL0 σ ∈ hΦL0 i −1 −1 .Hence . So we have that . By Schur's lemma we have that σ ΦL0 σΦL0 ∈ hΦL0 i [σ, ΦL0 ] ∈ H ∩ hΦL0 i [σ, ΦL0 ] = 1 τ(ΦL0 ) = λ idd. Dene χ to be

• χ(IK ) = 1 √ f • χ(ΦK ) = λ Set −1. So f . ρ := τ ⊗ χ ρ(ΦL0 ) = ρ(ΦK σ) = 1

9 4.2 Innite image of inertia Denition 4.3.

√ t (σ) √ 1. Let be the character dened in the following way: where ln l ln . tl : IK → Zl σ 7→ tl(σ) σ( πK ) = ζln πK n (Where ζln is a primitive l th root of unity) This is called the l-adic tame character 2. For any n ≥ 0, 1 t t2/2! . . . tn/n!  ..  0 1 .  .  . .. ..  sp(n)(σ) = . . .    . ..  . . t  0 1

where t = tl(σ), σ ∈ Ik. And we dene 1 0   q    sp(n) (ΦK ) =  ..   .  0 qm where q = #FK I0 Theorem 4.4. Let τ : GK → GLd(F) be an l-adic Galois representation such hat ΦK acts semisimply on τ , for 0 every nite index subgroup I ⊆ I , and for every choice of ΦK . Then

τ = ⊕i (ρi ⊗ sp(ni))

(after a nite extension) where ρi is an l-adic Galois representation such that ρi(IK ) is nite and with Frobenius acting semisimply.

Remark. By continuity, we can nd a nite Galois extension L/K such that τ(I ) = τ(H), where H = Gal(L /Lrn) =∼ √ L l , where ∞ nr ln
. Zp Ll = ∪n−1L πL Note that and is a Frobenius element, then q where . σ ∈ H ΦK σΦL = ΦLσ q = #FL Proof.

Case 1. d = 1 Let . Then q q −1 −1 . Hence q−1 , so σ ∈ H τ(σ) = τ(σ ) = τ(ΦL σΦL) = τ(ΦL )τ(σ)τ(ΦL) = τ(σ) τ(σ) = 1 τ(σ) ∈ µq−1. Case 2. d = 2 Pick σ ∈ H which is a topological generator of H. By extending F if necessary, we can assume that ∗ ∗ τ(σ) = . We have three cases: 0 ∗

λ 0 Case i. τ(σ) = . This is the same as the case d = 1. 0 λ λ 0 Case ii. τ(σ) = , λ 6= µ. Let V be the subrepresentation spanned by the ith vector. We use 0 µ i q the above note. Let v1 ∈ V1, then σΦK (v1) = ΦK σ (v1), hence ΦK V1 is a subrepresentation of τ|H . Similarly, we can conclude that ΦK V2 is a subrepresentation of τ|H . If ΦK V1 = V2, q q q then µ(ΦK v1) = σ(ΦK vi) = ΦK (σ v1) = λ ΦK v1. Similarly λ(ΦK v2) = µ ΦK v2. Hence λ, µ are roots of unity so the image of inertia is nite.

10 λ ∗ Case iii. τ(σ) = and ∗ 6= 0. Φ V is a subrepresentation of τ| implies that Φ V = V . We 0 λ K 1 H K 1 1 1 ∗ can write τ 0 = τ ⊗ χ−1, τ 0(σ) = with σ ∈ H. 0 1 nr nr Claim. For any Gal(Ll/L ) and θ ∈ Gal(Ll/K ) we have σθ = θσ.

5 l-adic representations of Elliptic curves (Heline) 5.1 Denition Notation. Let or • K = Q Qp

• GK := Gal(K/K) • E/K an elliptic curve

• 2 ≤ m ∈ Z
2 • E[m] = {P ∈ K : mP = 0} =∼ (Z/mZ)

• For σ ∈ GK and P ∈ E[m], we have mσ(P ) = σ(mP ) = 0, hence GK acts on E[m].

• Pick a basis P1,Q1 for E[m], then for σ ∈ GK we have σ(P1) = aP1 + cQ1 and σ(Q1) = bP1 + dQ1 for some a b a, b, c, d ∈ . Hence we have G → Aut(E[m]) =∼ GL ( /m ) dened by σ 7→ . If gcd(m, n0) = 1 Z K 2 Z Z c d then E[mn0] =∼ E[m] × E[n0]. • We are going to be taking m = ln with l a prime distinct from p. [l] [l] Note. We have natural maps E[ln] → E[ln−1] → · · · → E[l] → 0

Denition 5.1. For E an elliptic curve and l a prime, we dene the l-adic Tate module of E to be TlE := lim E[ln] ∼ ( )2. ←− = Zl We also dene ∼ 2. VlE := TlE ⊗Zl Ql = (Ql) Note that GK acts on both TlE and VlE. Denition 5.2. The mod representation of is ∼ . l E ρE,l : GK → Aut(E[l]) = GL2(Z/lZ) The -adic representation is ∼ or depending of reference l ρE,l : GK → Aut(Tl(E)) = GL2(Zl) ρE,l : GK → ∼ Aut(VlE) = GL2(Ql) ,→ GL2(C) Recall the cyclotomic character ∗ dened by, for l(σ). l : Gk → Zl σ ∈ GK : σ(ζl) = ζl We have the Weil pairing: e[ , ]: E[m] × E[m] → µm (Where µm is the m-th root of unity), which is bilinear, alternating, Galois invariant, non-degenerate and computable. a b Given σ ∈ G with ρ (σ) = , and P,Q ∈ E[m] be a basis, we have that K E,l c d σe[P,Q] = e[σP, σQ] = e[aP + cQ, bP + dQ] = e[P,P ]abe[P,Q]ade[Q, P ]cbe[Q, Q]cd e[P,Q]ad−bc

 (σ) But from σ(ζ) = ζ l , we see that ad − bc = l(σ). Hence

l(σ) = det ρ(σ) ∀σ ∈ GK

11 5.2 Local invariants Let and consider the short exact sequence where GFp = Gal(Fp/Fp) 1 → I → GQp → GFp → 1 I = {σ ∈ GQp : σ = . Let be any elements of that reduces to p. Recall that a module is unramied if acts 1} Frobp GQp x 7→ x GQp M I trivially on M. √ Example. Let (note and , unramied), 2 3 and 2 K = Q5 −1 ∈ Q5 Q5(ζ8) = Q5(ζ3) E1 : y = x − 1 E2 : y = (x − 1)(x2 − 5). Over we get 2 3 (curve of good reduction) and 2 3 2 (multiplicative reduction, and F5 Ee1 : y = x −√1 √ Ee2 : y = x − x note that it is equivalent to (y + −1x)(y − −1x) = x3) We consider n with . So 2 so is unramied E[l ] √ l = 2√ E1[2] = {0, (1√, 0), (ζ3, 0), (ζ3 , 0)} Q5(E1[2]) so in ramied. E2[2] = {0, (1, 0), ( 5, 0), (− 5, 0)} Q5( 5) √ 4 Q5(E[4]) Q5(ζ16) Q5( 5) √ Q5(E[2]) Q5(ζ8) Q5( 5)

Q5 Q5 Q5 Recall the denition of the local polynomial −1 ∗ I Pp(ρE,l,T ) = det(1 − Frobp T |(VlE ) ) Good Reduction: Theorem 5.3 (Neron-Ogg-Shaferevich). If is an elliptic curve, . Then has good reduction at if E/Qp l 6= p E p and only if E[ln] is unramied for all n (if and only if I acts trivially on E[ln] for all n) Proof. Silverman pg 201

1 0 From this we know that I → . Furthermore we want to know what Frob is, but  (Frob ) = p = 0 1 p ρ p det ρ(Frobp). Hence Frobp is a 2 × 2 matrix with determinant p. Fact. , . But is separable implies that • Q ∈ E(Fp) ⇐⇒ Frobp(Q) = Q #E(Fp) = # ker(1 − Frobp) 1 − Frobp ker(1 − Frobp) = deg(1 − Frobp) If , then . Hence . So the • ψ ∈ End(E) tr(ψ) = 1 + deg ψ − deg(1 − ψ) tr(Frobp) = 1 + p − #E(Fp) =: ap 2 characteristic polynomial of Frobp is T − aT + p

∗ I ∗ 2 Now (VlE ) = VlE , so Pp(T ) = 1 − aT + pT . Example. 2 3 , , hence , we have 2. So in E! : y = x − 1 E1(F5) = {0, (±2, 0), (1, 0), (3, ±1)} #E1(F5) = 6 P5 = 1 + 5T 1 0 0 −5 some basis I → and Frob → . 0 1 p 1 0 Multiplicative Reduction: exp Suppose the reduction is split multiplicative. Recall E/C =∼ C/(Z + τZ) → C∗/qZ (where q = e2πiτ ) are isomorphic as complex Lie groups. Theorem 5.4 (Tate). Let has split multiplicative reduction, then there exists unique such E/Qp 0 6= q ∈ pZp that ∼ 2 3 where and are power series in which converges. E = Eq : y + xy = x + a4(q)x + a6(q) a4(q) q6(q) Z[[q]] ∗ Furthermore, and Q n 24. Hence ∼ ∼ Z (as j(Eq) = 1/q + 744 + 196884q + ... ∆(Eq) = q (1 − q ) E(Qp) = Eq(Qp) = Qp/q -modules) GQp

√ n √n Corollary 5.5. E[l] = ζl, l q and E[l ] = ζln , l q

So Q(E[ln]) has growing ramication for n ≥ 1 (it can be the same at each step, but it will slowly grow)

12 Example. , 2 2 . We get 14 and 10 , hence is a -unit. So E2/Q5√y = (x − 1)(x − 5) j(E2) = 2 /5 ∆ = 2 · 5 q 5 n 2n
(E[2 ]) =∼ 5, ζ n for all n ≥ 1. Q5 Q5 2 √ √ Action of on n , so consider and ln t ln , where -adic tame character. I E[l ] σ(ζln ) = ζln σ( q) = ζln q t = tl(σ) = l   1 t p Hence I 7→ . Now we look at the action of Frobenius. We saw that Frob (ζ n ) = ζ , and we know that the 0 1 p l ln p ∗ determinant is p, so Frob 7→ . To determine ∗, we can use the previous section: ρ = ρ ⊗ sp(1), but ρ is p 0 1 E p 0 trivial, so ∗ is trivial and Frob → . p 0 1 ∗ Now we calculate (VlE) and conclude that Pp(T ) = 1 − T .

In the non-split case, we nd that Pp(T ) = 1 + T . Putting all this together we get

1 − aT + pT 2 good reduction  1 − T split mult Pp(T ) = 1 + T non − split mult  1 additive

For an elliptic curve E over a number eld K we can dene

Y 1 L(ρE, s) = −s PP (ρE, Norm(P) ) P∈OK

6 Examples of l-adic representations for elliptic curves (Alejandro)

In this section ρE,l = ρl = ρ. Notation.

• l is a prime

2 • V = Ql 0 are lattices (i.e., rank -submodules of ) • L, L 2 Zl V • Λ, Λ0 are classes of lattices L, L0 with respect to homothety ∼ • ρ : Gk → GL(V ) = GL2(Ql) For a given l-adic Galois representation ρ, we are going to show that there exists a (non-canonical) lattice

GK / GL2(Ql) O & ? GL2(Zl) we are going to see proposition and examples. We will see Dictson's theorem and we will show that over Q for l ≥ 5, if ρ is surjective mod l then ρ is surjective. Denition 6.1. The Bichat-Tits tree is the graph T with: 1. Vertices, , where is the equivalence class of some lattice of 2 Λ := [l] Λ L Ql 0 2. There is an edge between two vertices v1, v2 of T if and only if there exists L and L such that v1 = Λ and 0 0 v2 = Λ and L ⊃ L ⊃ lL

13 Example. There are eight 2-isogony classes for the elliptic curves of conductor

• • • • ••• •

6.1 Stable lattices and Galois representations

ρ : GK → GL2(Ql)

Denition 6.2. A lattice L is GK -stable with respect to ρ if ρ(GK )(L) ⊆ L. This property only depends on the homothety class Λ of L. Proposition 6.3. Every representation ρ as at least one stable lattice. Sketch of proof. Let be any lattice of 2 and be the subgroup of such that for . This L Ql H GK ρ(σ)(L) ⊆ (L) σ ∈ H is an open subgroup since ??? nite index in GK because GK is compact. Hence the lattice generated by the sum is stable under GK . Denition 6.4. Two integral representations are isogeneous if they are conjugate as repres- ρj : GK → GL2(Zl) entations in , i.e., there exists such that −1 for all . GL2(Ql) U ∈ GL2(Ql) ρ2(σ) = Uρ1(σ)U σ ∈ GK Denition 6.5. Let be an integral representation. The Residual representation associated to ρ : GK → GL2(Zl) is the map obtained by composing with the reduction map. ρ ρ : GK → GL2(Fl) ρ

ρ GK / GL2(Zl) mod l ρ &  GL2(Fl)

Example. Let E1,E2 be two elliptic curve over K. Suppose there exists a K 2-rational isogeny E1 → E2. For each curve we have . The residual have image which is of order either (if has order ) or ρE1,2, ρE2,2 1 Ej(K)[2] 4 2 (if Ej(K)[2] has order 2). Proposition 6.6. The number of stable lattice (up to homothety) is nite if and only if ρ is irreducible.

Proposition 6.7. Let ρ be an integral representation. The number of stable lattices (up to homothety) if 1 if and only if the residual representation ρ is irreducible.

6.2 Dickson's Theorem Theorem 6.8. Let be a prime and a nite subgroup of . Then a conjugate of is one of the l ≥ 3 H PGL2(Fl) H following groups: 1. A nite subgroup of the upper triangular matrices (Borel subgroup) 2. or for some PSL2(Flr ) PGL2(Flr ) r ∈ Z>0 3. A dihedral group with and D2n n ∈ Z>1 (l, n) = 1

4. A subgroup isomorphic to either A4,S4 or A5.

14 6.3 Surjectivity l ≥ 5 and non-surjectivity for l = 2 or 3. Here we are only talking about representations attached to elliptic curves.

• Tim and Vlad published a paper showing that ρ2 is surjective mod 2 but not mod 4; and mod 4 but not mod 8

• Elkies showed that for l = 3, ρ3 is surjective mod 3 but not mod 9. Theorem 6.9. Let E : y2 = x3 + ax + b be an elliptic curve over Q with ∆ = −16(4a3 + 27b2) and j invariant 4a3 . Then −1728 ∆ 1. is surjective if and only if 3 irreducible over and ∗ 2 ρ2 x + ax + b Q ∆ ∈/ (Q ) 2. is surjective if and only if is surjective, ∗ 2 and 3 for any ρ4 ρ2 ∆ ∈/ −1 · (Q ) j 6= 4t (t + 8) t ∈ Q 3. is surjective if and only if is surjective and ∗ 2. ρ8 ρ4 ∆ ∈/ −2 · (Q ) 7 Galois Representations of Modular Curves (Chris Williams) 7.1 Modular Curves Let . Dene the (compactied) modular curve to be ∗ where Γ = Γ0(N) ≤ SL2(Z) X(Γ) = X0(N) := Γ\H H∗ = H ∪ P1(Q) Fact.

• X0(N) is a compact Hausdor Riemann surface

• g(X0(N)) = dimC S2(Γ0(N)) has a model as an algebraic curve over . (In fact it has a model as a scheme over  1 ) • X0(N) Q Z N 1 0 Hecke operators have a geometric interpretation. If we dene γ = , Γ0 = Γ∩γ−1Γγ and Γ00 = γ Γγ−1∩Γ p 0 p p p p p we then get x7→γ xγ−1 0 p p 00 Γ _ / Γ  _   ΓΓ This descent to α X(Γ0) / X(Γ00) π1  π2  X(Γ) X(Γ) To we get −1 . This extends linearly to x ∈ X(Γ) = X0(N) Tp(x) = π2 ◦ α ◦ π1 (x) ∈ Div(X(Γ)) Tp : Div(X(Γ)) → Div(X(Γ)).

7.2 Picard Groups

0 ∗ Denition 7.1. Let X be an algebraic curve over a eld K. The Picard group of X/K is Pic(X)K = Div (X/K)/K(X) . If φ is a nice map X → Y , then we get maps on the Picard group as follows: Pushforward: dened as P P • φ∗ : Pic(X) → Pic(Y ) x nx[x] 7→ x nx[φ(x)] Pullback: ∗ dened as P P P • φ : Pic(Y ) → Pic(X) y ny[y] 7→ y ny x∈φ−1(y) ex[x] ∗ Fact. As endomorphism of Pic(Y ) deg(φ) = φ∗ ◦ φ .

Remark. The action of Tp on Div(X0(N)) descend to Pic(X0(N)). is an abelian variety of dimension . Pic(X0(N)) g = genus(X0(X)) = dimC S2(Γ0(N))

15 7.3 Eichler-Schimura Recall that if is an elliptic curve over , a prime, a prime of . Then has characteristic E Q p - lNp P|p Z ρE,l(FrobP ) polynomial 2 . ∗ as endomorphism of x − qp(E)X + p Ee(Fp) = |ker(σp − 1)| = deg(σp − 1) = (σp − 1)∗ ◦ (σp − 1)

, hence ∗ ∗ ∗ . In particular, as endomorphism of Pic(E) Ee(Fp) = σp∗σp − (σp∗ + σp) + 1 = p + 1 − (σp∗ + σp) Pic(Ee) ∗. ap(E) = σp∗ + σp Fact. For , there exists a smooth projective curve dened over and a surjective map , • p - N X0(N) Fp X0(N) → X0(N) which we call the reduction of X0(N) mod p. Remark. This is base change of  1  to X0(N)/Z N Fp

• There is a map Tp on Pic(X0(N)) making the following commute:

Tp Pic(X0(N)) / Pic(X0(N))

 Tp  Pic(X0(N)) / Pic(X0(N))

Theorem 7.2 (Eichler - Shimura). ∗ as endomorphism of . Tp = σp∗ + σp Pic(X0(N))

Outline of proof. Igusa's theorem (See D-S Section 8.6) says that reduction of X0(N) as a curve is compatible with its interpretation as a moduli space. Then look at what Tp does at the level of moduli spaces.

7.4 The Galois representations of X0(N) Assume l - N Fact.

1. The natural inclusion n n ∼ n 2g is an isomorphism for all . Pic(X0(N)Q)[l ] ,→ Pic(X0(N)C)[l ] = (Z/l Z) n 2. The natural surjection (for ) n n is also an isomorphism. p - lN Pic(X0(N)Q)[l ]  Pic(X0(N))[l ] Hence from now on will be for . X0(N) X0(N)Q Denition 7.3. The l-adic Tate module of Pic(X (N)) is Ta Pic(X (N)) = lim Pic(X (N))[ln] =∼ 2g. 0 l 0 ←−n 0 Zl acts on the points of in the natural way. This gives a natural action of on G := Gal(Q/Q) X0(N) G Q P P Q Div(X0(N)), i.e., σ · nx[x] = nx[σ(x)]. This preserves degree 0 and principal divisors. Thus we get an action of on . The action is linear so preserves n for all and . This action is compatible GQ Pic(X0(N)) Pic(X0(N))[l ] l n n+1 n with the connecting maps: Pic(X0(N))[l ] → Pic(X0(N))[l ]. Thus we get an action on TalPic(X0(N)). Denition 7.4. For , dene ∼ . l - N ρX0(N),l : GQ → Aut(TalPic(X0(N)) = GL2g(Zl) Theorem 7.5. Let p - lN. 1. is unramied at ρX0(N),l p 2. If is a prime of , any Frobenius element, then satises 2 . P|p Z FrobP ρX0(N),l(FrobP ) X − TpX + p = 0 Proof.

16 1. We have a commutative diagram

ρX0(N),l Dp / Aut(TalPic(X0(N)))   GFp / Aut(TalPic(X0(N)))

Now, the inertia IP is in the kernel of the left hand map. The right hand map is an isomorphism (by fact 2. ) In particular, and hence is unramied at . IP ⊂ ker(ρX0(N),l) ρX0(N),l p 2. We have a commutative diagram

T n p n Pic(X0(N)[l ] / Pic(X0(N))[l ]  σ +σ∗  n p∗ p n Pic(X0(N))[l ] / Pic(X0(N))[l ]

we can describe the lifts of and ∗. is a lift of as is totally ramied of degree . While σp,∗ σp FrobP σp,∗ σp p ∗ P −1 so in particular, a lift is −1. So we get a commutative diagram: σp([x]) = y∈σ−1(x) ex[y] = p[σp [x] pFrobP

Frob +pFrob−1 n P P n Pic(X0(N)[l ] / Pic(X0(N))[l ] ∼ ∼ =  σ +σ∗ =  n p∗ p n Pic(X0(N))[l ] / Pic(X0(N))[l ]

Hence −1. This holds for all , hence it holds for . So 2 Tp = FrobP +pFrobP n TalPic(X0(N)) FrobP −TpFrobP +p = 0

8 Modular Galois Representations (Nicolas)

       Last week we had , a b a b ∗ ∗ N ∈ N Γ0(N) = ∈ SL2(Z) ≡ mod N c d c d 0 ∗        This week we use a b a b 1 ∗ . Note that , we Γ1(N) = ∈ SL2(Z) ≡ mod N Γ0(N) Γ1(N) c d c d 0 1 C a b have a map Γ (N)/Γ (N) → ( /N )∗ dened by 7→ d mod N. Dene X (N) = H ∪ 1( )
/Γ (N). 0 1 Z Z c d 1 P Q 1 We have acting on , which gives rise to the diamond operator for all ∗. Γ0  Γ0/Γ1 X1 hdi ∈ T d ∈ (Z/NZ) Let J (N) = Pic0(X (N)), let l ∈ be a prime. We dene T J (N) = lim J (N)[ln] and V J (N) = T J (N)⊗ 1 1 N l 1 ←−n 1 l 1 l 1 Q Theorem 8.1. aords (where genus of ) unramied at . GQ VlJ1(N) ρX1(N),l : GQ → GL2g(Ql) g = X1(N)) lN For all we have satises 2 . p - lN ρX1(N),l(Frobp) X − TqX + p hpi = 0 Actually is a free -module of rank , so and the characteristic VlJ1(N) (T ⊗ Ql) 2 ρX1(N),l : GQ → GL2(T ⊗ Ql) polynomial of is 2 . ρX1(N),l(Frobp) X − TqX + p hpi Let , and let . Reminder: a new form is an normalised eigenform k ∈ N Nk(N) = {new forms in Sk(Γ1(N))} which is genuinely of level N (i.e., does not come from lower level) ⊂ Remark. For all D|N , Nk(N) ⊂ Sk(Γ1(N)). P n For all f = q + anq ∈ Nk(N), we have that: is a number eld. • Kf = Q(an) • There exists  :(Z/NZ)∗ → C∗ such that for all d, hdi f = (d)f.

17 For all σ P n with . σ • σ ∈ GQ, f = q + σ(an)q σ(an) ∈ Nk(N) char = σ ◦  Pick with P n. Dene . We have the isomorphism f ∈ N2(N) f = anq If = {T ∈ T|T f = 0} ⊂ T (T ⊗ Q) /If → Kf dened by Tp 7→ ap, hdi 7→ (d) Dene . It is an abelian variety over of dimension . Af = J1(N)//If J1(N) Q d = [Kf : Q]

Theorem 8.2. Q σ0(N/D). And actually ∼ Q σ0(N/D) as J1(N) ∼ D|N,F ∈G \N (D) AF VlJ1(N) = D|N,F ∈G \N (D) VlAF -modules Q 2 Q 2 GQ ∼ Q , therefore Kl ⊗ Ql = `|l Kf,` Theorem 8.3. For all in , there exists unramied outside in . The characteristic `|l Kf ρf,` : GQ → GL2(Kf,`) lN polynomial of is 2 (for ). ρf,`(Frobp) X − apX + p(p) p - lN 8.1 Residual maps Let with nite. There exists 0 such that 0 . We want to dene ρ : GQ → GLd(K`) K`/Ql ρ ∼ ρ Imρ ⊆ GLd(ZK` ) ρ = ρ0 mod `. This is not well dened! χ ψ χ lψ χ ψ χ 0 Example. Let ρ = ∼ but reduced mod l we have 6∼ . 0 χ 0 χ 0 χ 0 χ

Denition 8.4. Let ρ : G → GL(V ) be a representation. Dene V ss = V if there is no W ⊂ V subrepresentation with V ss = (W ) ⊕ (V/W ) So we dene ρ = (ρ0 mod `)ss

ρ1 Theorem 8.5 (Brauer - Nabitt). Let G / GL(V ) be 2 semi-simple representation. If for all g ∈ G we / ρ2 have that the characteristic polynomial of ρ1(g) is equal to the characteristic polynomial of ρ2(g), then ρ1 ∼ ρ2.

8.2 Higher weights Theorem 8.6 (Deligne 1971). Let , for all P n , for all in , there exists k ≥ 2 f = anq ∈ Nk(N) `|l Kf ρf,` : GQ → 2 k−1 GL2(Kf,`) unramied outside lN. We have that the characteristic polynomial of ρf,`(Fobp) is X − apX + p (p) Remark. We have k−1 where is the -adic cyclotomic character. In particular, let be complex det ρf,` = χl  χl l c ∈ GQ conjugation we have k k−1 . Hence is odd det ρf,`(c) = χl (c)(c) = (−1) (−1) = −1 ρf,` a b −1  The last step relied on: for all γ = ∈ Γ (N), f(γz) = (d)(cz +d)kf(z). In particular γ = ∈ c d 0 −1 k Γ0 so (−1)(−1) = +1. k−1 k−1 Remark. For all and for all prime, we have 2 . For all we have 2 where Kf ,→ C p |ap| ≤ 2p n ∈ N |an| ≤ σ0(n)n σ0(n) = #{d|n}

8.3 Weight 1 Theorem 8.7 (Deligne - Serre, 1976). For all there exists , unramied outside . f ∈ N1(N) ρf : GQ → GL2(C) N 2 The characteristic polynomial of ρf (Frobp) is X − apX + (p). Actually ρf is irreducible and the conductor is N. Sketch of Proof. The steps for this proof are as follow 1. There exists for innitely many . ρf,l l

2. {ap, pprime} is almost nite

18 3. If then there exists constant for all such that Gl = Imρf,l ⊂ GL2(Fl) C l #Gl ≤ C 4. For , may be lifted to . This gives to representations in l  1 Gl GL2(C) ρf,l GL2(C) 5. Calculate characteristic polynomials

0 6. For all l, l , ρf,l ∼ ρf,l0 = ρf .

9 From l-adic to mod l representations. Serre's conjecture: the level (Samuele)

Let and be integers, . Let be an eigenform, P n. Let , N k k ≥ 2 f ∈ Sk(Γ1(N)) f(z) = q + n≥2 anq E = Q({an}) f a character of f, then hdi f = f (d)f. From the previous section we know there exists a family of continuous λ-adic representation where and is the completion of at . We have is ρf,λ : Gal(Q/Q) → GL2(Eλ) λ ⊂ OE Eλ E λ ρf,λ irreducible and , and k−1. To we can associate ∀p - N · Nm(λ) Tr(ρf,λ(Frobp)) = ap det(ρf,λ(Frobp)) = f (p) · p ρf,λ where and this representation is only dened up to semisimplication. ρf,λ : GQ → GL2(F) F = OE,λ/λ Let the question is when is ∼ ss? ρ : GQ → GL2(Fl) ρ = ρf,λ Serre: A necessary and sucient condition is that ρ is odd if ρ is semisimple. Let us forget about which are reducible ρ : GQ → GL2(Fl) Theorem 9.1 (Khane, Witenberger, Kism, Dieulefait. (Serre's conjecture)). Let
be a continuous, ρ : GQ → GL2 Fl irreducible, odd representation then ρ is modular Modular means that there exists integers such that ∼ where . There exists , N, k ρ = ρf,λ f ∈ Sk(Γ1(N)) N(ρ) k(ρ) minimal. is the Artin conductor of away from and is weight in terms of . N(ρ) ρ l k(ρ) ρ|Il

α Theorem 9.2 (Ribet). Assume l ≥ 3 and suppose that ρ arises from Γ1(M) when M = N · l , gcd(N, l) = 1.Then ρ arises from Γ1(N). Remark. Buzzard generalised the above for the case l = 2.

Theorem 9.3. Suppose that ρ arises from Sk(Γ1(N)) with gcd(N, l) = 1 and 2 ≤ k ≤ l + 1. Assume either l > 3 or N > 3, then ρ arises from S2(Γ1(Nl)). This theorem comes from Ash-Stevens under the condition that l ≥ 5 and Serre-Gross under the assumption that N ≥ 4.

Theorem 9.4 (Edixhoron). Let gcd(N, l) = 1 and assume ρ arises from Sk(Γ1(N)) then ρ arises from Sk(ρ)(Γ1(N)), where k(ρ) is Serre's weight, furthermore k ≡ k(ρ) mod l − 1 and k ≥ k(ρ) if l is odd. Corollary 9.5. If arises from and then there exists such that i arises from ρ Γ1(N) gcd(N, l) = 1 i ∈ Z ρ ⊗ χ Sk(Γ1(N)) for k ≤ l + 1, where χ is the mod l Cyclotomic character. Let be irreducible and consider the following four sets ρ : GQ → GL2(Fl)  •N1 = N gcd(N, l) = 1, ρ arises from Sk(ρ)(Γ1(N))  •N2 = N gcd(N, l) = 1, ρ arises from Γ1(N)

 α •N3 = N gcd(N, l) = 1, ρ arises from Γ1(Nl ), α > 0

 2 •N4 = N gcd(N, l) = 1, ρ arises from S2(Γ1(Nl ))

Theorem 9.6. If l ≥ 5 then the four sets N1, N2, N3 and N4 are equal

19 Proof. N1 = N2 are equal by Theorem 9.4 N2 = N3 are equal by Theorem 9.2 By denition N4 ⊆ N3 so we want to show that N3 ⊆ N4 or equivalently N2 ⊆ N4 . Assume that ρ arises i from Γ1(N) and choose i ≥ 0 by Corollary 9.5 such that ρ ⊗ χ arises from Sk(Γ1(N)) with 2 ≤ k ≤ l + 1. By i i Theorem 9.3 then ρ ⊗ χ arises from S2(Γ1(Nl)). Now tensoring with χ changes the level of a modular form but i ∼ i −i 2 not the weight. Look at χ as a Dirichlet character, f : ρf,λ = ρ ⊗ χ then consider f ⊗ χ ∈ S2(Γ1(Nl )). So i −i 2 ρ = (ρ ⊗ χ ) ⊗ χ arises from S2(Γ1(Nl )). We shall denote the four equal sets by N (ρ). So the question now becomes is N(ρ) ∈ N (ρ)?

Theorem 9.7 (Livne). Suppose ρ arises from Γ1(N) then N(ρ)|N. Let f be an eigenform giving rise to ρ. Then N the level of f is such that N(ρ)|N, or better N(ρ)|N 0 where N 0 is the prime-to-l part of N. The aim is: If N(ρ) 6= N 0 then we want to nd another form at level N(ρ) giving rise to ρ. 0 0 Note that we can replace f by a newform, f , giving the same eigensystem. ρf,λ = ρf 0,λ0 . We have N(ρ)|level(f ). So from now on, assume is a newform. Let us look at the conductors . f N(ρ) = N(ρf,λ) < N(ρf,λ) = level(f) Assume and consider and reduction. We look at the conductor exponents l 6= p ρp : Gal(Qp/Qp) → GL2(Ql) ρp   I and I . We know that nρp = np = dim(V ) − dim V + nρp,wild nρp = np = dim(V ) − dim V + nρp,wild nρp,wild = , we also know that I I , so . We want to study when . nρp,wild dim V ≥ dim V np ≤ np np < np

Theorem 9.8. The representation ρp which can degenerate (i.e., np < np) can be one of the following ∼ 1. Principal series: ρp = µ ⊕ ν such that nµ = 1 and nµ = 0, then np = nν + 1 and np = nν .

2. Special case (Steinberg I ): ρp = µ ⊗ sp(1) such that nµ = 0 (then np = 1 and np = 0)

3. Special case (Twist Steinberg): ρp = µ ⊗ sp(1) such that nµ = 1 and nµ = 0 (then np = 2 and np = 0) 4. (Super) Cuspidal case: such that and (then and ) ρp = Indζ nζ = 1 nζ = 0 np = 2 np = 0 Back to modular forms: Theorem 9.9 (Ribet level lowering). Assume that where is a newform of level and N(ρf,λ) < N f Γ1(N) . Then for every there exists a Dirichlet character of conductor and -power or- gcd(N, l) = 1 p N/N(ρf,λ) φ p l der such that the newform attached to has level dividing . In particular, is modular of level where f ⊗ φ N/p ρf,λ M Q where . M = N/ p p|N/N(ρf,λ)

20