Galois Representations

Transcription

1 Galois Representations Samir Siksek 12 July 2016

2 Representations of Elliptic Curves Crash Course E/Q elliptic curve; G Q = Gal(Q/Q); p prime. Fact: There is a τ H such that E(C) = C Z + τz = R Z R Z. Easy to see: E[p] = Z/pZ Z/pZ (2-dim l F p -vector space).

3 G Q acts on E[p]: Example ρ E,p : G Q GL 2 (F p ). Suppose P is a rational point of order p. Choose Q E[p] so that P, Q are an F p -basis. For any σ G Q, σ(p) = P, σ(q) = b σ P + d σ Q. Hence ( ) 1 bσ ρ E,p (σ) =. 0 d σ

4 Conclusion: E has p-torsion defined over Q iff ( ) 1 ρ E,p, (ψ : G 0 ψ Q F p is a character).

5 Example Suppose E has a p-isogeny φ : E E defined over Q. Then Ker(φ) = P is a cyclic subgroup of order p. Choose Q E[p] so that P, Q are an F p -basis. For any σ G Q, σ(p) = a σ P, σ(q) = b σ P + d σ Q. Hence ( ) aσ b ρ E,p (σ) = σ. 0 d σ

6 Conclusion: E has p-isogeny defined over Q iff ( ) ψ1 ρ E,p, (ψ 0 ψ 1, ψ 2 : G Q F p are characters). 2 i.e. ρ E,p is reducible.

7 Mazur s Isogeny Theorem Theorem (Mazur) Let E be an elliptic curve over Q. For p > 163, the elliptic curve E has no p-isogenies defined over Q. Equivalent theorem: Theorem (Mazur) Let E be an elliptic curve over Q. If p > 163, the mod p representation ρ E,p is irreducible.

8 Exercise Denote by ζ p a primitive p-th root of unity. Define the mod p cyclotomic character χ p : G Q F p, σ(ζ p ) = ζp χp(σ). (i) Show that χ p is a character (i.e. a homomorphism). (ii) Let σ G Q denote complex conjugation. Show that χ p (σ) = 1. (ii) Let l p be a prime, and let σ l be a Frobenius element at l. Show that χ p (σ l ) l (mod p). (iv) Show that for all σ G Q we have det(ρ E,p (σ)) = χ p (σ). Hint: Think about the Weil pairing.

9 Note that for complex conjugation σ we have We say that ρ E,p is odd. det(ρ E,p (σ)) = 1.

10 Galois Representations from Modular Forms Theorem (Eichler, Shimura, Deligne) Let f be a newform of level N and weight k 2, and write f = q + c n q n. Let K = Q(c 1, c 2,... ). Let p be a rational prime. Then there is some prime ideal P p of O K and a representation: ρ f,p : G Q GL 2 (F P ), F P = O K /P. Moreover, if l Np is a prime, and σ l G Q is the Frobenius element at l then Tr(ρ f,p (σ l )) c l (mod P).

11 Serre s Modularity Conjecture Theorem (Khare and Wintenberger) Let p be and odd prime and let ρ : G Q GL 2 (F q ) (here q = p r ) be an odd irreducible representation. Then there is a newform f of level N = N(ρ) (given by an explicit recipe) and a weight k = k(ρ) (given by an explicit recipe) and a prime ideal P p such that ρ ρ f,p.

12 Theorem (Khare and Wintenberger) Let p be and odd prime and let ρ : G Q GL 2 (F q ) (here q = p r ) be an odd irreducible representation. Then there is a newform f of level N = N(ρ) (given by an explicit recipe) and a weight k = k(ρ) (given by an explicit recipe) and a prime ideal P p such that ρ ρ f,p. We say ρ is unramified at l if ρ(i l ) = 1 where I l G Q is the inertia subgroup at l. The primes l N are the ramified primes l p. k = 2 if p #ρ(i p ).

13 Ribet s Theorem E/Q elliptic curve; p prime. We know that ρ E,p : G Q GL 2 (F p ) is odd. Suppose E has no p-isogenies. Then ρ E,p is irreducible. By Serre s modularity conjecture, there is a newform f of weight k p and level N p and a prime ideal P p such that ρ E,p ρ f,p (this is equivalent to E p f ). Goal: Demystifying the recipe for N p. In particular, we want to show that if l N and p ord l ( ) then ρ E,p is unramified at l.

14 Complex Parametrization Revisited Recall for E/C, we have E(C) = C Z + τz. Let q = exp(2πiτ). Then E(C) = C Z + τz = C /q Z, z + (Z + τz) exp(2πiz) q Z.

15 The Tate Curve Theorem (Tate) Let l be a prime, and E an elliptic curve with split multiplicative reduction at l. Then there is q lz l such that as G l -modules (G l = Gal(Q l /Q l )). E(Q l ) = Q l /q Z Corollary p l prime E[p] = ζ p q 1/p (mod q Z ) as G l -modules.

16 Corollary If σ G l then E[p] = ζ p q 1/p (mod q Z ) as G l -modules. σ(ζ p ) = ζ a p, σ(q 1/p ) = ζ b p q 1/p, a, b F p. Think of ζ p and q 1/p as an F p -basis for E[p]. The action of σ is given by Obtain a representation ρ p (σ) := ( ) a b. 0 1 ρ p : G l GL 2 (F p ).

17 Image of Inertia I l G l inertia subgroup As p l, the extension Q l (ζ p )/Q l is unramified, so σ(ζ p ) = ζ p, for all σ I l. So ρ p (I l ) {( ) } 1 b : b F 0 1 p (cyclic of order p). The extension Q l (q 1/p )/Q l is unramified if and only if p υ l (q). Lemma If p υ l (q) then #ρ p (I l ) = 1. If p υ l (q) then #ρ p (I l ) = p.

18 The discriminant of E is given by = q n 1(1 q n ) 24 (observe υ l (q) = υ l ( )). Lemma If p υ l ( ) then #ρ p (I l ) = 1. If p υ l ( ) then #ρ p (I l ) = p. Conclusion: Let E/Q of conductor N. Suppose ρ E,p is irreducible. By Serre s modularity conjecture, there is a newform f of level N p and a prime ideal P p such that ρ E,p ρ f,p. Let l N (i.e. E has multiplicative reduction at N) and p ord l ( ) then the above tells us that l N p.