Residual modular Galois representations: images and ... · PDF fileResidual modular Galois...

Residual modular Galois representations

Residual modular Galoisrepresentations: images and

applications

Samuele Anni

University of Warwick

London Number Theory Seminar

King’s College London, 20th May 2015


Mod ` modular forms

1 Mod ` modular forms

2 Residual modular Galois representations

3 Image: an algorithm

4 Local representation

5 Twist

6 Applications


Mod ` modular forms

Let n be a positive integer. The congruence subgroup Γ1(n) of SL2(Z )is the subgroup given by

Γ1(n) =

{(a bc d

)∈ SL2(Z ) : n | a−1 , n | c

}.

Γ1(n) acts on the complex upper half plane via fractional transformations.

Definition

We define the modular curve Y1(n)C to be the non-compact Riemannsurface obtained giving on Γ1(n)\H the complex structure induced by thequotient map. Let X1(n)C be the compactification of Y1(n)C.

The modular curve Y1(n)C can be defined algebraically over Z [1/n].


Mod ` modular forms

Let n ≥ 5 and k be positive integers and let ` be a prime not dividing n.Following Katz, we define the space of mod ` cusp forms as

mod ` cusp forms

S(n, k)F`= H0(X1(n)F`

, ω⊗k(−Cusps)).

S(n, k)F`is a finite dimensional F`-vector space, equipped with Hecke

operators Tn (n ≥ 1) and diamond operators 〈d〉 for every d∈(Z /nZ )∗.

The Z -subalgebra of EndC(S(Γ1(n), k)C) generated by Tp and diamondoperators 〈d〉 is the Hecke algebra T(n, k).

Analogous definition in characteristic zero and over any ring where n isinvertible.


Mod ` modular forms

One may think that mod ` modular forms come from reduction ofcharacteristic zero modular forms mod `:

S(n, k)Z [1/n] → S(n, k)F`.

Unfortunately, this map is not surjective for k = 1.

Even worse: let ε : (Z /nZ )∗ → C∗ be a character, whose image iscontained in OK , ring of integers of the number field K , then the map

S(n, k , ε)OK→ S(n, k , ε)F

is not always surjective even if k > 1, where F is the residue field at `and ε is the reduction of ε.

S(n, k, ε)OK: = {f ∈ S(n, k)OK

| ∀d ∈ (Z /nZ )∗, 〈d〉 f = ε(d)f }.

Tε(n, k) will denote the associated Hecke algebra.







5 Twist

6 Applications



Theorem (Deligne, Serre, Shimura)

Let n and k be positive integers. Let F be a finite field of characteristic`, with ` not dividing n, and f : T(n, k) � F a surjective morphism ofrings. Then there is a continuous semi-simple representation:

ρf : Gal(Q /Q )→ GL2(F),

unramified outside n`, such that for all p not dividing n` we have:

Trace(ρf (Frobp)) = f (Tp) and det(ρf (Frobp)) = f (〈p〉)pk−1 in F.

Such a ρf is unique up to isomorphism.



Computing ρf is “difficult”, but theoretically it can be done inpolynomial time in n, k,#F:

Edixhoven, Couveignes, de Jong, Merkl, Bruin, Bosman (#F ≤ 32):

Example: for n = 1, k = 22 and ` = 23, the number field correspondingto Pρf (Galois group isomorphic to PGL2(F23)) is given by:

x24 − 2x23 + 115x22 + 23x21 + 1909x20 + 22218x19 + 9223x18 + 121141x17

+ 1837654x16 − 800032x15 + 9856374x14 + 52362168x13 − 32040725x12

+ 279370098x11 + 1464085056x10 + 1129229689x9 + 3299556862x8

+ 14586202192x7 + 29414918270x6 + 45332850431x5 − 6437110763x4

− 111429920358x3 − 12449542097x2 + 93960798341x − 31890957224

Mascot, Zeng, Tian (#F ≤ 41).



Question

Can we compute the image of a residual modular Galois representationwithout computing the representation?


Image: an algorithm



3 Image: an algorithmFinite subgroup of PGL2(F`)Serre’s ConjectureThe algorithm


5 Twist

6 Applications


Image: an algorithm

Main ingredients:

1 Classification of possible images

2 Serre’s conjecture


Image: an algorithm

Finite subgroup of PGL2(F`)

1 Classification of possible images

Theorem (Dickson)

Let ` be an odd prime and H a finite subgroup of PGL2(F`). Then aconjugate of H is one of the following groups:

a finite subgroup of the upper triangular matrices;

PSL2(F`r ) or PGL2(F`r ) for r ∈ Z>0;

a dihedral group D2n with n ∈ Z>1, (`, n) = 1;

or it is isomorphic to A4, S4 or A5.

Definition

If G := ρf (Gal(Q /Q )) has order prime to ` we call the imageexceptional.


Image: an algorithm

Finite subgroup of PGL2(F`)

The field of definition of the representation is the smallest field F ⊂ F`over which ρf is equivalent to all its conjugate. The image of therepresentation ρf is then a subgroup of GL2(F).

Let Pρf : Gal(Q /Q )→ PGL2(F) be the projective representationassociated to the representation ρf :

Gal(Q /Q )ρf //

Pρf &&

GL2(F)

π��

PGL2(F).

The representation Pρf can be defined on a different field than the fieldof definition of the representation. This field is called the Dickson’s fieldfor the representation.


Image: an algorithm

Serre’s Conjecture

2 Serre’s conjecture

Theorem (Khare, Wintenberger, Dieulefait, Kisin),Serre’s Conjecture

Let ` be a prime number and let ρ : Gal(Q /Q )→ GL2(F`) be an odd,absolutely irreducible, continuous representation. Then ρ is modular oflevel N(ρ), weight k(ρ) and character ε(ρ).

N(ρ) (the level) is the Artin conductor away from `.

k(ρ) (the weight) is given by a recipe in terms of ρ|I` .

ε(ρ) : (Z /N(ρ)Z )∗ → F∗` is given by:

det ρ = ε(ρ)χk(ρ)−1` ,

where χ` is the cyclotomic character mod `.


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;

` a prime not dividing n and such that 2 ≤ k ≤ `+ 1;

a character ε : (Z /nZ )∗ → C∗;a morphism of ring f : Tε(n, k)→ F`, i.e. the images of all diamondoperators and of the Tp operators up to a bound B(n, k),

and gives as output the image of the associated Galois representation ρf ,up to conjugacy as subgroup of GL2(F`) without computing ρf .


Image: an algorithm

The algorithm

In almost all cases, the bound B(n, k) is the Sturm Bound for Γ0(n) andweight k :

B(n, k) =k

12· n ·

∏p|n prime

(1 +

1

p

)� k

12· n log log n

In the cases when this bound is not enough, then the Sturm Bound forΓ0(nq2) 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

ρf can arise from lower level or weight:there exists g ∈ S(m, j)F`

with m ≤ n orj ≤ k such that

ρg ∼= ρf

ρf can arise as twist of a representation oflower level or weight: there exist acharacter χ and g ∈ S(m, j)F`

with m ≤ nor j ≤ k such that

ρg ⊗ χ ∼= ρf

Algorithm

Iteration “down to top”,i.e. considering alldivisors of n: creation ofa database

Determine minimalitywith respect to level andwith respect to weight.

Determine minimalityup to twisting.


Image: an algorithm

The algorithm

Algorithm

Step 1 Iteration “down totop”

Step 2 Determine minimalitywith respect to level andweight.

Step 3 Determine whetherreducible or irreducible.

Step 4 Determine minimalityup to twisting.

Step 5 Compute theprojective image

Step 6 Compute the image

Remarks

Reducibility is checked by comparingwith systems of eigenvalues comingfrom specific Eisenstein series.

Compute the field of definition of therepresentation: this can be done usingcoefficients up to a finite bound.

Compute the Dickson’s field: this isobtained twisting.

The projective image is determined byexcluding cases. Each case is relatedto an explicit equality of mod `modular forms or construction.


Image: an algorithm

The algorithm

Setting (∗)n and k positive integers;

` a prime not dividing n and such that 2 ≤ k ≤ `+ 1;

ε : (Z /nZ )∗ → C∗ a character;

f : Tε(n, k)→ F` a morphism of rings;

ρf : GQ → GL2(F`) the unique, up to isomorphism, continuoussemi-simple representation attached to f ;

ε : (Z /nZ )∗ → F∗` be the character defined by det(ρf ) = εχk−1` .

Let p be a prime dividing n`. Let us denote by

Gp = Gal(Q p/Q p) ⊂ GQ the decomposition subgroup at p;

Ip the inertia subgroup;

Gi,p, with i ∈ Z>0, the higher ramification subgroups;

Np(ρ) the valuation at p of the Artin conductor of ρ, where ρ is aresidual representation.


Local representation




4 Local representationLocal representation and conductor

5 Twist

6 Applications



Theorem (Gross-Vigneras-Fontaine, Serre:Conjecture 3.2.6?)

Let ρ : GQ → GL(V ) be a continuous, odd, irreducible representation ofthe absolute Galois group over Q to a 2-dimensional F`-vector space V .Let f ∈ S(N(ρ), k(ρ))F`

be an eigenform such that ρf ∼= ρ.Let p be a prime divisor of `n.

(1) If f (Tp) 6= 0, then there exists a stable line D ⊂ V for the action ofGp, such that Ip acts trivially on V /D. Moreover, f (Tp) is equal tothe eigenvalue of Frobp which acts on V /D.

(2) If f (Tp) = 0, then there exists no stable line D ⊂ V as in (1).

(1) ⇒ ρf |Gp is reducible;

(2) ⇒ ρf |Gp is irreducible.



Local representation and conductor

From now on we will assume setting (∗).

Proposition

Let us suppose that ρf is irreducible and it does not arise from lowerlevel. Let p be a prime dividing n such that f (Tp) 6= 0.Then ρf |Gp is decomposable if and only if ρf |Ip is decomposable.

This proposition is proved using representation theory.

Can we deduce a computational criterion from this?




Proposition

Assume that ρf is irreducible and it does not arise from lower level. Let pbe a prime dividing n, such that f (Tp) 6= 0. Then:

(a) ρf |Ip is decomposable if and only if Np(ρf ) = Np(ε);

(b) ρf |Ip is indecomposable if and only if Np(ρf ) = 1 + Np(ε).




Sketch of the proof

The valuation of N(ρf ) at p is given by:

Np(ρf ) =∑i≥0

1

[G0,p : Gi,p]dim(V /V Gi,p ) = dim(V /V Ip ) + b(V ),

where V is the two-dimensional F`-vector space underlying therepresentation, V Gi,p is the subspace of invariants under Gi,p, and b(V ) isthe wild part of the conductor.Since f (Tp) 6= 0, the representation restricted to the decompositiongroup at p is reducible. Hence, after conjugation,

ρf |Gp∼=(ε1χ

k−1` ∗

0 ε2

), ρf |Ip ∼=

(ε1|Ip ∗

0 1

)where ε1 and ε2 are characters of Gp with ε2 unramified, χ` is the mod `cyclotomic character and ∗ belongs to F`.




Sketch of the proof

ρf |Ip ∼=(ε1|Ip ∗

0 1

).

If ρf |Ip is indecomposable (∗ 6= 0) then V Ip is either {0} if ε1 is ramified,

or F` · ( 10 ) if ε1 is unramified. The wild part of the conductor is equal to

the wild part of the conductor of ε1. Hence, we have that

Np(ρf ) =

{1 = 1 + Np(ε1) if ε1 is unramified,

2 + b(ε1) = 1 + Np(ε1) if ε1 is ramified.

Since det(ρf ) = εχk−1` , then det(ρf )|Ip = ε|Ip , so ε1|Ip = ε|Ip . Therefore,

we have that if ρf |Ip is indecomposable then Np(ρf ) = 1 + Np(ε).

The other case is analogous.


Twist





5 TwistTwisting by Dirichlet charactersThe conductor of a twistThe system of eigenvalues of a twistExampleFields of definition

6 Applications


Twist

Twisting by Dirichlet characters

Assume setting (∗) and that ρf is irreducible and it does not arise fromlower level or weight.

Then ρf can still arise as twist of a representation of lower level orweight, i.e. there exist g ∈ S(m, j)F`

with m ≤ n or j ≤ k and acharacter χ such that ρg ⊗ χ ∼= ρf .

Question

Can we compute twists?

For this talk, we limit ourselves to study twists of modular Galoisrepresentations by Dirichlet characters with prime power conductorcoprime 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 fortwisting.


Twist


Proposition

Let p be a prime not dividing n`. Let χ : (Z /piZ )∗ → F∗` , for i > 0, bea non-trivial character. Then

Np(ρf ⊗ χ) = 2Np(χ).


Twist


Proposition

Assume that ρf is irreducible and it does not arise from lower level.

Let p be a prime dividing n and suppose that f (Tp) 6= 0 .

Let χ : (Z /piZ )∗ → F∗` , for i > 0, be a non-trivial character. Then

Np(ρf ⊗ χ) = Np(χε) + Np(χ).


Twist

The system of eigenvalues of a twist

It is also possible to know what is the system of eigenvalues associatedto the twist:

Proposition

Suppose that ρf is irreducible and that N(ρf ) = n. Let p be a prime

dividing n and suppose that f (Tp) 6= 0 . Let χ from (Z /piZ )∗ to F∗` ,

with i > 0, be a non-trivial character. Then

(a) if ρf |Ip is decomposable then the representation ρf ⊗ χ, restricted toGp, admits a stable line with unramified quotient if and only ifNp(ρf ⊗ χ) = Np(ρf );

(b) if ρf |Ip is indecomposable then the representation ρf ⊗ χ, restrictedto Gp, does not admit any stable line with unramified quotient.

(a) ⇒ (ρf ⊗ χ)(Frobp) 6= 0;

(b) ⇒ (ρf ⊗ χ)(Frobp) = 0;


Twist

The system of eigenvalues of a twist

Analogous result holds when f (Tp) = 0, where p | N(ρf ):

Proposition

Suppose that ρf is irreducible and that N(ρf ) = n. Let p be a prime

dividing n and suppose that f (Tp) = 0 . Then:

(a) if ρf |Gp is reducible then there exists a mod ` modular form g ofweight k and level at most np and a non-trivial characterχ : (Z /piZ )∗ → F∗` with i > 0 such that g(Tp) 6= 0 andρg ∼= ρf ⊗ χ;

(b) if ρf |Gp is irreducible then for any non-trivial character

χ : (Z /piZ )∗ → F∗` with i > 0 the representation ρf ⊗ χ restrictedto Gp does not admit any stable line with unramified quotient.


Twist

Example

Example: let n = 135 = 335. Let ε be the Dirichlet character modulo135 of conductor 5 mapping 56→ 1, 82→ ζ936.

S(135, 3, ε)newC two Galois orbits, the two Hecke eigenvalue fields are:Q (x16 + 217x12 + 9264x8 + 59497x4 + 28561) andQ (x16 + 286x12 + 16269x8 + 85684x4 + 62500).

We want to compute the image of the mod 7 representations attached tothe reduction of the forms in the first Galois orbit.

Applying the reduction map for all prime ideals above 7, we obtain thefollowing eigenvalue systems defined over F7[x ]/[x2 + 6x + 4] ∼= F7[α]:


Twist

Example

p f1(Tp) f2(Tp) f3(Tp)2 α 6α α + 63 0 0 05 5α + 4 2α + 3 5α + 57 0 0 011 α + 3 6α + 4 α + 313 6α 6α α + 617 6α α 6α + 119 6α + 4 6α + 4 α + 323 3α + 4 4α + 3 3α

f1,f2 and f3 are mod 7 modular forms of level 135 and weight 3.

It is easy to verify that the corresponding representations are of minimallevel and weight.


Twist

Example

Let us focus on f1.

p f1(Tp) f2(Tp) f3(Tp)2 α 6α α + 63 0 0 05 5α + 4 2α + 3 5α + 57 0 0 011 α + 3 6α + 4 α + 313 6α 6α α + 617 6α α 6α + 119 6α + 4 6α + 4 α + 323 3α + 4 4α + 3 3α


Twist

Example

Since 5 divides the level and f1(T5) 6= 0:

ρf1 |G5∼=(ε1χ

27 ∗

0 ε2

)∼=(ε−12 εf1χ

27 0

0 ε2

)The character εf1 attached to f1 has conductor 5, hence therepresentation is reducible and decomposable, so ∗ = 0.Moreover ε2(5) = f1(T5) = 5α + 4.

If we twist by ε−1f1then we have:

(ρf1 ⊗ ε−1f1)|G5∼=(ε−12 χ2

7 00 ε2ε

−1f1

)The conductor of the twist is 135, the eigenvalues at primes p 6= 5 aregiven by f (Tp)ε−1f1

(p) while the eigenvalue at 5 is

(ε−12 χ27)(5) = 4ε−12 (5) = 5α + 5.


Twist

Example

p f1(Tp) f2(Tp) f3(Tp) ρf1 ⊗ ε−1f12 α 6α α + 6 α + 63 0 0 0 05 5α + 4 2α + 3 5α + 5 5α + 57 0 0 0 011 α + 3 6α + 4 α + 3 α + 313 6α 6α α + 6 α + 617 6α α 6α + 1 6α + 119 6α + 4 6α + 4 α + 3 α + 323 3α + 4 4α + 3 3α 3α29 2 5 5 531 4 4 4 437 α + 6 α + 6 6α 6α41 5α + 1 2α + 6 5α + 1 5α + 1

ρf1 ⊗ ε−1f1∼= ρf3


Twist

Example

The level is also divisible by 3. But f1(T3) = 0.

The local representation at 3 is irreducible: to prove this claim we haveto 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 f1(Tp) g1(Tp) g2(Tp) g3(Tp) g4(Tp)2 α 6α + 6 α + 5 4α + 1 3α + 53 0 6α α + 6 α 6α + 15 5α + 4 3α + 5 4α + 1 2α + 1 5α + 37 0 4α + 5 3α + 2 2 211 α + 3 6α + 1 α α 6α + 113 6α 4α + 5 3α + 2 5 5


Twist

Example

It is possible to show that

Pρf1(GQ ) ∼= PSL2(F7) ⊂ PGL2(F72),

and that the image of the Galois representation up to conjugation is

ρf1(GQ ) ∼=< α > SL2(F7) ⊂ GL2(F72)


Twist

Fields of definition

An application of the study of twists is the computation of the field ofdefinition of the representation and of the projective representationassociated.


Twist


Field of definition of the representation:

Proposition

Suppose that ρf is irreducible and does not arise from lower level orweight. Then the field of definition of ρf is the smallest extension of F`containing the elements of the set

S := {f (Tp)| p prime, p 6= ` and p ≤ B(n, k)}∪{f (〈d〉)| d ∈ (Z /nZ )∗} ,

where B(n, k) is the Sturm bound for cusp forms for Γ0(n).


Twist


Field of definition of the projective representation Pρf

Proposition

Let us suppose that ρf is irreducible, it does not arise from lower level orweight and it is realized over F. Then the field of definition of Pρf is thesubfield of F fixed by

A : ={σ ∈ Aut(F)| ∃τ : Gal(Q /Q )→ F∗ : ρσf

∼= ρf ⊗ τ}.


Twist


Proposition

Assume that ρf is irreducible, does not arise from lower level or weightand is realized over F. Then the field of definition of Pρf is the smallestextension of F` containing

(f (Tp))2/f (〈p〉)pk−1

for all p prime, p - n`.

This proposition is not effective: there is no bound for the number oftraces needed. Anyway, combining it with the previous, there is analgorithm to compute the Dickson’s field.


Applications





5 Twist

6 ApplicationsDiophantine equationsGraphs and modularity lifting


Applications

Diophantine equations

Proposition

Let p ≥ 5 be a prime and r ≥ 1 be an integer. Then there are nosolutions to

17rxp + 2yp = z2

for non-zero coprime integers x , y , z.

Proof

If xy = ±1, then there is no solution (reduce modulo 8).If xy 6= ±1, then, without loss of generality, we can assume that 17rx , 2yand z are coprime, 1 ≤ r < p, and xy is odd. The solutions are thenrelated to the Frey curve

E : Y 2 + XY = X 3 + 2zX 2 + 2ypX

which has minimal discriminant 2817r (xy2)p and conductor 27Rad(17xy).


Applications

Diophantine equations

Proof.

It is possible to show that:

E does not have complex multiplication

E arises mod ` from some newform f ∈ S(2717, 2) for primes `dividing a constant given by an explicit recipe by Mazur. Forexample ` = 3 works.

There are 16 Galois orbits of newforms of level 2717 weight 2 which allare irrational.

The mod 3 representation cannot arise from a lower level: this wouldcontradict Ribet’s Level-Lowering Theorem. It is irreducible because theFrey curve does not admit 3 isogenies.

The field of definitions of the mod 3 Galois representations associated toeach of these newforms is larger than F3: for all the newforms thereduction of f (T5) /∈ F3. 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 enrichthese 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


What kind of relations we are going to consider?

R1: congruence modulo some prime `;

R2: isomorphic mod ` Galois representations for some prime `;

R3: isomorphic mod ` projective Galois representations for someprime `.

until now we implemented an algorithm which draws graphs for R1.


Applications



Applications


Residual modular Galoisrepresentations: images and

applications

Samuele Anni

University of Warwick

London Number Theory Seminar

King’s College London, 20th May 2015

Thanks!

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