analytic construction of points on modular elliptic curves

Analytic construction of pointson modular elliptic curves

Congreso de Jovenes InvestigadoresUniversidad de Murcia

Xavier Guitart 1 Marc Masdeu 2 Mehmet Haluk Sengun 3

1Universitat de Barcelona

2University of Warwick

3University of Sheffield

September 10, 2015

Marc Masdeu Analytic construction of pointson modular elliptic curves September 10, 2015 0 / 26

Elliptic Curves

Let E be an elliptic curve defined over Q:

Y 2 + a1Y + a3XY = X3 + a2X2 + a4X + a6, ai ∈ Z

Let K/Q be a number field, and consider the abelian group

E(K) = (x, y) ∈ K2 : y2 +a1y+a3xy = x3 +a2x2 +a4x+a6∪O.

Theorem (Mordell–Weil)E(K) is finitely generated: E(K) ∼= (Torsion)⊕ Zr.

The integer r = rkZE(K) is called the algebraic rank of E(K).I Open problem: Given E and K, find r.


The Hasse-Weil L-function

Suppose that K = Q(√D) is quadratic.

Can reduce the coefficients a1, . . . , a6 modulo primes p.I For almost all primes, the reduction is an elliptic curve (nonsingular).I Obviously E(Fp) is finite.

The conductor of E is an integer N encoding the shape of E whenthis reduction is singular.

I Assume that N is square-free, coprime to disc(K/Q).

The L-function of E/K (Re(s) > 3/2)

L(E/K, s) =∏p|N

(1− ap|p|−s

)−1 ×∏p-N

(1

ap(E) = 1 + |p| −#E(Fp).

− ap|p|−s + |p|1−2s)−1.

Modularity (Wiles, Taylor–Wiles, Breuil–Conrad–Diamond–Taylor)=⇒

I Analytic continuation of L(E/K, s) to C.I Functional equation relating s↔ 2− s.


The BSD conjecture

Bryan Birch Sir Peter Swinnerton-Dyer

BSD conjecture (coarse version)

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

So L(E/K, 1) = 0BSD=⇒ ∃PK ∈ E(K) of infinite order.

Open problem: construct such PK .Marc Masdeu Analytic construction of pointson modular elliptic curves September 10, 2015 3 / 26

Modular forms

Let N > 0 be an integer and consider

Γ0(N) = (a bc d

)∈ SL2(Z) : N | c.

Γ0(N) acts on the upper-half plane H = z ∈ C : Im(z) > 0:I Via

(a bc d

)· z = az+b

cz+d .A cusp form of level N is a holomorphic map f : H→ C such that:

1 f(γz) = (cz + d)2f(z) for all γ =(a bc d

)∈ Γ0(N).

2 Cuspidal: limz→i∞ f(z) = 0.Since ( 1 1

0 1 ) ∈ Γ0(N), have Fourier expansions

f(z) =∞∑n=1

an(f)e2πinz.

The (finite) vector space of all cusp forms is denoted by S2(Γ0(N)).There is a family of commuting linear operators (Hecke algebra)acting on S2(Γ0(N)), indexed by integers coprime to N .

I A newform is a simultaneous eigenvector for the Hecke algebra. (. . . )


Modularity

Theorem (Modularity, automorphic version)Given an elliptic curve E, there exists a newform fE ∈ S2(Γ0(N)) s.t.

ap(fE) = 1 + p−#E(Fp), for all p - N.

The complex manifold Y0(N)(C) = Γ0(N)\H can be compactified byadding a finite set of points (cusps), yielding X0(N)(C).Shimura proved that X0(N)(C) is the set of C-points of an algebraic(projective) curve X0(N) defined over Q.

Theorem (Modularity, geometric version)Given an elliptic curve E, there exists a surjective morphism

φE : X0(N)/Q→ E.


Plan

1 Heegner points

2 After Heegner

3 Quadratic ATR points

4 Cubic (1, 1) points

5 Example


The main tool for BSD: Heegner points

Kurt Heegner

Only available when K = Q(√D) is imaginary: D < 0.

I will define Heegner points under the additional condition:I Heegner hypothesis: p | N =⇒ p split in K.

This ensures that ords=1 L(E/K, s) is odd (so ≥ 1).Modularity is crucial in the construction.


Heegner Points (K/Q imaginary quadratic)Modularity =⇒ ∃ modular form fE attached to E.

ωE = 2πifE(z)dz = 2πi∑n≥1

an(f)e2πinzdz ∈ Ω1Γ0(N)\H.

Given τ ∈ K ∩H, set Jτ =

∫ τ

i∞ωE ∈ C.

Well-defined up to ΛE =∫

γ ωE | γ ∈ H1 (X0(N),Z)

.

Theorem (Weierstrass Uniformization)There exists a computable complex-analytic group isomorphism

η : C/ΛE → E(C), ΛE = lattice of rank 2.

Theorem (Shimura, Gross–Zagier, Kolyvagin)1 Pτ = η(Jτ ) ∈ E(Kab) ⊂ E(C).2 PK = Tr(Pτ ) is nontorsion ⇐⇒ L′(E/K, 1) 6= 0.3 If ords=1 L(E/Q, s) ≤ 1 then BSD holds for E(Q).


Heegner Points: why did this work?

Why did this work?

1 The Riemann surface Γ0(N)\H has an algebraic model X0(N)/Q.

2 Existence of the morphism φE defined over Q:

φE : X0(N)→ E. (geometric modularity)

3 CM theory shows that τ ∈ Γ0(N)\H is defined on X0(N)(Kab).An explicit description of φ shows that:

Pτ = φE(τ) ∈ E(Kab).


Generalization

One can replace Q with any totally real field F .I i.e. the defining polynomial of F factors completely over R.

Consider an elliptic curve E defined over F , of conductor NE .The field K/F needs then to be a CM extension.

I i.e. the defining polynomial for K over Q has no linear terms over R.

Suppose that NE is coprime to the discriminant of K/F .The Heegner hypothesis can be relaxed to:

Heegner Hypothesis: [F : Q] + #p | NE : p inert in K is odd.

I This still ensures that ords=1 L(E/K, s) is odd.


Plan

1 Heegner points

2 After Heegner



5 Example


Darmon’s Idea

Henri Darmon

What if K/F is not CM?I Simplest case: F = Q, K real quadratic.I Or what if F is not totally real?

. . . this may get us in trouble!

1 Algebraic model X/F .

2 Geometric modularity: φE : X → E.

3 CM points τ ∈ X(Kab).


History

Henri Darmon Adam Logan Xevi Guitart Jerome Gartner

H. Darmon (2000): F totally real.

I Darmon-Logan (2003): F quadratic norm-euclidean, NE trivial.I Guitart-M. (2011): F quadratic norm-euclidean, NE trivial.I Guitart-M. (2012): F quadratic norm-euclidean, NE trivial.

J. Gartner (2010): F totally real, relaxed Heegner hypothesis.

I ?


Notation

Consider a number field F .I If v is an infinite real place of F , then:

1 It may extend to two real places of K (splits), or2 It may extend to one complex place of K (ramifies).

I If v is complex, then it extends to two complex places of K (splits).

n = #v | ∞F : v splits in K.

K/F is CM ⇐⇒ n = 0.Can compute only n ≤ 1, although construction works in general.

Let E be an elliptic curve over F .

S(E,K) =v | NE∞F : v not split in K

.

Sign of functional equation for L(E/K, s) should be (−1)#S(E,K).I All constructions assume that #S(E,K) is odd.I In this talk: assume that S(E,K) = ν, with ν an infinite place.


Plan

1 Heegner points

2 After Heegner



5 Example


Darmon’s quadratic ATR points

Let E be defined over a real quadratic F (with h+F = 1).

DefinitionA Hilbert modular form (HMF) of level N is a holomorphic functionf : H×H→ C, such that

f(γ1z1, γ2z2) = (c1z1 + d1)2(c2z2 + d2)2f(z1, z2), γ ∈ Γ0(N).

Have also Fourier expansions

fE(z1, z2) =∑n>>0

an(fE)e2πi(n1z1/δ1+n2z2/δ2).

Theorem (Freitas–Le-Hung–Siksek)There is a HMF fE of level NE such that ap(fE) = ap(E) for all p.


Darmon’s quadratic ATR points (II)

Suppose that K/F is an ATR extension:I The 1st embedding v1 of F extends to one complex place of K.I The 2nd embedding v2 of F extends to two real places of K.

Suppose that p | NE =⇒ p is split in K ( =⇒ S(E,K) = v1).Let τ ∈ K \ F . One has StabΓ0(NE)(τ) = 〈γτ 〉. Set τ1 = v1(τ).Given τ2 ∈ H, consider the geodesic joining τ2 with τ ′2 = γττ2.

×H H

τ1τ2

τ ′2

γτ

Γ0(NE)

X0(NE)

Fact: τ1 × γτ in Z1(Γ0(NE)\(H×H),Z) is null-homologous.; ∆τ a 2-chain such that ∂∆τ = τ1 × γτ .

I ∆τ is well-defined up to H2(X0(NE),Z).


Darmon’s quadratic ATR points (III)

Need to symmetrize the HMF fE to account for units of F .I Yields fE , which is no longer holomorphic.

Integration yields an element Jτ =∫∫

∆τfEdz1dz2 ∈ C.

Well-defined up to a lattice

L =∫∫

∆fEdz1dz2 : ∆ ∈ H2(X0(NE),Z).

Conjecture 1 (Oda)There is an isogeny β : C/L→ E(C).

Pτ = β(Jτ ) ∈ E(C).

Conjecture 2 (Darmon)1 The local point Pτ is global, and belongs to E(Kab).2 Pτ is nontorsion if and only if L′(E/K, 1) 6= 0.


Plan

1 Heegner points

2 After Heegner



5 Example


Cubic Darmon points (I)

Let F be a cubic field of signature (1, 1) (and h+F = 1).

Let E/F be an elliptic curve, of conductor NE .Consider the arithmetic group

Γ0(NE) ⊂ SL2(OF ) ⊂ SL2(R)× SL2(C).

H3 = C× R>0 = hyperbolic 3-space, on which SL2(C) acts:(a bc d

)· (x, y) =

((ax+ b)(cx+ d) + acy2

|cx+ d|2 + |cy|2 ,y

|cx+ d|2 + |cy|2

).

Get an action of Γ0(NE) on the symmetric space H×H3.

R>0

C

×

H H3


Cubic Darmon points (II)

Assume that E is modular.E ; an automorphic form ωE with Fourier-Bessel expansion:

ωE(z, x, y) =∑

α∈δ−1OFα0>0

a(δα)(E)e2πi(α0z+α1x+α2x)yH (α1y) ·(−dx∧dzdy∧dzdx∧dz

)

H(t) =

(− i

2eiθK1(4πρ),K0(4πρ),

i

2e−iθK1(4πρ)

)t = ρeiθ.

I K0 and K1 are hyperbolic Bessel functions of the second kind:

K0(x) =

∫ ∞0

e−x cosh(t)dt, K1(x) =

∫ ∞0

e−x cosh(t) cosh(t)dt.

ωE descends to a harmonic 2-form on Γ0(NE)\ (H×H3).


Cubic Darmon points (III)

Let K be a totally complex quadratic extension of F .I Suppose that p | NE =⇒ p is split in K. ( =⇒ S(E,K) = v1).

Choose τ ∈ K \ F .R>0

C

×

H H3

τ1

γτ

τ2

τ ′2

Γ0(NE)

τ ; ∆τ ∈ C2(Γ0(NE),Z).

Jτ =

∫∆τ

ωE ∈ C.

Jτ ; Pτ via C/ΛE → E(C).Conjecture: Pτ is defined over a finite abelian extension of K.


Remarks

Conjectures say the exact extension Hτ/K over which Pτ is defined.

When n > 1 construct analogous cycles, but of higher dimension.I Need to develop computational (co)homology of arithmetic groups.

When #S(E,K) > 1 the group Γ0(NE) is replaced with the(norm-one) units of a certain quaternion algebra over F , and X0(N)is replaced with a Shimura curve.

I No computations have been done in this setting.

There is a p-adic counterpart to all these constructions, where therole of the place v1 is substituted with a (finite) prime.

I See tomorrow’s talk by Carlos de Vera Piquero!


Plan

1 Heegner points

2 After Heegner



5 Example


Example (I)

Let F = Q(r) with r3 − r2 + 1 = 0.

F signature (1, 1) and discriminant −23.

Consider the elliptic curve E/F given by the equation:

E/F : y2 + (r − 1)xy +(r2 − r

)y = x3 +

(−r2 − 1

)x2 + r2x.

E has prime conductor NE =(r2 + 4

)of norm 89.

K = F (α), with α2 + (r + 1)α+ 2r2 − 3r + 3 = 0.

I K has class number 1, thus we expect the point to be defined over K.


Example (II)

Take τ = α. This gives

γτ =

(−4r − 3 −r2 + 2r + 3

−2r2 − 4r − 3 −r2 + 4r + 2

)Finding ∆τ with ∂∆τ = τ × γτ amounts to decomposing γτ into aproduct of elementary matrices.

I Effective version of congruence subgroup problem.

Jτ =∑i

∫ si

ri

∫ γi·O

OωE(z, x, y).

We obtain, summing over all ideals (α) of norm up to 400, 000:

Jτ = 0.1419670770183− 0.0550994633√−1 ∈ C/ΛE ; Pτ ∈ E(C).

Numerically (up to 32 decimal digits) we obtain:

Pτ?= 10×

(r − 1, α− r2 + 2r

)∈ E(K).


Thank you !

“The fun of the subject seems to me to be in the examples.

B. Gross, in a letter to B. Birch, 1982”Bibliography, code and slides at:http://www.warwick.ac.uk/mmasdeu/


http://www.warwick.ac.uk/mmasdeu/

analytic construction of points on modular elliptic curves

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