Residual modular Galois representations: images and applications Samuele Anni University of Warwick London Number Theory Seminar King s College London, 20 th May 2015
Mod l modular forms 1 Mod l modular forms 2 Residual modular Galois representations 3 Image: an algorithm 4 Local representation 5 Twist 6 Applications
Mod l modular forms Let n be a positive integer. The congruence subgroup Γ 1 (n) of SL 2 (Z ) is the subgroup given by {( ) } a b Γ 1 (n) = SL c d 2 (Z ) : n a 1, n c. Γ 1 (n) acts on the complex upper half plane via fractional transformations. Definition We define the modular curve Y 1 (n) C to be the non-compact Riemann surface obtained giving on Γ 1 (n)\h the complex structure induced by the quotient map. Let X 1 (n) C be the compactification of Y 1 (n) C. The modular curve Y 1 (n) C can be defined algebraically over Z [1/n].
Mod l modular forms Let n 5 and k be positive integers and let l be a prime not dividing n. Following Katz, we define the space of mod l cusp forms as mod l cusp forms S(n, k) Fl = H 0 (X 1 (n) Fl, ω k ( Cusps)). S(n, k) Fl is a finite dimensional F l -vector space, equipped with Hecke operators T n (n 1) and diamond operators d for every d (Z /nz ). The Z -subalgebra of End C (S(Γ 1 (n), k) C ) generated by T p and diamond operators d is the Hecke algebra T(n, k). Analogous definition in characteristic zero and over any ring where n is invertible.
Mod l modular forms One may think that mod l modular forms come from reduction of characteristic zero modular forms mod l: S(n, k) Z [1/n] S(n, k) Fl. Unfortunately, this map is not surjective for k = 1. Even worse: let ɛ: (Z /nz ) C be a character, whose image is contained in O K, ring of integers of the number field K, then the map S(n, k, ɛ) OK S(n, k, ɛ) F is not always surjective even if k > 1, where F is the residue field at l and ɛ is the reduction of ɛ. S(n, k, ɛ) OK : = {f S(n, k) OK d (Z /nz ), d f = ɛ(d)f }. T ɛ (n, k) will denote the associated Hecke algebra.
Residual modular Galois representations 1 Mod l modular forms 2 Residual modular Galois representations 3 Image: an algorithm 4 Local representation 5 Twist 6 Applications
Residual modular Galois representations Theorem (Deligne, Serre, Shimura) Let n and k be positive integers. Let F be a finite field of characteristic l, with l not dividing n, and f : T(n, k) F a surjective morphism of rings. Then there is a continuous semi-simple representation: ρ f : Gal(Q /Q ) GL 2 (F), unramified outside nl, such that for all p not dividing nl we have: Trace(ρ f (Frob p )) = f (T p ) and det(ρ f (Frob p )) = f ( p )p k 1 in F. Such a ρ f is unique up to isomorphism.
Residual modular Galois representations Computing ρ f is difficult, but theoretically it can be done in polynomial time in n, k, #F: Edixhoven, Couveignes, de Jong, Merkl, Bruin, Bosman (#F 32): Example: for n = 1, k = 22 and l = 23, the number field corresponding to Pρ f (Galois group isomorphic to PGL 2 (F 23 )) is given by: x 24 2x 23 + 115x 22 + 23x 21 + 1909x 20 + 22218x 19 + 9223x 18 + 121141x 17 + 1837654x 16 800032x 15 + 9856374x 14 + 52362168x 13 32040725x 12 + 279370098x 11 + 1464085056x 10 + 1129229689x 9 + 3299556862x 8 + 14586202192x 7 + 29414918270x 6 + 45332850431x 5 6437110763x 4 111429920358x 3 12449542097x 2 + 93960798341x 31890957224 Mascot, Zeng, Tian (#F 41).
Residual modular Galois representations Question Can we compute the image of a residual modular Galois representation without computing the representation?
Image: an algorithm 1 Mod l modular forms 2 Residual modular Galois representations 3 Image: an algorithm Finite subgroup of PGL 2 (F l ) Serre s Conjecture The algorithm 4 Local representation 5 Twist 6 Applications
Image: an algorithm Main ingredients: 1 Classification of possible images 2 Serre s conjecture
Image: an algorithm Finite subgroup of PGL 2 (F l ) 1 Classification of possible images Theorem (Dickson) Let l be an odd prime and H a finite subgroup of PGL 2 (F l ). Then a conjugate of H is one of the following groups: a finite subgroup of the upper triangular matrices; PSL 2 (F l r ) or PGL 2 (F l r ) for r Z >0 ; a dihedral group D 2n with n Z >1, (l, n) = 1; or it is isomorphic to A 4, S 4 or A 5. Definition If G := ρ f (Gal(Q /Q )) has order prime to l we call the image exceptional.
Residual modular Galois representations Image: an algorithm Finite subgroup of PGL 2 (F l ) The field of definition of the representation is the smallest field F F l over which ρ f is equivalent to all its conjugate. The image of the representation ρ f is then a subgroup of GL 2 (F). Let Pρ f : Gal(Q /Q ) PGL 2 (F) be the projective representation associated to the representation ρ f : Gal(Q /Q ) ρ f GL 2 (F) Pρ f π PGL 2 (F). The representation Pρ f can be defined on a different field than the field of definition of the representation. This field is called the Dickson s field for the representation.
Image: an algorithm Serre s Conjecture 2 Serre s conjecture Theorem (Khare, Wintenberger, Dieulefait, Kisin), Serre s Conjecture Let l be a prime number and let ρ: Gal(Q /Q ) GL 2 (F l ) be an odd, absolutely irreducible, continuous representation. Then ρ is modular of level N(ρ), weight k(ρ) and character ɛ(ρ). N(ρ) (the level) is the Artin conductor away from l. k(ρ) (the weight) is given by a recipe in terms of ρ Il. ɛ(ρ): (Z /N(ρ)Z ) F l is given by: det ρ = ɛ(ρ)χ k(ρ) 1 l, where χ l is the cyclotomic character mod l.
Image: an algorithm The algorithm The algorithm Theorem There is a polynomial time algorithm which takes as input: n and k positive integers, n is given with its factorization; l a prime not dividing n and such that 2 k l + 1; a character ɛ : (Z /nz ) C ; a morphism of ring f : T ɛ (n, k) F l, i.e. the images of all diamond operators and of the T p operators up to a bound B(n, k), and gives as output the image of the associated Galois representation ρ f, up to conjugacy as subgroup of GL 2 (F l ) without computing ρ f.
Image: an algorithm The algorithm In almost all cases, the bound B(n, k) is the Sturm Bound for Γ 0 (n) and weight k: B(n, k) = k 12 n ( 1 + 1 ) k n log log n p 12 p n prime In the cases when this bound is not enough, then the Sturm Bound for Γ 0 (nq 2 ) and weight k, where q is the smallest odd prime not dividing n, is the required bound.
Image: an algorithm The algorithm For an input as in the previous slide we run into the following problem: Problems ρ f can arise from lower level or weight: there exists g S(m, j) Fl with m n or j k such that ρ g = ρf ρ f can arise as twist of a representation of lower level or weight: there exist a character χ and g S(m, j) Fl with m n or j k such that ρ g χ = ρ f Algorithm Iteration down to top, i.e. considering all divisors of n: creation of a database Determine minimality with respect to level and with respect to weight. Determine minimality up to twisting.
Image: an algorithm The algorithm Algorithm Step 1 Iteration down to top Step 2 Determine minimality with respect to level and weight. Step 3 Determine whether reducible or irreducible. Step 4 Determine minimality up to twisting. Step 5 Compute the projective image Step 6 Compute the image Remarks Reducibility is checked by comparing with systems of eigenvalues coming from specific Eisenstein series. Compute the field of definition of the representation: this can be done using coefficients up to a finite bound. Compute the Dickson s field: this is obtained twisting. The projective image is determined by excluding cases. Each case is related to an explicit equality of mod l modular forms or construction.
Image: an algorithm The algorithm Setting ( ) n and k positive integers; l a prime not dividing n and such that 2 k l + 1; ɛ : (Z /nz ) C a character; f : T ɛ (n, k) F l a morphism of rings; ρ f : G Q GL 2 (F l ) the unique, up to isomorphism, continuous semi-simple representation attached to f ; ɛ : (Z /nz ) F l be the character defined by det(ρ f ) = ɛχ k 1 l. Let p be a prime dividing nl. Let us denote by G p = Gal(Q p /Q p ) G Q the decomposition subgroup at p; I p the inertia subgroup; G i,p, with i Z >0, the higher ramification subgroups; N p (ρ) the valuation at p of the Artin conductor of ρ, where ρ is a residual representation.
Local representation 1 Mod l modular forms 2 Residual modular Galois representations 3 Image: an algorithm 4 Local representation Local representation and conductor 5 Twist 6 Applications
Local representation Theorem (Gross-Vignéras-Fontaine, Serre: Conjecture 3.2.6? ) Let ρ : G Q GL(V ) be a continuous, odd, irreducible representation of the absolute Galois group over Q to a 2-dimensional F l -vector space V. Let f S(N(ρ), k(ρ)) Fl be an eigenform such that ρ f = ρ. Let p be a prime divisor of ln. (1) If f (T p ) 0, then there exists a stable line D V for the action of G p, such that I p acts trivially on V /D. Moreover, f (T p ) is equal to the eigenvalue of Frob p which acts on V /D. (2) If f (T p ) = 0, then there exists no stable line D V as in (1). (1) ρ f Gp is reducible; (2) ρ f Gp is irreducible.
Local representation Local representation and conductor From now on we will assume setting ( ). Proposition Let us suppose that ρ f is irreducible and it does not arise from lower level. Let p be a prime dividing n such that f (T p ) 0. Then ρ f Gp is decomposable if and only if ρ f Ip is decomposable. This proposition is proved using representation theory. Can we deduce a computational criterion from this?
Local representation Local representation and conductor Proposition Assume that ρ f is irreducible and it does not arise from lower level. Let p be a prime dividing n, such that f (T p ) 0. Then: (a) ρ f Ip is decomposable if and only if N p (ρ f ) = N p (ɛ); (b) ρ f Ip is indecomposable if and only if N p (ρ f ) = 1 + N p (ɛ).
Local representation Local representation and conductor Sketch of the proof The valuation of N(ρ f ) at p is given by: N p (ρ f ) = i 0 1 [G 0,p : G i,p ] dim(v /V G i,p ) = dim(v /V Ip ) + b(v ), where V is the two-dimensional F l -vector space underlying the representation, V G i,p is the subspace of invariants under G i,p, and b(v ) is the wild part of the conductor. Since f (T p ) 0, the representation restricted to the decomposition group at p is reducible. Hence, after conjugation, ( ) ( ) ρ f Gp ɛ1 χ k 1 = l, ρ 0 ɛ f Ip ɛ1 Ip = 2 0 1 where ɛ 1 and ɛ 2 are characters of G p with ɛ 2 unramified, χ l is the mod l cyclotomic character and belongs to F l.
Local representation Local representation and conductor Sketch of the proof ( ) ρ f Ip ɛ1 Ip =. 0 1 If ρ f Ip is indecomposable ( 0) then V Ip is either {0} if ɛ 1 is ramified, or F l ( 1 0 ) if ɛ 1 is unramified. The wild part of the conductor is equal to the wild part of the conductor of ɛ 1. Hence, we have that { 1 = 1 + N p (ɛ 1 ) if ɛ 1 is unramified, N p (ρ f ) = 2 + b(ɛ 1 ) = 1 + N p (ɛ 1 ) if ɛ 1 is ramified. Since det(ρ f ) = ɛχ k 1 l, then det(ρ f ) Ip = ɛ Ip, so ɛ 1 Ip = ɛ Ip. Therefore, we have that if ρ f Ip is indecomposable then N p (ρ f ) = 1 + N p (ɛ). The other case is analogous.
Twist 1 Mod l modular forms 2 Residual modular Galois representations 3 Image: an algorithm 4 Local representation 5 Twist Twisting by Dirichlet characters The conductor of a twist The system of eigenvalues of a twist Example Fields of definition 6 Applications
Twist Twisting by Dirichlet characters Assume setting ( ) and that ρ f lower level or weight. is irreducible and it does not arise from Then ρ f can still arise as twist of a representation of lower level or weight, i.e. there exist g S(m, j) Fl with m n or j k and a character χ such that ρ g χ = ρ f. Question Can we compute twists? For this talk, we limit ourselves to study twists of modular Galois representations by Dirichlet characters with prime power conductor coprime with the characteristic.
Twist The conductor of a twist Question What is the conductor of the twist? Shimura gave an upper bound: lcm(cond(χ) 2, n) where n is the level of the form and χ is the Dirichlet character used for twisting.
Twist The conductor of a twist Proposition Let p be a prime not dividing nl. Let χ : (Z /p i Z ) F l, for i > 0, be a non-trivial character. Then N p (ρ f χ) = 2N p (χ).
Twist The conductor of a twist Proposition Assume that ρ f is irreducible and it does not arise from lower level. Let p be a prime dividing n and suppose that f (T p ) 0. Let χ : (Z /p i Z ) F l, for i > 0, be a non-trivial character. Then N p (ρ f χ) = N p (χɛ) + N p (χ).
Twist The system of eigenvalues of a twist It is also possible to know what is the system of eigenvalues associated to the twist: Proposition Suppose that ρ f is irreducible and that N(ρ f ) = n. Let p be a prime dividing n and suppose that f (T p ) 0. Let χ from (Z /p i Z ) to F l, with i > 0, be a non-trivial character. Then (a) if ρ f Ip is decomposable then the representation ρ f χ, restricted to G p, admits a stable line with unramified quotient if and only if N p (ρ f χ) = N p (ρ f ); (b) if ρ f Ip is indecomposable then the representation ρ f χ, restricted to G p, does not admit any stable line with unramified quotient. (a) (ρ f χ)(frob p ) 0; (b) (ρ f χ)(frob p ) = 0;
Twist The system of eigenvalues of a twist Analogous result holds when f (T p ) = 0, where p N(ρ f ): Proposition Suppose that ρ f is irreducible and that N(ρ f ) = n. Let p be a prime dividing n and suppose that f (T p ) = 0. Then: (a) if ρ f Gp is reducible then there exists a mod l modular form g of weight k and level at most np and a non-trivial character χ : (Z /p i Z ) F l with i > 0 such that g(t p ) 0 and ρ g = ρf χ; (b) if ρ f Gp is irreducible then for any non-trivial character χ : (Z /p i Z ) F l with i > 0 the representation ρ f χ restricted to G p does not admit any stable line with unramified quotient.
Twist Example Example: let n = 135 = 3 3 5. Let ɛ be the Dirichlet character modulo 135 of conductor 5 mapping 56 1, 82 ζ 9 36. S(135, 3, ɛ) new C two Galois orbits, the two Hecke eigenvalue fields are: Q (x 16 + 217x 12 + 9264x 8 + 59497x 4 + 28561) and Q (x 16 + 286x 12 + 16269x 8 + 85684x 4 + 62500). We want to compute the image of the mod 7 representations attached to the reduction of the forms in the first Galois orbit. Applying the reduction map for all prime ideals above 7, we obtain the following eigenvalue systems defined over F 7 [x]/[x 2 + 6x + 4] = F 7 [α]:
Twist Example p f 1 (T p ) f 2 (T p ) f 3 (T p ) 2 α 6α α + 6 3 0 0 0 5 5α + 4 2α + 3 5α + 5 7 0 0 0 11 α + 3 6α + 4 α + 3 13 6α 6α α + 6 17 6α α 6α + 1 19 6α + 4 6α + 4 α + 3 23 3α + 4 4α + 3 3α f 1,f 2 and f 3 are mod 7 modular forms of level 135 and weight 3. It is easy to verify that the corresponding representations are of minimal level and weight.
Twist Example Let us focus on f 1. p f 1 (T p ) f 2 (T p ) f 3 (T p ) 2 α 6α α + 6 3 0 0 0 5 5α + 4 2α + 3 5α + 5 7 0 0 0 11 α + 3 6α + 4 α + 3 13 6α 6α α + 6 17 6α α 6α + 1 19 6α + 4 6α + 4 α + 3 23 3α + 4 4α + 3 3α
Twist Example Since 5 divides the level and f 1 (T 5 ) 0: ( ) ( ρ f1 G5 ɛ1 χ 2 = 7 ɛ 1 2 = ɛ ) f 1 χ 2 7 0 0 ɛ 2 0 ɛ 2 The character ɛ f1 attached to f 1 has conductor 5, hence the representation is reducible and decomposable, so = 0. Moreover ɛ 2 (5) = f 1 (T 5 ) = 5α + 4. If we twist by ɛ 1 f 1 then we have: ( ) (ρ f1 ɛ 1 f 1 ) G5 ɛ 1 = 2 χ2 7 0 0 ɛ 2 ɛ 1 f 1 The conductor of the twist is 135, the eigenvalues at primes p 5 are given by f (T p )ɛ 1 f 1 (p) while the eigenvalue at 5 is (ɛ 1 2 χ2 7)(5) = 4ɛ 1 2 (5) = 5α + 5.
Twist Example p f 1 (T p ) f 2 (T p ) f 3 (T p ) ρ f1 ɛ 1 f 1 2 α 6α α + 6 α + 6 3 0 0 0 0 5 5α + 4 2α + 3 5α + 5 5α + 5 7 0 0 0 0 11 α + 3 6α + 4 α + 3 α + 3 13 6α 6α α + 6 α + 6 17 6α α 6α + 1 6α + 1 19 6α + 4 6α + 4 α + 3 α + 3 23 3α + 4 4α + 3 3α 3α 29 2 5 5 5 31 4 4 4 4 37 α + 6 α + 6 6α 6α 41 5α + 1 2α + 6 5α + 1 5α + 1 ρ f1 ɛ 1 f 1 = ρf3
Twist Example The level is also divisible by 3. But f 1 (T 3 ) = 0. The local representation at 3 is irreducible: to prove this claim we have to check all lover levels and possible twist. In this case it is easy, since the newforms space is empty in most cases. The argument is similar for all levels, let us what happens for level 15. p f 1 (T p ) g 1 (T p ) g 2 (T p ) g 3 (T p ) g 4 (T p ) 2 α 6α + 6 α + 5 4α + 1 3α + 5 3 0 6α α + 6 α 6α + 1 5 5α + 4 3α + 5 4α + 1 2α + 1 5α + 3 7 0 4α + 5 3α + 2 2 2 11 α + 3 6α + 1 α α 6α + 1 13 6α 4α + 5 3α + 2 5 5
Twist Example It is possible to show that Pρ f1 (G Q ) = PSL 2 (F 7 ) PGL 2 (F 7 2), and that the image of the Galois representation up to conjugation is ρ f1 (G Q ) =< α > SL 2 (F 7 ) GL 2 (F 7 2)
Twist Fields of definition An application of the study of twists is the computation of the field of definition of the representation and of the projective representation associated.
Twist Fields of definition Field of definition of the representation: Proposition Suppose that ρ f is irreducible and does not arise from lower level or weight. Then the field of definition of ρ f is the smallest extension of F l containing the elements of the set S := {f (T p ) p prime, p l and p B(n, k)} {f ( d ) d (Z /nz ) }, where B(n, k) is the Sturm bound for cusp forms for Γ 0 (n).
Twist Fields of definition Field of definition of the projective representation Pρ f Proposition Let us suppose that ρ f is irreducible, it does not arise from lower level or weight and it is realized over F. Then the field of definition of Pρ f is the subfield of F fixed by A: = { σ Aut(F) τ : Gal(Q /Q ) F : ρ σ f = ρ f τ }.
Twist Fields of definition Proposition Assume that ρ f is irreducible, does not arise from lower level or weight and is realized over F. Then the field of definition of Pρ f is the smallest extension of F l containing for all p prime, p nl. (f (T p )) 2 /f ( p )p k 1 This proposition is not effective: there is no bound for the number of traces needed. Anyway, combining it with the previous, there is an algorithm to compute the Dickson s field.
Applications 1 Mod l modular forms 2 Residual modular Galois representations 3 Image: an algorithm 4 Local representation 5 Twist 6 Applications Diophantine equations Graphs and modularity lifting
Applications Diophantine equations Proposition Let p 5 be a prime and r 1 be an integer. Then there are no solutions to 17 r x p + 2y p = z 2 for non-zero coprime integers x, y, z. Proof If xy = ±1, then there is no solution (reduce modulo 8). If xy ±1, then, without loss of generality, we can assume that 17 r x, 2y and z are coprime, 1 r < p, and xy is odd. The solutions are then related to the Frey curve E : Y 2 + XY = X 3 + 2zX 2 + 2y p X which has minimal discriminant 2 8 17 r (xy 2 ) p and conductor 2 7 Rad(17xy).
Applications Diophantine equations Proof. It is possible to show that: E does not have complex multiplication E arises mod l from some newform f S(2 7 17, 2) for primes l dividing a constant given by an explicit recipe by Mazur. For example l = 3 works. There are 16 Galois orbits of newforms of level 2 7 17 weight 2 which all are irrational. The mod 3 representation cannot arise from a lower level: this would contradict Ribet s Level-Lowering Theorem. It is irreducible because the Frey curve does not admit 3 isogenies. The field of definitions of the mod 3 Galois representations associated to each of these newforms is larger than F 3 : for all the newforms the reduction of f (T 5 ) / F 3. Contradiction.
Applications Graphs and modularity lifting Work on progress, joint with Vandita Patel (University of Warwick). Idea: build graphs of congruences between modular forms and enrich these graphs using datas coming form residual representations. Vertices: newforms of levels and weights in a given set; Edges: an eadge between two verices is drawn if a relation R holds.
Applications Graphs and modularity lifting What kind of relations we are going to consider? R 1 : congruence modulo some prime l; R 2 : isomorphic mod l Galois representations for some prime l; R 3 : isomorphic mod l projective Galois representations for some prime l. until now we implemented an algorithm which draws graphs for R 1.
Applications Graphs and modularity lifting
Applications Graphs and modularity lifting
Applications Graphs and modularity lifting
Applications Graphs and modularity lifting
Applications Graphs and modularity lifting
Applications Graphs and modularity lifting
Applications Graphs and modularity lifting Residual modular Galois representations: images and applications Samuele Anni University of Warwick London Number Theory Seminar King s College London, 20 th May 2015 Thanks!