numerical evidence for darmon points

Numerical Evidence for Darmon PointsMetodes Efectius en Geometria Algebraica

Xavier Guitart 1 Marc Masdeu 2

1Universitat Politecnica de Catalunya

2Columbia University

June 4, 2013

Marc Masdeu Numerical Evidence for Darmon Points June 4, 2013 1 / 28

The Problem

ProblemGiven an algebraic curve C defined over Q, find points onC, defined on prescribed algebraic extensions K of Q.


Mordell-Weil

Louis Mordell Andre Weil


Basic Setup

E = elliptic curve defined over Q, and K = Q(√dK).

I If dK > 0 then K is called real quadratic, andI If dK < 0 then K is called imaginary quadratic.

Theorem (Mordell–Weil)

E(K) = (torsion)⊕ Zrkalg(E,K)

The algebraic rank rkalg(E,K) is hard to determine.Attach to E and K an L-function L(E/K, s) as follows.

I Let N = conductor(E).I If p is a rational prime, ap(E) = 1 + p−#E(Fp).I If v is a closed point of SpecOK , |v| = size of the residue field κ(v).

L(E/K, s) =∏v|N

(1− a|v||v|−s

)−1 ×∏v-N

(1− a|v||v|−s + |v|1−2s

)−1Modularity (Wiles et al) =⇒ analytic continuation of L(E/K, s) to C.


Birch and Swinnerton-Dyer

Brian Birch Peter Swinnerton-Dyer


Birch and Swinnerton-Dyer

Conjecture (BSD, rough version)

ords=1 L(E/K, s) = rkalg(E,K).

The left-hand side is the analytic rank, rkan(E,K).

Consequencerkan(E,K) ≥ 1 =⇒ ∃PK ∈ E(K) of infinite order.

FactDefine S(K,N) = ` | N : ` inert in K.Then rkan(E,K) is ≥ 1 when:

1 K is imaginary quadratic and #S(K,N) is even, or2 K is real quadratic and #S(K,N) is odd.


Heegner vs Darmon (points)

Kurt Heegner Henri Darmon



FactThe analytic rank rkan(E,K) is ≥ 1 when:

1 K is imaginary quadratic and #S(K,N) is even, or

2 K is real quadratic and #S(K,N) is odd.

In case 1 , one can construct Heegner points on E(K).Attached to elements τ ∈ K ∩H.Gross–Zagier formula =⇒ they have infinite order if rkan(E,K) = 1.If S(K,N) = ∅, they can be obtained by C-analytic methods.If S(K,N) 6= ∅, can use p-adic methods.



FactThe analytic rank rkan(E,K) is ≥ 1 when:

1 K is imaginary quadratic and #S(K,N) is even, or

2 K is real quadratic and #S(K,N) is odd.

In case 2 , Heegner points are not available.I Note that in this case, K ∩H = ∅!

In 2001, Darmon gave a construction (a.k.a “Stark-Heegner”).I p-adic analytic =⇒ a priori they lie in E(Kp).I Proof of their algebraicity completely open so far.I Darmon’s construction restricted to S(K,N) = p.I Greenberg extended to S(K,N) of arbitrary odd size.I We call them Darmon points.


Goals

In this talk we will:1 Explain what Darmon Points are,2 Explain how to calculate them, and

“ The fun of the subject seems to me to be in the examples.B. Gross, in a letter to B. Birch, 1982”

3 See some fun examples!


Integration on the p-adic upper-half plane

DefinitionThe p-adic upper half plane is the rigid-analytic space

Hp = P1(Cp) \ P1(Qp) ( = Cp \Qp ).

This is the p-adic analogue of H± = C \ R.Ω1Hp = space of rigid-analytic one-forms on Hp.

Coleman integral: allows to make sense of∫ τ2

τ1

ω, for τ1, τ2 ∈ Hp and ω ∈ Ω1Hp .

If the residues of ω are all integers, have a multiplicative refinement:

×∫ τ2

τ1

ω = lim−→U

∏U∈U

(tU − τ2tU − τ1

)µ(U)

where µ(U) = resA(U) ω.


The p-adic upper half plane

×∫ τ2

τ1

ω = lim−→U

∏U∈U

(tU − τ2tU − τ1

)µ(U)

where µ(U) = resA(U) ω.

Bruhat-Tits tree ofGL2(Qp) with p = 2.Hp having theBruhat-Tits as retract.Annuli A(U) for U acovering of size p−3.tU is any point inU ⊂ P1(Qp).


Darmon points a la Greenberg

Henri DarmonMatthew Greenberg


Darmon points a la Greenberg

Assumption1 K is real quadratic.2 S(K,N) = ` : ` | pD has odd cardinality.

Write N = pDM , and B/Q = quaternion algebra of discriminant D.Fix ιp : B →M2(Qp).

I B× → GL2(Qp) acts on Hp by:(a bc d

)τ =

aτ + b

cτ + d.

R an Eichler order of level M (and such that ιp(R) is “nice”

ιp(R) ⊂(

a bc d

)∈M2(Zp) : vp(c) ≥ 1

.

).

Γ =(R[1p ]

)×1

(elements in R[1p ] of reduced norm 1) → SL2(Qp).

If D = 1, then B = M2(Q) and Γ =(

a bc d

)∈ SL2(Z[1/p]) : M | c

.


Cohomology

Let Ω1Hp,Z = rigid-analytic differentials having integral residues.

Can attach to E a unique (up to sign) class [ΦE ] ∈ H1(

Γ,Ω1Hp,Z

).

Uses Hecke action and Jacquet–Langlands correspondence.

Theorem (M. Greenberg)

There exists a unique class [ΦE ] ∈ H1(Γ,Ω1Hp,Z) satisfying:

1 T`[ΦE ] = a`[ΦE ] for all primes ` - pDM ;

2 U`[ΦE ] = a`[ΦE ] for all ` | DM ;

3 Wp[ΦE ] = ap[ΦE ] (Atkin-Lehner involution);

4 W∞[ΦE ] = [ΦE ] (Involution at∞).

Hecke action can be made explicitI Suitable for computation.


Homology

Start with an embedding ψ : K → B.

ψ induces an action of K× on Hp via ιp.1 Let τψ ∈ Hp be the unique fixed point of K×

2 Set γψ = ψ(ε2), where O×K = ±1 × 〈ε〉.

ψ ; Θψ = [γψ ⊗ τψ] ∈ H1(Γ,DivHp).

Key exact sequence:

H1(Γ,Div0Hp) // H1(Γ,DivHp)deg // H1(Γ,Z)

Θψ ? // Θψ

// torsion

Challenge: pull

a multiple of

Θψ back to Θψ ∈ H1(Γ,Div0Hp).I Requires algorithms adapted to the nature of Γ.


Conjecture

Recall our data1 Θψ = [

∑γ γ ⊗Dγ ] ∈ H1(Γ,Div0Hp)

2 ΦE ∈ Z1(Γ,Ω1Hp,Z)

Jψ =∏γ

×∫Dγ

(ΦE)γ ∈ K×p .

Jψ is well defined modulo powers of the Tate parameter qE

Conjecture (Darmon (D = 1), Greenberg (D > 1))1 Pψ = ΨTate(Jψ) belongs to E(Kab)2 If rkan(E,K) = 1, then Pψ has infinite order.


Numerical Evidence

Conjecture (Darmon (D = 1), Greenberg (D > 1))1 Pψ = ΨTate(Jψ) belongs to E(Kab)

2 If rkan(E,K) = 1, then Pψ has infinite order.

Numerical EvidenceM = 1 M > 1

D = 1 Darmon–Green (2002) Guitart–M. (2012)(B = M2(Q)) Darmon–Pollack (2006)

D > 1 Guitart–M. (2013)


Other directions

E defined over other number field F 6= Q.

1 F totally real: Greenberg (2009).F No numerical evidence.

2 F quadratic imaginary: Trifkovic (2006).F Numerical evidence when F is Euclidean.

3 F arbitrary: Guitart–Sengun–M. (in progress).

Lift to Jac(X0(N)): Rotger-Longo-Vigni (2012).I No numerical evidence.

Higher weight modular forms: Rotger-Seveso (2012).I No numerical evidence.


Calculating the Cycle

Key Exact Sequence

H1(Γ,Div0Hp) // H1(Γ,DivHp)deg // H1(Γ,Z)

Θψ ? // Θψ

// torsion

M = 1 M > 1

D = 1 Adjacent Cusps Elementary Matrix Decomposition

(B = M2(Q)) (Manin Trick) Go Go

D > 1 Commutator Decomposition Go


Commutator decomposition example

G = R×1 , R maximal order on B = B6.

F = 〈X,Y 〉 G = 〈x, y | x2 = y3 = 1〉.

Goal: write g ⊗ τ as∑gi ⊗Di, with Di of degree 0.

Take for instance g = yxyxy. Note that wt(x) = 2 and wt(y) = 3.We trivialize on Fab: g = yxyxy(x−2)(y−3).To simplify g ⊗ τ0 in H1(Γ,DivHp), use:

1 γ1γ2 ⊗D = γ1 ⊗D + γ2 ⊗γ−11 D.2 γ−1 ⊗D = −γ ⊗γD.

g ⊗ τ0= yxyxyx−2y−3 ⊗ τ0= y ⊗ τ0

+ xyxyx−2y−3 ⊗y−1τ0 = y ⊗ τ0+ xyxyx−2y−3 ⊗ τ1 = y ⊗ τ0 + x⊗ τ1+ yxyx−2y−3 ⊗x−1τ1 = y ⊗ τ0 + x⊗ τ1+ yxyx−2y−3 ⊗ τ2 = y ⊗ τ0 + x⊗ τ1+ y ⊗ τ2 + xyx−2y−3 ⊗y−1τ2 = y ⊗ (τ0 + τ2) + x⊗ τ1+ xyx−2y−3 ⊗ τ3 = y ⊗ (τ0 + τ2) + x⊗ τ1+ x⊗ τ3 + yx−2y−3 ⊗x−1τ3 = y ⊗ (τ0 + τ2) + x⊗ (τ1 + τ3)

+ yx−2y−3 ⊗ τ4 = y ⊗ (τ0 + τ2) + x⊗ (τ1 + τ3)

+ y ⊗ τ4 + x−2y−3 ⊗y−1τ4 = y ⊗ (τ0 + τ2 + τ4) + x⊗ (τ1 + τ3)

+ x−2y−3 ⊗ τ5 = y ⊗ (τ0 + τ2 + τ4) + x⊗ (τ1 + τ3)

+ x−1 ⊗ τ5 + x−1y−3 ⊗xτ5 = y ⊗ (τ0 + τ2 + τ4) + x⊗ (τ1 + τ3)

− x⊗xτ5 + x−1y−3 ⊗xτ5 = y ⊗ (τ0 + τ2 + τ4) + x⊗ (τ1 + τ3)

− x⊗ τ6 + x−1y−3 ⊗ τ6 = y ⊗ (τ0 + τ2 + τ4) + x⊗ (τ1 + τ3 − τ6)+ x−1 ⊗ τ6 + y−3 ⊗xτ6 = y ⊗ (τ0 + τ2 + τ4) + x⊗ (τ1 + τ3 − τ6)− x⊗xτ6 + y−3 ⊗xτ6 = y ⊗ (τ0 + τ2 + τ4) + x⊗ (τ1 + τ3 − τ6)− x⊗ τ7 + y−3 ⊗ τ7 = y ⊗ (τ0 + τ2 + τ4) + x⊗ (τ1 + τ3 − τ6 − τ7)+ y−3 ⊗ τ7 = y ⊗ (τ0 + τ2 + τ4) + x⊗ (τ1 + τ3 − τ6 − τ7)+ y−1 ⊗ τ7 + y−2 ⊗yτ7 = y ⊗ (τ0 + τ2 + τ4) + x⊗ (τ1 + τ3 − τ6 − τ7)− y ⊗yτ7 + y−2 ⊗yτ7 = y ⊗ (τ0 + τ2 + τ4) + x⊗ (τ1 + τ3 − τ6 − τ7)− y ⊗ τ8 + y−2 ⊗ τ8 = y ⊗ (τ0 + τ2 + τ4 − τ8) + x⊗ (τ1 + τ3 − τ6 − τ7)+ y−2 ⊗ τ8 = y ⊗ (τ0 + τ2 + τ4 − τ8) + x⊗ (τ1 + τ3 − τ6 − τ7)+ y−1 ⊗ τ8 + y−1 ⊗yτ8 = y ⊗ (τ0 + τ2 + τ4 − τ8) + x⊗ (τ1 + τ3 − τ6 − τ7)− y ⊗yτ8 + y−1 ⊗yτ8 = y ⊗ (τ0 + τ2 + τ4 − τ8) + x⊗ (τ1 + τ3 − τ6 − τ7)− y ⊗ τ9 + y−1 ⊗ τ9 = y ⊗ (τ0 + τ2 + τ4 − τ8 − τ9) + x⊗ (τ1 + τ3 − τ6 − τ7)+ y−1 ⊗ τ9 = y ⊗ (τ0 + τ2 + τ4 − τ8 − τ9) + x⊗ (τ1 + τ3 − τ6 − τ7)− y ⊗yτ9 = y ⊗ (τ0 + τ2 + τ4 − τ8 − τ9) + x⊗ (τ1 + τ3 − τ6 − τ7)− y ⊗ τ10 = y ⊗ (τ0 + τ2 + τ4 − τ8 − τ9 − τ10) + x⊗ (τ1 + τ3 − τ6 − τ7)


Implementation

We have written SAGE code to do all of the above.

Depend on:

1 Overconvergent method for D = 1 (R.Pollack).

2 Finding a presentation for units of orders in B (J.Voight).

F Currently depends on MAGMA.

Overconvergent methods (adapted to D > 1)I Efficient (polynomial time) integration algorithm.I Apart from checking the conjecture, can use the method to actually

finding the points.


Examples

Please show themthe examples !


p = 5, Curve 15A1

E : y2 + xy + y = x3 + x2 − 10x− 10

dK h P+

13 1(−√

13 + 1, 2√

13− 4)

28 1(−15√

7 + 43, 150√

7− 402)

37 1(−5

9

√37 + 5

9 ,2527

√37− 70

27

)73 1

(−17

32

√73 + 77

32 ,187128

√73− 1199

128

)88 1

(−17

9 ,1427

√22 + 4

9

)97 1

(− 25

121

√97 + 123

121 ,3752662

√97− 4749

2662

)133 1

(1039 ,

9227

√133− 56

9

)172 1

(−1923

1681 ,1178168921

√43 + 121

1681

)193 1

(1885288

√193 + 25885

288 ,2921753456

√193 + 4056815

3456

)


p = 3, Curve 21A1

E : y2 + xy = x3 − 4x− 1

dK h P+

8 1(−9√

2 + 11, 45√

2− 64)

29 1(− 9

25

√29 + 32

25 ,63125

√29− 449

125

)44 1

(− 9

49

√11− 52

49 ,54343

√11 + 557

343

)53 1

(− 37

169

√53 + 184

169 ,5552197

√53− 5633

2197

)92 1

(53346 ,

173252116

√23− 533

92

)137 1

(− 1959

11449

√137 + 242

11449 ,2958092450086

√137− 162481

2450086

)149 1

(− 261

2809

√149 + 2468

2809 ,8091

148877

√149− 101789

148877

)197 1

(− 79135143

209961032

√197 + 977125081

209961032 ,14395473863131075630366936

√197− 9297639417941

537815183468

)D h hD(x)

65 2 x2 +(618516241

√65− 491926

6241

)x− 403782

6241

√65 + 3256777

6241


p = 11, Curve 33A1

E : y2 + xy = x3 + x2 − 11x

dK h P+

13 1(− 1

2

√13 + 3

2, 12

√13− 7

2

)28 1

(227, 5549

√7− 11

7

)61 1

(− 1

2

√61 + 5

2,√61− 11

)73 1

(− 53339

49928

√73 + 324687

49928, 31203315

7888624

√73− 290996167

7888624

)76 1

(−2,√19 + 1

)109 1

(− 143

2

√109 + 1485

2, 5577

2

√109− 58223

2

)172 1

(− 51842

21025, 20651473048625

√43 + 25921

21025

)184 1

(5948821609

, 1092523176523

√46− 29744

21609

)193 1

(94663533349261678412148664608

√193 + 1048806825770477

678412148664608,

14777895792093129931712494688311813553741184

√193 + 30862934493092416035541

12494688311813553741184

)D h hD(x)

40 2 x2 +(

28491681

√10− 6347

1681

)x− 5082

1681

√10 + 16819

1681

85 2 x2 +(

119361

√85− 1022

361

)x− 168

361

√85 + 1549

361

145 4 x4 +(

16901600345383168215321

√145− 1621540207320

83168215321

)x3

+(− 1534717557538

83168215321

√145 + 18972823294799

83168215321

)x2 +

(553340519048983168215321

√145− 66553066916820

83168215321

)x

+− 641491338945683168215321

√145 + 77248348177561

83168215321


p = 13, Curve 78A1

Exclusive for MEGAFirst examples of Quaternionic Darmon Points.

78 = 2 · 3 · 13, we take p = 13 and D = 6.

E : y2 + xy = x3 + x2 − 19x+ 685

dK h P+

5 1(10, 18

√5− 5

)149 1

(− 10654790

1138489 ,12293960701214767763

√149 + 5327395

1138489

)197 1

(964090121 ,− 67449270

1331

√197− 482045

121

)293 1

(666267325 ,−

2318837410731125

√293− 33313

7325

)317 1

(36974414388325 ,− 226154303712

4308465875

√317− 18487207

388325

)437 1

(971110339889 ,−

244302600198155287

√437− 485555

339889

)461 1

(− 146849210

23609881 ,132062657760114720411779

√461 + 73424605

23609881

)509 1

(866260300904613262241 ,−

1193915313019350313337384670961

√509− 43313015045

4613262241

)557 1

(3943121444298862528440578125 , 785894731464535585238494887001960802734375

√557− 197156072214943

2528440578125

)Marc Masdeu Numerical Evidence for Darmon Points June 4, 2013 27 / 28

Thank you !

Find the slides at: http://www.math.columbia.edu/∼masdeu/


Bibliography

Henri Darmon and Peter Green.Elliptic curves and class fields of real quadratic fields: Algorithms and evidence.Exp. Math., 11, No. 1, 37-55, 2002.

Henri Darmon and Robert Pollack.Efficient calculation of Stark-Heegner points via overconvergent modular symbols.Israel J. Math., 153:319–354, 2006.

Xavier Guitart and Marc Masdeu.Elementary matrix Decomposition and the computation of Darmon points with higher conductor.arXiv.org, 1209.4614, 2012.

Matthew Greenberg.Stark-Heegner points and the cohomology of quaternionic Shimura varieties.Duke Math. J., 147(3):541–575, 2009.

David Pollack and Robert Pollack.A construction of rigid analytic cohomology classes for congruence subgroups of SL3(Z).Canad. J. Math., 61(3):674–690, 2009.

trifkovic Mak Trifkovic,Stark-Heegner points on elliptic curves defined over imaginary quadratic fields.Duke Math. J., 135, No. 3, 415-453, 2006.


Differential forms and measures

Theorem (Teitelbaum?)The assignment

µ 7→ ω =

∫P1(Qp)

dz

z − tdµ(t)

induces an isomorphism Meas0(P1(Qp),Cp) ∼= Ω1Hp .

Theorem (Teitelbaum)∫ τ2

τ1

ω =

∫P1(Qp)

log

(t− τ1t− τ2

)dµ(t).

Proof sketch.∫ τ2

τ1

ω =

∫ τ2

τ1

∫P1(Qp)

dz

z − tdµ(t) =

∫P1(Qp)

log

(t− τ2t− τ1

)dµ(t).


Continued Fractions

Theorem (Manin)The cusps∞ and γ∞ can always be connected by a chain of adjacentcusps∞ = c0 ∼ c1 ∼ · · · ∼ cr = γ∞.

Effective version: continued fractions.

(γ∞−∞)⊗ τ =∑i

(gi∞− gi0)⊗ τ

=∑i

(∞− 0)⊗ τi

=∑i

(∞− 1)⊗ τi + (1− 0)⊗ τi τi = g−1i τ

=∑i

(∞− 0)⊗ t−1τi + (0−∞)⊗ tsτi

=∑i

(∞− 0)⊗ (t−1τi − tsτi)

Back Marc Masdeu Numerical Evidence for Darmon Points June 4, 2013 3 / 4

Elementary Matrices

Theorem (Vaserstein (1972), Guitart–M. (2012) (effective version))Any matrix γ ∈ Γ1(M) can be decomposed as

γ = l1u1 · · · lrur, ui =(1 xi0 1

), li =

(1 0yi 1

).

Obtain the following:

(ug∞−∞)⊗ τ = (g∞−∞)⊗u−1τ.

and (lg∞−∞)⊗ τ simplifies to:

= (lg∞− 0)⊗ τ + (0−∞)⊗ τ= (g∞− 0)⊗ l−1τ + (0−∞)⊗ τ= (g∞−∞)⊗ l−1τ + (∞− 0)⊗ l−1τ + (0−∞)⊗ τ= (g∞−∞)⊗ l−1τ + (∞− 0)⊗ (l−1τ − τ).

Iterating this process we reduce to a degree-0 divisor. Back


numerical evidence for darmon points

Science

heegner points

darmon pointsme

points onc

padic analytic

factdefine sk

real quadratic

imaginary quadratic

n of arbitrary odd size