Galois representations - · PDF filePlan for the course IIntroduction, review of algebraic number theory IArtin and ‘-adic representations, L-functions ILocal Galois representations,

Galois representations

London Taught Course Centre

Lecture 1, 16 January 2017

Plan for the course

I Introduction, review of algebraic number theoryI Artin and `-adic representations, L-functionsI Local Galois representations, compatible systemsI `-adic monodromy theorem, Weil–Deligne representationsI p-adic Hodge theory (a user’s perspective)

1. Introduction

1.1. What is a Galois representation?

A Galois representation is a representation of a Galois group

Gal(L/K )→ GLn(E).

Here E , K and L are fields, L/K Galois

In number theory, especially interested in:

I K = Q, or Fp(T ), for a prime p,I a finite extension of the above (i.e., a global field),I a completion of the above (i.e., a local field).

Allow L/K to be infinite (e.g., L = Q = {algebraic numbers}).

E is often C, a finite field, Qp, or might allow rings (e.g., Zp).

1.2. First examples

1.2.1. Dirichlet characters

Let K = Q, L = Q(ζN), ζN = e2πi/N . Then

Gal(L/K ) ∼= (Z/NZ)×

(σa : ζN 7→ ζaN) ←→ (a mod N)

So any Dirichlet character, i.e., homomorphism:

χ : (Z/NZ)× → C×

defines a Galois representation

ρχ : Gal(L/K )→ C×,

where ρχ(σa) = χ(a).E.g., N = 3, L = Q(ζ3) = Q(

√−3),

Gal(L/Q) ∼= (Z/3Z)× ∼= {±1} ⊂ C×.

1.2.2. A 2-dimensional example

Let K = Q, L = Q(ζ3, α) = splitting field of x3 − 2, α =3√

2.

Gal(L/K ) ∼= S3 = {permutations of {α, ζ3α, ζ23α} }.

Generated by σ (of order 2), τ (of order 3), defined by

σ(ζ3) = ζ23 , σ(α) = α, τ(ζ3) = ζ3, τ(α) = ζ3α.

Define ρ : Gal(L/Q) ∼= S3 → GL2(C), by

σ 7→(

0 11 0

),

τ 7→(

0 1−1 −1

)

1.2.3. Mod p examples (E = Z/pZ)Take the preceding example, but note in fact:

ρ : Gal(L/K )→ GL2(Z).

Compose with reduction mod p to get:

ρp : Gal(L/K )→ GL2(Z/pZ).

E.g., for p = 2, Gal(L/Q) ∼= S3∼= GL2(Z/2Z).

For p = 3, conjugating by(

1 0−1 1

)makes this:

σ 7→(−1 0

0 1

), τ 7→

(1 10 1

)

This is Gal(L/Q) acting on:

〈ζ3, α〉 /〈2〉 ∼= (Z/3Z)2

(where 〈·〉 is the subgroup generated in L×),a two-dimensional vector space over Z/3Z.

1.3 Why study Galois representations?

1.3.1. Answer 1: To study Galois groups.

E.g., for G finite, E = C:I G abelian⇔ all representations are one-dimensional;I # irreducible representations = # conjugacy classes.

Representation theory of G ↔ structure of E [G].

Functorial constructions, e.g. induction, restriction, ⊗, etc

Lacking good description of Gal(Q/Q), study representations.

1.3.2. Answer 2. To study varieties, e.g. over number fields.

Galois representations arise naturally from (étale) cohomology,

carrying arithmetic information about varieties, detecting e.g.:I rational points (e.g., Weil Conjectures, BSD)I algebraic cycles (Tate conjecture)I reduction properties (e.g., Néron–Ogg-Shafarevich)I other invariants (e.g., Selmer groups, regulators).

For an elliptic curve A : y2 = x3 + ax + b over Q,

étale cohomology ↔ torsion points.

Concretely, A[N] ∼= (Z/NZ)2, with action of Gal(Q/Q).

For N = ` (prime), get Gal(Q/Q)→ GL2(E), E = F`.

Taking N = `r , r →∞ gives the `-adic Tate module T`(A):

ρA,` : Gal(Q/Q)→ GL2(E), E = Q`.

I knows: #A(Fp) for all p;I isogeny class of A (Serre, Faltings);I reduction properties of A (e.g. conductor);I Selmer group, Tate–Shafarevich group, . . .

1.3.3. Answer 3: Relation to modular (and automorphic) forms

Class field theory describes abelian extensions of K ,i.e., one-dimensional representations of Gal(K̄/K ).(E.g.,↔ Dirichlet characters if K = Q, E = C.)

More generally, for n ≥ 1, Langlands Programme relates

{automorphic forms} ↔ {Galois representations}.

Many cases of→ known, some cases of←, e.g. every

ρ : Gal(L/Q)→ GL2(C), s.t. det(ρ(cx conj)) = −1

arise from a weight one modular form (Khare–Wintenberger).

If A is an elliptic curve over Q, then ρA,` arises froma weight two modular form (STW Conj, Wiles, TW, BCDT).

1.3.4. Answer 4: Surprising(?) applications

I Mordell Conj (Faltings), via cases of Tate ConjI Fermat’s Last Theorem (Wiles), via modularityI Holomorphy of L(A, s) for e.c.’s A over Q, via modularity

Truth in advertising: The course will focus on how to studyGalois representations (answer: locally), not why.

1.4 Examples revisited:

Let A : y2 = x3 − 108, L = Q(ζ3, α), α =3√

2, as before. Then

A[2] = {∞, (6ζ r3/α,0)}.

A[3] = {∞, (0,±6i√

3), (6ζ r3α,±18)}.

Action of Gal(Q/Q) on A[`] gives earlier examples:

Gal(L/Q)→ GL2(F`), ` = 2,3.

A has CM by K = Q(ζ3), so representations on T`(A)(and A[`] for ` > 3) are induced from characters:

χ : Gal(Q/K )→ E×.

(May need to enlarge E .) In particular image has the form:{(∗ 00 ∗

)}∪{(

0 ∗∗ 0

)}.

2. Algebraic number theory review

2.1. Algebraic integers

Suppose L is a number field, i.e., finite extension of Q, so

L = Q(β), for some algebraic number β,

minimal polynomial h(x), degree d = [L : Q].(Everything generalizes to L/K .)ν ∈ L an algebraic integer if its minimal polynomial is in Z[x ].

OL = {algebraic integers in L},

OL is a ring, rank d as abelian group.Examples:I L = Q(ζN), d = φ(N), OL = Z[ζN ].I L = Q(ζ3, α), α =

3√

2, d = 6, OL = Z[ζ3, α].

2.2 Factorization of ideals, primes

Let I be a non-zero ideal of OL.I I has finite index in OL

I Pe11 Pe2

2 . . .Pekk for prime ideals P1, . . . ,Pk (uniquely)

In particular, this applies to I = pOL for p prime.Suppose for simplicity that OL = Z[β] = Z[x ]/(h(x)), so

OL/pOL∼= Fp[x ]/(h̄(x)).

Then factorizaton of h̄(x) ∈ Fp[x ] gives that of pOL.

In particular, ifh̄ = he1

1 he22 . . . hek

k

for distinct monic irreducible h1, . . . ,hk , then:

pOL = Pe11 Pe2

2 . . .Pekk

for distinct primes ideals (“primes”) P1, . . . ,Pk of OL.I k = # primes over p = # factors, h1, . . . ,hk ;I pOL is prime⇔ h̄(x) irreducible;I #(OL/Pi) = pfi where fi = deg(hi);I∑

i ei fi = d ;I ei = multiplicity of hi ;I all ei = 1⇔ p - Disc(L), call p unramified in L.

Example: L = Q(ζq), q prime, d = q − 1, OL = Z[ζq].

h(x) =xq − 1x − 1

= xq−1 + xq−2 + · · ·+ x + 1.

In Fq[x ], we have h(x) = (x − 1)q−1, so

qOL = Qq−1, Q = (ζq − 1)OL.

If p 6= q, then p is unramified (Disc(L) = ±qq−2), so

pOL = P1P2 · · ·Pk , distinct Pi .

For each i , OL/Pi∼= Fp[ζq] ∼= Fpf ,

where f is the least positive integer s.t. q|(pf − 1), i.e.,

f = order of p in (Z/qZ)×, k = (q − 1)/f .

E.g., q = 3, L = Q(ζ3) = Q(√−3), OL = Z[ζ3].

I 3OL = Q2, where Q = (√−3)OL;

I pOL = P1P2 ⇔ p ≡ 1 mod 3;I pOL is prime⇔ p ≡ 2 mod 3.

Example: L = Q(α), α =3√

2, OL = Z[α], h(x) = x3 − 2.I p = 2: h̄(x) = x3, so 2OL = P3 = (α)3.I p = 3: h̄(x) = x3 + 1 = (x + 1)3, 3OL = Q3 = (1 + α)3.

If p > 3, then p is unramified (discriminant = −108.)If p ≡ 2 mod 3, then order of F×

p not divisible by 3,soh̄(x) = x3 − 2 has a unique root mod p

=⇒ pOL = P1P2, with #(OL/Pi) = pi .

If p ≡ 1 mod 3, then either:I 2 is a cube mod p, so h̄ has 3 roots, pOL = P1P2P3;I 2 is not a cube mod p, so h̄ irreducible, pOL prime.

If L is Galois over Q, then ei and fi are independent of i .Writing ei = e, fi = f , we have d = efk .

Example: L = Q(ζ3, α), OL = Z[ζ3, α].

Combining the previous two examples gives:

I 2OL = P3, P = αOL, f = 2;I 3OL = Q6; otherwise:I if p ≡ 2 mod 3, then pOL = P1P2P3, f = 2;I if p ≡ 1 mod 3, 2 ∈ (F×

p )3, then f = 1, k = 6;I if p ≡ 1 mod 3, 2 6∈ (F×

p )3, then pOL = P1P2, f = 3.

2.3 Decomposition and inertia groups

Assume from now on L is Galois over Q, G = Gal(L/Q),

pOL = Pe1 Pe

2 · · ·Pek , #(OL/Pi) = pf , d = efk .

G acts on OL, hence on {P1, . . . ,Pk}, in fact transitively.Let GPi = stabilizer of Pi = {g ∈ G |g(Pi) = Pi}.Call GPi the decomposition group of Pi .Then GPi acts on OL/Pi , so get:

GPi → Aut(OL/Pi) ∼= Gal(Fpf /Fp).

In fact, this map is surjective.

Recall Gal(Fpf /Fp) is cyclic of order f , generated by:

Frobp : ν 7→ νp.

So we have

GPi � Gal(Fpf /Fp) = 〈Frobp〉 ∼= Z/fZ.

Let IPi = kernel of GPi � Gal(Fpf /Fp).Call IPi the inertia group of Pi .Transitivity of G on {Pi} and surjectivity of GPi → Z/fZ imply:I GPi/IPi

∼= Gal(Fpf /Fp) ∼= Z/fZ;I #GPi = ef and #IPi = e;I GP1 , . . . ,GPk are conjugate in G, as are IP1 , . . . , IPk .

If e = 1, then GPi∼= Gal(Fpf /Fp), define FrobPi ↔ Frobp.

{FrobP1 , . . . ,FrobPk} forms a conjugacy class: [Frobp] .

Cebotarev Density Theorem: If g ∈ G, then

{p - Disc(L) such that [Frobp] = [g]}

has density #[g]/#G.

Example: L = Q(ζq), q prime, p 6= q.

G = Gal(L/Q) ∼= (Z/qZ)×, σa ↔ a mod q.

pOL = P1P2 . . .Pk and OL/Pi∼= Fpf = Fp[ζq],

where f = order of p mod q.

σp(ζq) = ζpq =⇒ Frobp = σp (↔ p ∈ (Z/qZ)×).

(Since G is abelian, no need for conjugacy class.)

GPi = 〈σp〉 ←→ 〈p〉 ⊂ (Z/qZ)× (independent of i).

IPi is trivial, and GQ = IQ = (Z/qZ)× for Q = (ζq − 1)OL.

This generalizes to L = Q(ζN) for any N: Frobp ←→ p mod N.

Cebotarev Density Theorem⇐⇒Dirichlet’s Theorem on primes in arithmetic progressions.

Aside: Proof of Quadratic Reciprocity.

Return to N = q prime, G ∼= (Z/qZ)×, cyclic order q − 1.⇒ L = Q(ζq) has a unique quadratic subfield, say K .

L ramified only at q ⇒ K ramified only at q, so

K = Q(√±q), where q ≡ ±1 mod 4.

Suppose q ≡ 1 mod 4. Then:

p a square mod q ⇐⇒ its order f | (q − 1)/2 ⇐⇒σp ∈ Gal(L/K ) ⇐⇒ pOK = P1P2 ⇐⇒

x2 − q factors in Fp[x ] ⇐⇒ q a square mod p.

Similarly if q ≡ −1 mod 4, get

p a square mod q ⇐⇒ −q a square mod p.

Example: L = Q(ζ3, α), G = Gal(L/Q) = 〈σ, τ〉.I p = 2, GP = G, IP = 〈τ〉;I p = 3, GQ = IQ = G; otherwise:I p ≡ 2 mod 3 =⇒ [Frobp] = [σ];I p ≡ 1 mod 3 and 2 ∈ (F×

p )3 =⇒ [Frobp] = [e];I p ≡ 1 mod 3 and 2 6∈ (F×

p )3 =⇒ [Frobp] = [τ ].

Cebotarev Density in this case says:

I 1/2 of all primes p are ≡ 2 mod 3;I 2 is a cube for 1/3 of all primes p ≡ 1 mod 3.