cubic 4-folds with an eckardt point · outline 1 1. cubic fourfolds 1.a. rationality questions 1.b....

Cubic 4-folds with an Eckardt point

R. Laza

Stony Brook University

September 13, 2018

R. Laza (Stony Brook University) Cubics with an Eckardt point September 13, 2018 1 / 21

Outline

1 1. Cubic fourfolds1.A. Rationality Questions1.B. Cubics and Hyperkahlers

2 2. Cubics with an Eckardt point [LPZ17]2.A. Generalities2.B. The GIT model2.C. The D/Γ model2.D. The main result of [LPZ17]

3 3. Eckardt points and rationality [L18]

4 4. Back to hyper-Kahlers

1. Cubic Fourfolds

ConsiderX ⊂ P5

a smooth cubic hypersurface (/C).

The middle cohomology H4(X,C) is a Hodge structure of K3 type

0 1 21 1 0

Consequently, cubic 4-folds are very relevant for

Rationality QuestionsConstruction of HK manifolds

1. Cubic Fourfolds

ConsiderX ⊂ P5

0 1 21 1 0

1. Cubic Fourfolds

ConsiderX ⊂ P5

0 1 21 1 0

1.A. Rationality Questions

cubic surfaces S are rational

cubic threefolds Y are irrational (Clemens-Griffiths’ 72)

some cubic fourfolds are rational (Morin, Iskovskikh, Tregub,Hassett...)

but conjecturally, the very general cubic fourfold is not rational.

A lot of progress on rationality questions (Voisin, Colliot-Thelene,Totaro, Hassett, Tschinkel, Schreieder, etc.), but one case thatremains open is cubic fourfolds.

1.A. Harris, Hassett, Kuznetsov, Kulikov, etc.

Modeled by the Clemens-Griffiths proof, it is natural to expect

Rationality Conjecture

Let X be a smooth cubic fourfold. If H4(X)tr does not embed into theK3 lattice (e.g. Hodge general cubic, or Hodge general cubiccontaining a plane), then X is not rational.

For convenience

Definition (Potentially irrational)

Let X be a smooth cubic fourfold. If H4(X)tr does not embed into theK3 lattice, we call X potentially irrational.

All known rational examples are not potentially irrational.

1.A. Harris, Hassett, Kuznetsov, Kulikov, etc.

Modeled by the Clemens-Griffiths proof, it is natural to expect

Rationality Conjecture

Let X be a smooth cubic fourfold. If H4(X)tr does not embed into theK3 lattice (e.g. Hodge general cubic, or Hodge general cubiccontaining a plane), then X is not rational.

For convenience

Definition (Potentially irrational)

Let X be a smooth cubic fourfold. If H4(X)tr does not embed into theK3 lattice, we call X potentially irrational.

All known rational examples are not potentially irrational.

1.B. HK Manifolds

A hyper-Kahler (HK) manifold (or IHSM) Z is a simply connectedKahler manifold admitting a holomorphic 2-form ω which isnon-degenerate, and unique

H2,0(Z) ∼= C[ω]

Example: K3 surfaces.

They are building blocks in AG (KZ ≡ 0, BB decomposition:abelian varieties, Calabi-Yau, or HK)

Few deformation classes known (Beauville, Mukai, O’Grady):

2 infinite series: K3[n], Kumn (dim = 2n)2 exceptional cases (O’Grady): OG10, OG6 (dim = 10, 6 resp.)

Are there any other examples?

1.B. HK Manifolds

H2,0(Z) ∼= C[ω]

1.B. HK Manifolds

H2,0(Z) ∼= C[ω]

1.B. HK Manifolds

H2,0(Z) ∼= C[ω]

1.B. Cubics and hyper-Kahlers

Cubic 4-folds give many interesting examples of hyper-Kahlermanifolds

Beauville-Donagi’ 85: X cubic fourfold =⇒ F (X) is HK 4-fold(deformation equivalent to K3[2])

[LLSvS’17] - twisted cubics on X =⇒ HK 8-fold (deformationequivalent to K3[4])

cubics are better than K3s - 20 dimensional moduli, leads tolocally complete examples of families of HKs.

recently, we (L–Sacca–Voisin [LSV17]) gave a new construction ofthe OG10 exceptional example

1.B. LSV’s Construction of OG10 example

Theorem [LSV17]

Let X ⊂ P5 be a Hodge general cubic fourfold, and B = (P5)∨. LetY/B the universal hyperplane section, with J /U the associatedrelative intermediate Jacobian bundle (with U = B \X∨). Then

i) There exists a compactification Z/B of J /U such that Z issmooth, and Z/B is flat.

ii) Z is a 10-dimensional HK manifold (and Z/B is a Lagrangianfibration).

iii) Z is deformation equivalent to OG10.

(A lot of previous work: Donagi-Markman, O’Grady–Rapagnetta,Markushevich, etc.)

Theorem [LSV17]

2. New title - “Pseudo-cubics”

Question

Are there any other cubic like fourfolds?

joint work with G. Pearlstein (TAMU) and Z. Zhang (U.Colorado)

what we are really hunting for are VHS of K3 (and Calabi-Yau)type with geometric origin

inspired by M. Reid list of 95 weighted K3 surfaces

Specifically, we wantX ⊂WP4

quasi-smooth with H4(X) of K3 type.

Answer - “Pseudo-Cubics”

17∗ cases, 14 induced from Reid’s list, 3 new(∗ assuming the existence of a Fermat type member)

Question

2.A. Cubics with an Eckardt point

Main example: cubic 4-folds (20 moduli),

but with 14 moduli, weget another new(ish) example:

X6 ⊂ P(1, 2, 2, 2, 2, 3)

[Un]fortunately, X6 is birational to a cubic with an Eckardt point

X = V(f(x0, . . . , x4) + x0x

)⊂ P5

Question

A smooth cubic hypersurface X (of dimension n) has an Eckardt pointp ∈ X if TpX ∩X = CS , where CS is the cone over an (n− 2)dimensional cubic S.

Main example: cubic 4-folds (20 moduli),but with 14 moduli, weget another new(ish) example:

X6 ⊂ P(1, 2, 2, 2, 2, 3)

X = V(f(x0, . . . , x4) + x0x

)⊂ P5

Question

X6 ⊂ P(1, 2, 2, 2, 2, 3)

X = V(f(x0, . . . , x4) + x0x

)⊂ P5

Question

X6 ⊂ P(1, 2, 2, 2, 2, 3)

X = V(f(x0, . . . , x4) + x0x

)⊂ P5

Question

2.A. Cubics with an Eckardt point (2)

Two other natural characterization:

X has an Eckardt point iff X admits an involution fixing ahyperplane H0.

ι : x5 → −x5(fixes H0 = V (x5) and p = (0 : · · · : 0 : 1) the Eckardt point)

(X, p) ⇐⇒ (Y, S) a n− 1 dimensional cubic pair (withY = X ∩H0, S = X ∩H0 ∩ TpX)

going from (Y, S) to (X, p) was inspired by Allcock-Carlson-Toledo:

Y = V (f3(x0, . . . , x4)), S = V (f3, `) =⇒ X = V (f3 + x25`)

(Recall ACT: Y = V (f3) =⇒ X = V (f3 + x35))

2.A. Cubics with an Eckardt point (2)

Two other natural characterization:

X has an Eckardt point iff X admits an involution fixing ahyperplane H0.

ι : x5 → −x5(fixes H0 = V (x5) and p = (0 : · · · : 0 : 1) the Eckardt point)

(X, p) ⇐⇒ (Y, S) a n− 1 dimensional cubic pair (withY = X ∩H0, S = X ∩H0 ∩ TpX)

going from (Y, S) to (X, p) was inspired by Allcock-Carlson-Toledo:

Y = V (f3(x0, . . . , x4)), S = V (f3, `) =⇒ X = V (f3 + x25`)

(Recall ACT: Y = V (f3) =⇒ X = V (f3 + x35))

2.B. Moduli - The GIT Model

GIT model: X6 ⊂WP5 ??? (non-reductive group)

We have a GIT model for pairs

Mpairs = P55 × (P5)∨//SL(6)

rk Pic(P55 × (P5)∨) = 2 =⇒ choice of linearization, VGIT(Thaddeus, Dolgachev-Hu)...

Family of compactifications for pairs

M(t) = P55 × (P5)∨//O(1,t) SL(6), t ∈ Q+.

2.B. Moduli - The GIT Model

GIT model: X6 ⊂WP5 ??? (non-reductive group)

We have a GIT model for pairs

Mpairs = P55 × (P5)∨//SL(6)

rk Pic(P55 × (P5)∨) = 2 =⇒ choice of linearization, VGIT(Thaddeus, Dolgachev-Hu)...

Family of compactifications for pairs

M(t) = P55 × (P5)∨//O(1,t) SL(6), t ∈ Q+.

2.B. Moduli - The GIT Model (2)

There are two forgetful maps:

M(ε) → M3,3

(Y, S) → Y

and at the opposite end

M(3/4− ε) → M3,2

(Y, S) → S

and flips M(ε) 99KM(3/4− ε) giving a wall crossing situation.

see my thesis [L09] (related to deformations of singularities, N16,O16)

M(ε) → M3,3

(Y, S) → Y

M(3/4− ε) → M3,2

(Y, S) → S

M(ε) → M3,3

(Y, S) → Y

M(3/4− ε) → M3,2

(Y, S) → S

2.C. Moduli (2) - The D/Γ model

A cubic surface S contains 27 lines, thus the cone CS contains 27planes.

If X has an Eckardt point p, CS ⊂ X, and then there are 27planes on X

X is an M -polarized cubic fourfold in the sense of Dolgachev, withM a rank 7 lattice (=⇒ 14 moduli)

In fact,

7 3 3 3 3 3 33 3 1 1 1 1 13 1 3 1 1 1 13 1 1 3 1 1 13 1 1 1 3 1 13 1 1 1 1 3 13 1 1 1 1 1 3

with M ⊂ H2,2 ∩H4(X,Z).

In fact,

7 3 3 3 3 3 33 3 1 1 1 1 13 1 3 1 1 1 13 1 1 3 1 1 13 1 1 1 3 1 13 1 1 1 1 3 13 1 1 1 1 1 3

with M ⊂ H2,2 ∩H4(X,Z).

In fact,

7 3 3 3 3 3 33 3 1 1 1 1 13 1 3 1 1 1 13 1 1 3 1 1 13 1 1 1 3 1 13 1 1 1 1 3 13 1 1 1 1 1 3

with M ⊂ H2,2 ∩H4(X,Z).

2.C. The D/Γ model (2)

Easier to remember, the primitive algebraic part:

Mprim∼= E6(2)

(corresponding to H2(S)prim ∼= E6)

Giving the transcendental lattice

T ∼= (D4)3 ⊕ U2

signature (14, 2)

As usually (e.g. see M -polarized K3s), we get a 14 dimensionallocally symmetric variety D/Γ as model for the moduli space,where

D = {z ∈ P(TC) | z2 = 0, z.z > 0}◦

Γ ∼ O(T )

Mprim∼= E6(2)

T ∼= (D4)3 ⊕ U2

signature (14, 2)

D = {z ∈ P(TC) | z2 = 0, z.z > 0}◦

Γ ∼ O(T )

Mprim∼= E6(2)

T ∼= (D4)3 ⊕ U2

signature (14, 2)

D = {z ∈ P(TC) | z2 = 0, z.z > 0}◦

Γ ∼ O(T )

2.D. Main result of [LPZ17]

Theorem [LPZ17]

With notations as above

M(1/3) ∼= (D/Γ)∗

compare

Theorem (L./Looijenga 2017; Voisin ’86, Hassett ’96)

M3,4∼= (D20 \ H2 \ H6)/Γ

Furthermore

K−blow−up��

flip// D20/Γ

L−factorialization��

M3,4GIT P // (D20/Γ)∗

Theorem [LPZ17]

M(1/3) ∼= (D/Γ)∗

compare

M3,4∼= (D20 \ H2 \ H6)/Γ

Furthermore

K−blow−up��

flip// D20/Γ

M3,4GIT P // (D20/Γ)∗

Theorem [LPZ17]

M(1/3) ∼= (D/Γ)∗

compare

M3,4∼= (D20 \ H2 \ H6)/Γ

Furthermore

K−blow−up��

flip// D20/Γ

M3,4GIT P // (D20/Γ)∗

2.X. Commercial Break “HKL Program”

with K. O’Grady, we have a “Hassett–Keel–LooijengaProgram” to systematically understand

P :M 99K (D/Γ)∗

GIT vs. Baily-Borel compactifications (Type IV case, or ballquotients)

Thus, we understand well the “compare” above.

“Modularity principle”: The tautological models of D/Γ areobtained by arithmetic modifications of the Baily-Borelcompactification.

The upshot, the complexity of P is determined by the complexityof the “discriminant” hyperplane arrangement ∆

deg 2 K3 surfaces, cubic fourfolds ... Easy.deg 4 K3 surfaces, EPW sextics, DV ... Hard

P :M 99K (D/Γ)∗

3. Rationality of cubics revisited

It is expected that the Hodge general cubic fourfold is not rational.

In fact, even the Hodge general cubic fourfold containing a planeshould not be rational.

But a cubic fourfold containing two disjoint planes is rational.

Natural to ask

Question (Most algebraic irrational cubic)

Assuming that there exist irrational cubic fourfolds, which are themost algebraic cubic fourfolds X for which the rationality fails?

Two issues

define “most algebraic”avoid rationality, and replace by “potentially rational” (orconjecturally rational)

Natural to ask

Two issues

Natural to ask

Two issues

Natural to ask

Two issues

Natural to ask

Two issues

3. Maximally algebraic Cubics

Inspired by Vinberg “The two most algebraic K3 surfaces” (Math.Ann. ’83), and Morrison (Inv. Math. ’84)

Clearly, we want

ρX := rank(H2,2(X) ∩H4(X,Z)

)to be as large as possible

but we also want

dX := det(H2,2(X) ∩H4(X,Z)

)to be as small as possible

combine them in a single index

κX :=2dX

The most algebraic cubic fourfolds have the same transcendentallattices (A2 or A1 +A1) as in Vinberg, so κX = 221

3 or 219 resp.

Clearly, we want

but we also want

dX := det(H2,2(X) ∩H4(X,Z)

κX :=2dX

3 or 219 resp.

Clearly, we want

but we also want

dX := det(H2,2(X) ∩H4(X,Z)

κX :=2dX

3 or 219 resp.

Clearly, we want

but we also want

dX := det(H2,2(X) ∩H4(X,Z)

κX :=2dX

3 or 219 resp.

Clearly, we want

but we also want

dX := det(H2,2(X) ∩H4(X,Z)

κX :=2dX

3 or 219 resp.

3. Maximally algebraic potentially irrational Cubics

Theorem [L18]

Let X be a cubic fourfold.

(0) If X is potentially irrational, then κX ≤ 1.

(1) A Hodge general cubic fourfold X containing an Eckardt point ispotentially irrational with κX = 1.

(2) Conversely, any potentially irrational X with κX = 1 is a cubicfourfold with an Eckardt point.

If X has two Eckardt points, then X is rational.

Cubic threefolds Y with an Eckardt points should be relevant tothe question of stable rationality.

Theorem [L18]

4. Cubics with an Eckardt point and HK Manifolds

Question

Is there an exotic hyper-Kahler 8-fold?

[LPZ18+] – Maybe?

cubic 4-folds with an eckardt point · outline 1 1. cubic fourfolds 1.a. rationality questions 1.b....

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