Moduli spaces of genus 2 curves with split Jacobians through K3 - - PowerPoint PPT Presentation

Moduli spaces of genus 2 curves with split Jacobians through K3 surfaces

Abhinav Kumar

Stony Brook

October 22, 2015

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Introduction

Let C be a genus 2 curve over a field k, and J = Jac(C) its Jacobian. Recall that we say J is simple if it does not contain a proper abelian subvariety. Otherwise we say that J is reducible or decomposable or split.

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Introduction

Let C be a genus 2 curve over a field k, and J = Jac(C) its Jacobian. Recall that we say J is simple if it does not contain a proper abelian subvariety. Otherwise we say that J is reducible or decomposable or split. The only possibility for a genus 2 curve is that J is isogenous to a product E × E ′ of elliptic curves. Equivalently, there is a degree n map C → E to some elliptic curve, for some natural number n.

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Introduction

Let C be a genus 2 curve over a field k, and J = Jac(C) its Jacobian. Recall that we say J is simple if it does not contain a proper abelian subvariety. Otherwise we say that J is reducible or decomposable or split. The only possibility for a genus 2 curve is that J is isogenous to a product E × E ′ of elliptic curves. Equivalently, there is a degree n map C → E to some elliptic curve, for some natural number n. Goal: give explicit examples of genus 2 curves with split Jacobians, and explicit descriptions of moduli spaces of such (C, E).

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Examples

We give a few examples of split (over Q) Jacobians. Example Classical example: X0(37) is a genus 2 curve with equation y2 = x6 + 8x5 − 20x4 + 28x3 − 24x2 + 12x − 4 Its Jacobian J0(37) is (2, 2)-isogenous to a product of elliptic curves with j-invariants 37/542 and −372/543.

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Examples II

Example Let C be the curve y2 = (x3 + 420x − 5600)(x3 + 42x2 + 1120) Then J(C) is (3, 3)-isogenous to a product of elliptic curves with j-invariants −27 · 72 and −25 · 7 · 173.

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Examples II

Example Let C be the curve y2 = (x3 + 420x − 5600)(x3 + 42x2 + 1120) Then J(C) is (3, 3)-isogenous to a product of elliptic curves with j-invariants −27 · 72 and −25 · 7 · 173.

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Examples II

Example Let C be the curve y2 = (x3 + 420x − 5600)(x3 + 42x2 + 1120) Then J(C) is (3, 3)-isogenous to a product of elliptic curves with j-invariants −27 · 72 and −25 · 7 · 173. A morphism of degree 3 to the elliptic curve y2

1 = x3 1 + 4900x2 1 + 7031500x1 + 2401000000

is given by x1 = −882000(x − 14) x3 + 420x − 5600, y1 = 49000(x3 − 21x2 − 140)y (x3 + 420x − 5600)2 .

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Examples III

Example The genus two curve defined by y2 = (x3 − 10x2 − 816x − 204)(x3 + 4x2 − 893x − 13861) has Jacobian (7, 7)-isogenous to a product of elliptic curves with j-invariants 21932/55 and 2832721393/52.

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Examples III

Example The genus two curve defined by y2 = (x3 − 10x2 − 816x − 204)(x3 + 4x2 − 893x − 13861) has Jacobian (7, 7)-isogenous to a product of elliptic curves with j-invariants 21932/55 and 2832721393/52. Example The genus two curve defined by y2 = (x3+496x2+52302x−2673552)(x3+584x−271740x+24634652) has Jacobian (11, 11)-isogenous to a product of elliptic curves with j-invariants 212293

365243 and 21231317093153913 355343210911

.

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A moduli space example

Theorem A birational model for the surface parametrizing (C, E) related by a degree 7 map is given by z2 = −16s4r4 + 2s(20s2 + 17s − 1)r3 − (44s3 + 57s2 + 18s − 1)r2 + 2s(15s + 17)r + s2. It is a singular (i.e. of Picard number 20) elliptic K3 surface.

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A tautological curve example

Example Over the moduli space L3, which has equation z2 = 11664r2 − 8(54s3 + 27s2 − 72s + 23)r + (s − 1)4(2s − 1)2, a tautological family of genus 2 curves is given by y2 =

• x3 + 3(3s − 1)x2 − 2(m − 27)(9s − 5)2

(m + 27)

• ×
• x3 − 3(m − 27)(9s − 5)2

4(m + 27) x + (m − 27)(9s − 5)3 4(m + 27)

• where m =
• z − (s − 1)2(2s − 1)
• /(4r).

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Background

Clasically, such pairs (C, E) arose in the reduction of hyperelliptic integrals to elliptic ones. Studied by Legendre, Jacobi, Bolza, Humbert, etc. Example

• dx

√ x6 + 2x4 + 5x2 + 1 =

• dy

2

• y(y3 + 2y2 + 5y + 1)

=

• −dz

2 √ z3 + 5z2 + 2z + 1 More recently, their moduli spaces were studied from a computational perspective by Shaska and his collaborators, and also from an arithmetic standpoint by Frey, Kani, etc.

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Facts

1 Let C → E of degree n. Then J(C) → E × E ′ by an

(n, n)-isogeny. And conversely.

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Facts

1 Let C → E of degree n. Then J(C) → E × E ′ by an

(n, n)-isogeny. And conversely.

2 Then J(C) → E × E ′ an (n, n)-isogeny over a field k. It gives

rise to E[n] ∼ = E ′[n] an anti-isometry (under Weil pairing), Galois invariant.

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Facts

1 Let C → E of degree n. Then J(C) → E × E ′ by an

(n, n)-isogeny. And conversely.

2 Then J(C) → E × E ′ an (n, n)-isogeny over a field k. It gives

rise to E[n] ∼ = E ′[n] an anti-isometry (under Weil pairing), Galois invariant.

3 There is a map of degree n induced on the hyperelliptic

quotients C E P1 P1

φ ψ

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Usual method

The standard approach to studying the moduli space is by looking at the covers P1 → P1 of degree n which lift to a degree n map from a genus 2 curve to an elliptic curve.

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Usual method

The standard approach to studying the moduli space is by looking at the covers P1 → P1 of degree n which lift to a degree n map from a genus 2 curve to an elliptic curve. (This becomes some condition on the branch points of ψ. Without any conditions we’d have a Hurwitz space).

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Usual method

The standard approach to studying the moduli space is by looking at the covers P1 → P1 of degree n which lift to a degree n map from a genus 2 curve to an elliptic curve. (This becomes some condition on the branch points of ψ. Without any conditions we’d have a Hurwitz space). Then do some invariant theory to mod out by choices. The previous state of the art: knew explicit models of the moduli spaces for n ≤ 5.

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New approach

Space of J(C) reducible by (n, n)-isogeny is a hypersurface in A2, the Humbert surface Hn2 of discriminant n2.

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New approach

Space of J(C) reducible by (n, n)-isogeny is a hypersurface in A2, the Humbert surface Hn2 of discriminant n2. Reason: decomposition corresponds to real multiplication by quadratic ring of discriminant n2. The choice of E corresponds to a double cover, the Hilbert modular surface Y−(n2) of discriminant n2.

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New approach

Space of J(C) reducible by (n, n)-isogeny is a hypersurface in A2, the Humbert surface Hn2 of discriminant n2. Reason: decomposition corresponds to real multiplication by quadratic ring of discriminant n2. The choice of E corresponds to a double cover, the Hilbert modular surface Y−(n2) of discriminant n2. Previously, in joint work with Elkies, we developed a method to parametrize Hilbert modular surfaces of fundamental discriminant D as moduli spaces of K3 surfaces. I’ll go over this method next, and then describe the additional ideas in the new work.

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Elliptic K3 surfaces with Shioda-Inose structure

We say that a K3 surface X has a Shioda-Inose structure if it has a symplectic involution ι such that the quotient map is a Kummer surface Km(A) and the 2 : 1 quotient maps X

• ❅

❅ ❅ ❅

A

⑦ ⑦ ⑦ ⑦

Y induce a Hodge isometry of transcendental lattices TX ∼ = TA.

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Elliptic K3 surfaces with Shioda-Inose structure

We say that a K3 surface X has a Shioda-Inose structure if it has a symplectic involution ι such that the quotient map is a Kummer surface Km(A) and the 2 : 1 quotient maps X

• ❅

❅ ❅ ❅

A

⑦ ⑦ ⑦ ⑦

Y induce a Hodge isometry of transcendental lattices TX ∼ = TA. Building on work of Dolgachev and Naruki, I showed that the moduli space of elliptic K3 surfaces with E8 and E7 reducible fibers is isomorphic to A2, with corresponding K3 surfaces and abelian surfaces related by a Shioda-Inose structure.

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Explicit map

Let C be a genus 2 curve and (I2 : I4 : I6 : I10) its Igusa-Clebsch invariants. Theorem (K,2006) The elliptic K3 surface given by y2 = x3 − t3 I4 12t + 1

• x + t5

I10 4 t2 + I2I4 − 3I6 108 t + I2 24

• has fibers of type II∗ at t = ∞ and III∗ at t = 0 and is related to

Jac(C) by a Shioda-Inose structure.

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Explicit map

Let C be a genus 2 curve and (I2 : I4 : I6 : I10) its Igusa-Clebsch invariants. Theorem (K,2006) The elliptic K3 surface given by y2 = x3 − t3 I4 12t + 1

• x + t5

I10 4 t2 + I2I4 − 3I6 108 t + I2 24

• has fibers of type II∗ at t = ∞ and III∗ at t = 0 and is related to

Jac(C) by a Shioda-Inose structure.

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Explicit map

Let C be a genus 2 curve and (I2 : I4 : I6 : I10) its Igusa-Clebsch invariants. Theorem (K,2006) The elliptic K3 surface given by y2 = x3 − t3 I4 12t + 1

• x + t5

I10 4 t2 + I2I4 − 3I6 108 t + I2 24

• has fibers of type II∗ at t = ∞ and III∗ at t = 0 and is related to

Jac(C) by a Shioda-Inose structure. Another way to think of this theorem : an isomorphism of moduli spaces between A2 and the moduli space of K3 surfaces lattice polarized by U ⊕ E8(−1) ⊕ E7(−1).

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Explicit map

Let C be a genus 2 curve and (I2 : I4 : I6 : I10) its Igusa-Clebsch invariants. Theorem (K,2006) The elliptic K3 surface given by y2 = x3 − t3 I4 12t + 1

• x + t5

I10 4 t2 + I2I4 − 3I6 108 t + I2 24

• has fibers of type II∗ at t = ∞ and III∗ at t = 0 and is related to

Jac(C) by a Shioda-Inose structure. Another way to think of this theorem : an isomorphism of moduli spaces between A2 and the moduli space of K3 surfaces lattice polarized by U ⊕ E8(−1) ⊕ E7(−1). The isomorphism is Galois invariant and leads to several arithmetic applications.

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Igusa-Clebsch invariants

Write f (x) =

6

• i=1

fixi =

6

• i=1

(x − αi). Then I2 = f 2

6

• (12)2(34)2(56)2 := f 2

6

• (α1−α2)2(α3−α4)2(α5−α6)2

where the sum is over all partitions of {1, . . . , 6} into three subsets

• f two elements. Similarly,

I4 = f 4

6

• (12)2(23)2(31)2(45)2(56)2(64)2

I6 = f 6

6

• (12)2(23)2(31)2(45)2(56)2(64)2(14)2(25)2(36)2

I10 = f 10

6

• (ij)2

These transform “covariantly” under the action of GL2(¯ k), so that the resulting point in (1, 2, 3, 5)-weighted dimensional projective 3-space only depends on C.

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Hilbert modular surfaces

(joint with Elkies) To use this theorem to describe Hilbert modular surfaces, we

• bserve that abelian surfaces with real multiplication by OD

correspond to K3 surfaces with Picard lattice of signature (1, 17) and discriminant −D containing U ⊕ E8(−1) ⊕ E7(−1). In fact abstractly this lattice is LD := E8(−1)2 ⊕ OD.

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Hilbert modular surfaces

(joint with Elkies) To use this theorem to describe Hilbert modular surfaces, we

• bserve that abelian surfaces with real multiplication by OD

correspond to K3 surfaces with Picard lattice of signature (1, 17) and discriminant −D containing U ⊕ E8(−1) ⊕ E7(−1). In fact abstractly this lattice is LD := E8(−1)2 ⊕ OD. Step 0: Find a sublattice Λ = U ⊕ R of small codimension in LD, such that R is a root lattice and Λ⊥ has small discriminant.

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Hilbert modular surfaces

(joint with Elkies) To use this theorem to describe Hilbert modular surfaces, we

• bserve that abelian surfaces with real multiplication by OD

correspond to K3 surfaces with Picard lattice of signature (1, 17) and discriminant −D containing U ⊕ E8(−1) ⊕ E7(−1). In fact abstractly this lattice is LD := E8(−1)2 ⊕ OD. Step 0: Find a sublattice Λ = U ⊕ R of small codimension in LD, such that R is a root lattice and Λ⊥ has small discriminant. Step 1: Compute the corresponding family of elliptic K3 surfaces, where U comes from ZO + ZF and R from reducible fibers (the rest comes from sections). This involves reverse engineering Tate’s algorithm, to get started.

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Hilbert modular surfaces

Step 2: Use a sequence of “elliptic hops” (neighbor method on the lattices) to compute an alternate elliptic fibration with E8 and E7 fibers.

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Hilbert modular surfaces

Step 2: Use a sequence of “elliptic hops” (neighbor method on the lattices) to compute an alternate elliptic fibration with E8 and E7 fibers. Step 3: Use the formulas from the theorem read out the Igusa-Clebsch invariants of the associated genus 2 curve. This gives the map to A2 (birational to M2).

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Hilbert modular surfaces

Step 2: Use a sequence of “elliptic hops” (neighbor method on the lattices) to compute an alternate elliptic fibration with E8 and E7 fibers. Step 3: Use the formulas from the theorem read out the Igusa-Clebsch invariants of the associated genus 2 curve. This gives the map to A2 (birational to M2). Step 4: Write down a set of candidates for components of the branch locus for the double cover Y−(D) → HD (the rank of the K3 jumps to 19 and we can control the discriminant). Use various arithmetic tricks to compute the double cover.

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Square discriminants

We first generalized the theorems from Elkies-K, so that the method is applicable to any discriminant (not just fundamental).

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Square discriminants

We first generalized the theorems from Elkies-K, so that the method is applicable to any discriminant (not just fundamental). In addition, we need two extra ingredients:

1 Eigenform location algorithm ([K-Mukamel]) Abhinav Kumar Genus 2 curves with split Jacobians 17 / 33

Square discriminants

We first generalized the theorems from Elkies-K, so that the method is applicable to any discriminant (not just fundamental). In addition, we need two extra ingredients:

1 Eigenform location algorithm ([K-Mukamel]) 2 Integration of eigenforms (suggested by Elkies) Abhinav Kumar Genus 2 curves with split Jacobians 17 / 33

Square discriminants

We first generalized the theorems from Elkies-K, so that the method is applicable to any discriminant (not just fundamental). In addition, we need two extra ingredients:

1 Eigenform location algorithm ([K-Mukamel]) 2 Integration of eigenforms (suggested by Elkies) Abhinav Kumar Genus 2 curves with split Jacobians 17 / 33

Square discriminants

We first generalized the theorems from Elkies-K, so that the method is applicable to any discriminant (not just fundamental). In addition, we need two extra ingredients:

1 Eigenform location algorithm ([K-Mukamel]) 2 Integration of eigenforms (suggested by Elkies)

to exhibit the family of C → E over the whole moduli space.

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Square discriminants

We first generalized the theorems from Elkies-K, so that the method is applicable to any discriminant (not just fundamental). In addition, we need two extra ingredients:

1 Eigenform location algorithm ([K-Mukamel]) 2 Integration of eigenforms (suggested by Elkies)

to exhibit the family of C → E over the whole moduli space. Current state of knowledge: we can compute the moduli space Ln for n ≤ 11. It is rational for n ≤ 5, K3 for n = 6, 7, honestly elliptic for 8 ≤ n ≤ 10, and general type for n = 11.

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Tautological family

Next, let’s see how to exhibit a normalized tautological family of genus 2 curves over the moduli space. First, we numerically compute: say we have a point p ∈ Hn2. Then we can

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Tautological family

Next, let’s see how to exhibit a normalized tautological family of genus 2 curves over the moduli space. First, we numerically compute: say we have a point p ∈ Hn2. Then we can

1 write the corresponding curve C by constructing it from

Igusa-Clebsch invariants (work over C).

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Tautological family

Next, let’s see how to exhibit a normalized tautological family of genus 2 curves over the moduli space. First, we numerically compute: say we have a point p ∈ Hn2. Then we can

1 write the corresponding curve C by constructing it from

Igusa-Clebsch invariants (work over C).

2 use Analytic Jacobian (Magma code - van Wamelen etc.) to

calculate endomorphisms.

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Tautological family

Next, let’s see how to exhibit a normalized tautological family of genus 2 curves over the moduli space. First, we numerically compute: say we have a point p ∈ Hn2. Then we can

1 write the corresponding curve C by constructing it from

Igusa-Clebsch invariants (work over C).

2 use Analytic Jacobian (Magma code - van Wamelen etc.) to

calculate endomorphisms.

3 take σ /

∈ Z, say √ D (we’re assuming p is generic, so endomorphism ring is exactly OD). Compute its action on differentials and find the eigenforms.

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Tautological family

Next, let’s see how to exhibit a normalized tautological family of genus 2 curves over the moduli space. First, we numerically compute: say we have a point p ∈ Hn2. Then we can

1 write the corresponding curve C by constructing it from

Igusa-Clebsch invariants (work over C).

2 use Analytic Jacobian (Magma code - van Wamelen etc.) to

calculate endomorphisms.

3 take σ /

∈ Z, say √ D (we’re assuming p is generic, so endomorphism ring is exactly OD). Compute its action on differentials and find the eigenforms.

4 do a Weierstrass transformation to normalize the curve so

that the eigenforms are dx/y and x dx/y and the coefficient

• f x5 is 1.

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Recognizing algebraic numbers

Let z2 = f (r, s) be the equation of the Hilbert modular surface. The point p corresponds to a choice (r0, s0) ∈ Q2 (preferably). Choose a square root z0 =

• f (r0, s0).

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Recognizing algebraic numbers

Let z2 = f (r, s) be the equation of the Hilbert modular surface. The point p corresponds to a choice (r0, s0) ∈ Q2 (preferably). Choose a square root z0 =

• f (r0, s0).

We use LLL to recognize the coefficients (ai) of the Weierstrass equation as elements of Q(z0). Their trace and norm are rational numbers.

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Recognizing algebraic numbers

Let z2 = f (r, s) be the equation of the Hilbert modular surface. The point p corresponds to a choice (r0, s0) ∈ Q2 (preferably). Choose a square root z0 =

• f (r0, s0).

We use LLL to recognize the coefficients (ai) of the Weierstrass equation as elements of Q(z0). Their trace and norm are rational numbers. We run this process for a large number of points p, and then interpolate Tr(ai) and Nm(ai) to recognize these in Q(r, s). Finally, we solve for ai ∈ Q(r, s)[z].

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Recognizing algebraic numbers

Let z2 = f (r, s) be the equation of the Hilbert modular surface. The point p corresponds to a choice (r0, s0) ∈ Q2 (preferably). Choose a square root z0 =

• f (r0, s0).

We use LLL to recognize the coefficients (ai) of the Weierstrass equation as elements of Q(z0). Their trace and norm are rational numbers. We run this process for a large number of points p, and then interpolate Tr(ai) and Nm(ai) to recognize these in Q(r, s). Finally, we solve for ai ∈ Q(r, s)[z]. We have two choices for each coefficient a4, . . . , a0 giving a total

• f 32 possibilities. Of these only two will work (i.e. give a

Weierstass equation whose Igusa-Clebsch invariants match the

• nes we started from). We pick one - this gives us a normalized

tautological family of genus 2 curves C.

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Proving it

Of course, this is not a proof, because of numerical computation. But

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Proving it

Of course, this is not a proof, because of numerical computation. But We can directly compute Igusa-Clebsch invariants and see that we have a tautological family.

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Proving it

Of course, this is not a proof, because of numerical computation. But We can directly compute Igusa-Clebsch invariants and see that we have a tautological family. To check that the eigenforms are dx/y and x dx/y, we can use the Eigenform Location algorithm from K-Mukamel. (This uses a pairing between the space of quadratic differentials on C and the tangent space to Teichmuller space. Well known, e.g. see McMullen’s work.)

Abhinav Kumar Genus 2 curves with split Jacobians 20 / 33

Proving it

Of course, this is not a proof, because of numerical computation. But We can directly compute Igusa-Clebsch invariants and see that we have a tautological family. To check that the eigenforms are dx/y and x dx/y, we can use the Eigenform Location algorithm from K-Mukamel. (This uses a pairing between the space of quadratic differentials on C and the tangent space to Teichmuller space. Well known, e.g. see McMullen’s work.)

Abhinav Kumar Genus 2 curves with split Jacobians 20 / 33

Proving it

Of course, this is not a proof, because of numerical computation. But We can directly compute Igusa-Clebsch invariants and see that we have a tautological family. To check that the eigenforms are dx/y and x dx/y, we can use the Eigenform Location algorithm from K-Mukamel. (This uses a pairing between the space of quadratic differentials on C and the tangent space to Teichmuller space. Well known, e.g. see McMullen’s work.) Thus we can prove that we have a normalized tautological family. This is useful for the next step: to describe the map C → E.

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Integrating eigenforms

We have C as y2 = x6 + . . . . Write E as y2

1 = x3 1 + ax2 1 + bx1 + c

where a, b, c are unknown. Assume ∞+ on C goes to ∞ on E under the map φ.

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Integrating eigenforms

We have C as y2 = x6 + . . . . Write E as y2

1 = x3 1 + ax2 1 + bx1 + c

where a, b, c are unknown. Assume ∞+ on C goes to ∞ on E under the map φ. Write φ∗(ω) = dx/y, expand formally about ∞+ and integrate.

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Integrating eigenforms

We have C as y2 = x6 + . . . . Write E as y2

1 = x3 1 + ax2 1 + bx1 + c

where a, b, c are unknown. Assume ∞+ on C goes to ∞ on E under the map φ. Write φ∗(ω) = dx/y, expand formally about ∞+ and integrate. We get a system of equations in the undetermined coefficients a, b, c and the coefficients of the map φ of degree n. Solve. (I will illustrate with an example later.)

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Discriminant 49

Let’s see how this method goes for n = 7. We start with elliptic K3 with E8, A4, A2 and A1 fibers. These give a root sublattice of rank 15 and discriminant 30.

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Discriminant 49

Let’s see how this method goes for n = 7. We start with elliptic K3 with E8, A4, A2 and A1 fibers. These give a root sublattice of rank 15 and discriminant 30. So we need a section of height 49 30 = 4 − 6 5 − 2 3 − 1 2.

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Discriminant 49

Let’s see how this method goes for n = 7. We start with elliptic K3 with E8, A4, A2 and A1 fibers. These give a root sublattice of rank 15 and discriminant 30. So we need a section of height 49 30 = 4 − 6 5 − 2 3 − 1 2.

Abhinav Kumar Genus 2 curves with split Jacobians 22 / 33

Discriminant 49

Let’s see how this method goes for n = 7. We start with elliptic K3 with E8, A4, A2 and A1 fibers. These give a root sublattice of rank 15 and discriminant 30. So we need a section of height 49 30 = 4 − 6 5 − 2 3 − 1 2. Some work yields the following Weierstrass equation over A2

r,s.

y2 = x3 −

• (4r2s2 + 4rs2 − s2 + 4r2s + 2rs − r2)t2 − 2(2r3s4

+ 4r3s3 − 4r2s3 + rs3 + 6r3s2 − 5r2s2 + rs2 − s2 + 4r3s − 7r2s + 2rs + 2s − 3r2 + 2r)t − (rs2 + rs − s − r)2 x2 + 8t

• (s + 1)t − rs − r − s
• rst − (r − 1)(rs − 1)
• ·
• (4r3s4 + 8r3s3 − 6r2s3 + rs3 + 6r3s2 − 7r2s2 + 3rs2

− s2 + 2r3s − 3r2s + 2rs − r2)t + (rs2 + rs − s − r)2 x + 16t2((s + 1)t − rs − r − s)2(rst − (r − 1)(rs − 1))2· (4r2s2(s + 1)(rs + r − 1)t + (rs2 + rs − s − r)2). Abhinav Kumar Genus 2 curves with split Jacobians 22 / 33

Neighbor steps

First, we take a 2-neighbor step. The darkened portion of the diagram indicates a new A7 fiber F ′. Note that its intersection with the class of the original fiber is (O + P) · F = 2. So this corresponds to a 2-neighbor step on the lattice level.

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Neighbor steps II

In practice, this step means we have to find a suitable section of O(P + O) on the elliptic curve over the base, with suitable poles along the various components of the bad fibers.

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Neighbor steps II

In practice, this step means we have to find a suitable section of O(P + O) on the elliptic curve over the base, with suitable poles along the various components of the bad fibers. We find that an elliptic parameter is given by 1 t(t − t0) y + y0 x − x0 + α0 + α1t

• where

α0 = rs2 + (r − 1)s − r, α1 = −

• 2r2s2 + (2r2 − r + 1)s + r2 − r
• /(r − 1).

where P = (x0, y0) is the section of height 49/30.

Abhinav Kumar Genus 2 curves with split Jacobians 24 / 33

Neighbor steps III

Finally, we take another 2-neighbor step to reach an E8E7 fibration.

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Neighbor steps III

Finally, we take another 2-neighbor step to reach an E8E7 fibration. We can then immediately read out the Igusa-Clebsch invariants (omitted here for brevity). After some work we get the double cover, which we displayed earlier in the talk: z2 = −16s4r4 + 2s(20s2 + 17s − 1)r3 − (44s3 + 57s2 + 18s − 1)r2 + 2s(15s + 17)r + s2.

Abhinav Kumar Genus 2 curves with split Jacobians 25 / 33

Neighbor steps III

Finally, we take another 2-neighbor step to reach an E8E7 fibration. We can then immediately read out the Igusa-Clebsch invariants (omitted here for brevity). After some work we get the double cover, which we displayed earlier in the talk: z2 = −16s4r4 + 2s(20s2 + 17s − 1)r3 − (44s3 + 57s2 + 18s − 1)r2 + 2s(15s + 17)r + s2. The rational point r = −10, s = −2/25, z = ±104/25 gives rise to

• ne of the examples at the beginning of the talk.

Abhinav Kumar Genus 2 curves with split Jacobians 25 / 33

Eigenform location

(joint work with Mukamel) Suppose we have a genus 2 curve C whose Jacobian corresponds to a point p on Y−(D), so that it has real multiplication by OD. Without knowing the explicit action of OD (which can be fairly complicated to describe), can we nevertheless exhibit the eigenforms for real multiplication on the Jacobian?

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Eigenform location

(joint work with Mukamel) Suppose we have a genus 2 curve C whose Jacobian corresponds to a point p on Y−(D), so that it has real multiplication by OD. Without knowing the explicit action of OD (which can be fairly complicated to describe), can we nevertheless exhibit the eigenforms for real multiplication on the Jacobian? Yes! The following theorem is a consequence of Ahlfors’s variational formula. Theorem Let p ∈ Y−(D) correspond to curve C and map to a smooth point

• n HD ⊂ A2. Then under the pairing of regular quadratic

differentials on C with the tangent space to M2 at p, the annihilator of the tangent space to HD is the span of the quadratic differential ω1ω2, where ω1 and ω2 are the eigenforms on C.

Abhinav Kumar Genus 2 curves with split Jacobians 26 / 33

Eigenform location algorithm

Explicitly, the calculation goes as follows: Consider the map φ : V = C6 → P3

(1:2:3:5) taking (a0, . . . , a5)

to (I2 : I4 : I6 : I10) of the curve Ca : y2 = x6 + a5x5 + · · · = a0. We can explicitly compute the pairing between TaV and the 3-dimensional space of quadratic differentials on Ca. Write this as a 3 × 6 matrix Ma.

Abhinav Kumar Genus 2 curves with split Jacobians 27 / 33

Eigenform location algorithm

Explicitly, the calculation goes as follows: Consider the map φ : V = C6 → P3

(1:2:3:5) taking (a0, . . . , a5)

to (I2 : I4 : I6 : I10) of the curve Ca : y2 = x6 + a5x5 + · · · = a0. We can explicitly compute the pairing between TaV and the 3-dimensional space of quadratic differentials on Ca. Write this as a 3 × 6 matrix Ma. On the other hand, we are given a point p ∈ Y−(D). Its Igusa-Clebsch invariants (functions of parameters r, s on Humbert surface) define a point q in M2. We compute a preimage a of q in V , and preimages of the tangent vectors ∂r(Ig − Cl)|p and ∂s(Ig − Cl)|p by φ∗. Call these vectors vr and vs.

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ELA II

Concatenating the column vectors vr and vs gives us a 6 × 2 matrix Na.

Abhinav Kumar Genus 2 curves with split Jacobians 28 / 33

ELA II

Concatenating the column vectors vr and vs gives us a 6 × 2 matrix Na. To compute the quadratic differential which is the product of the two eigenforms, we just need to take the left kernel of

• MaNa. Recall that the basis of quadratic differentials is

(dx2/y2, x dx2/y2, x2 dx2/y2). So to certify that dx/y and x dx/y are eigenforms, we simply need to verify the kernel is (0, 1, 0).

Abhinav Kumar Genus 2 curves with split Jacobians 28 / 33

ELA II

Concatenating the column vectors vr and vs gives us a 6 × 2 matrix Na. To compute the quadratic differential which is the product of the two eigenforms, we just need to take the left kernel of

• MaNa. Recall that the basis of quadratic differentials is

(dx2/y2, x dx2/y2, x2 dx2/y2). So to certify that dx/y and x dx/y are eigenforms, we simply need to verify the kernel is (0, 1, 0).

Abhinav Kumar Genus 2 curves with split Jacobians 28 / 33

ELA II

Concatenating the column vectors vr and vs gives us a 6 × 2 matrix Na. To compute the quadratic differential which is the product of the two eigenforms, we just need to take the left kernel of

• MaNa. Recall that the basis of quadratic differentials is

(dx2/y2, x dx2/y2, x2 dx2/y2). So to certify that dx/y and x dx/y are eigenforms, we simply need to verify the kernel is (0, 1, 0). So now we know how to exhibit a normalized family. Next: finding the map to the two elliptic curves in the square discriminant situation.

Abhinav Kumar Genus 2 curves with split Jacobians 28 / 33

Eigenform integration example

Let us take the example from earlier y2 = (x2 + 420x − 5600)(x3 + 42x + 1120) corresponding to a point on L3.

Abhinav Kumar Genus 2 curves with split Jacobians 29 / 33

Eigenform integration example

Let us take the example from earlier y2 = (x2 + 420x − 5600)(x3 + 42x + 1120) corresponding to a point on

• L3. For convenience, let’s rewrite as

y2 = (1 + 420x2 − 5600x3)(1 + 42x + 1120x3) after a simple change of variable.

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Eigenform integration example

Let us take the example from earlier y2 = (x2 + 420x − 5600)(x3 + 42x + 1120) corresponding to a point on

• L3. For convenience, let’s rewrite as

y2 = (1 + 420x2 − 5600x3)(1 + 42x + 1120x3) after a simple change of variable. This has a rational point (0, 1), which we’ll assume maps to the point (0, 1) on the elliptic curve y2

1 = Fx3 1 + Gx2 1 + Hx1 + 1.

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Eigenform integration II

Assume that the map φ : C → E satisfies φ∗(dx1/y1) = x dx/y, the latter being an eigenform on C.

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Eigenform integration II

Assume that the map φ : C → E satisfies φ∗(dx1/y1) = x dx/y, the latter being an eigenform on C. We have in the formal ring around (0, 1). x dx y = x dx

• (−5600x3 + 420x2 + 1)(1120x3 + 42x + 1)

1/2 = (x − 21x2 + 903x3/2 + . . . ) dx

Abhinav Kumar Genus 2 curves with split Jacobians 30 / 33

Eigenform integration II

Assume that the map φ : C → E satisfies φ∗(dx1/y1) = x dx/y, the latter being an eigenform on C. We have in the formal ring around (0, 1). x dx y = x dx

• (−5600x3 + 420x2 + 1)(1120x3 + 42x + 1)

1/2 = (x − 21x2 + 903x3/2 + . . . ) dx So z =

• x dx/y = x2/2 − 7x3 + . . .

Abhinav Kumar Genus 2 curves with split Jacobians 30 / 33

Eigenform integration II

Assume that the map φ : C → E satisfies φ∗(dx1/y1) = x dx/y, the latter being an eigenform on C. We have in the formal ring around (0, 1). x dx y = x dx

• (−5600x3 + 420x2 + 1)(1120x3 + 42x + 1)

1/2 = (x − 21x2 + 903x3/2 + . . . ) dx So z =

• x dx/y = x2/2 − 7x3 + . . .

Similarly, dx1 y1 = dx1 (Fx3

1 + Gx2 1 + Hx1 + 1)1/2

=

• 1 − H

2 x1 + −G 2 + 3H2 8

• x2

1 + . . .

• dx1,

Abhinav Kumar Genus 2 curves with split Jacobians 30 / 33

Eigenform integration III

so z1 =

• dx1/y1 = x1 − (H/4)x2

1 + . . . .

Now, since dz = φ∗(dz1), we have the equation z = φ∗(z1), which we can write using the above expressions as x2/2 − 7x3 + · · · = φ∗ x1 − (H/4)x2

1 + . . .

• = xE − (H/4)x2

E + . . . ,

where xE = φ∗(x1). We can invert the formal series on the right to write xE = x2/2 − 7x3 + (H/16 + 903/8)x4 + . . .

Abhinav Kumar Genus 2 curves with split Jacobians 31 / 33

Eigenform integration IV

We also know that xE = φ∗(x1) must be a rational function of degree 3 in x. Since its formal expansion starts x2/2 + . . . , it must equal xE = x2(1 + kx) 2(1 + mx + nx2 + px3) for some constants k, m, n, p which are also undetermined.

Abhinav Kumar Genus 2 curves with split Jacobians 32 / 33

Eigenform integration IV

We also know that xE = φ∗(x1) must be a rational function of degree 3 in x. Since its formal expansion starts x2/2 + . . . , it must equal xE = x2(1 + kx) 2(1 + mx + nx2 + px3) for some constants k, m, n, p which are also undetermined. Equating these two expressions, we get a system of equations for the unknown coefficients m, n, p, F, G, H. These have a unique solution, which gives us the Weierstrass equation of C, as well as the x-component of the map φ : C → E.

Abhinav Kumar Genus 2 curves with split Jacobians 32 / 33

References

Hilbert modular surfaces for square discriminants and elliptic subfields of genus 2 function fields, to appear in Research in the Mathematical Sciences. arXiv: 1412.2849 Algebraic models and arithmetic geometry of Teichm¨ uller curves in genus two (with R. Mukamel) submitted, arXiv: 1406.7057 K3 surfaces and equations for Hilbert modular surfaces (with N. Elkies) Algebra and Number Theory 8 (2014), no. 10, 2297-2411. arXiv: 1209.3527

Abhinav Kumar Genus 2 curves with split Jacobians 33 / 33

References

Hilbert modular surfaces for square discriminants and elliptic subfields of genus 2 function fields, to appear in Research in the Mathematical Sciences. arXiv: 1412.2849 Algebraic models and arithmetic geometry of Teichm¨ uller curves in genus two (with R. Mukamel) submitted, arXiv: 1406.7057 K3 surfaces and equations for Hilbert modular surfaces (with N. Elkies) Algebra and Number Theory 8 (2014), no. 10, 2297-2411. arXiv: 1209.3527

Thank you!

Abhinav Kumar Genus 2 curves with split Jacobians 33 / 33